people.math.sc.eduGalois groups of classical polynomials: •D. Hilbert (1892) used his now classical Hilbert’s Irreducibility Theorem to show that for each integer n ≥ 1, there
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NOT MY TITLE
How many positive integers n ≤ 1000
are such that 2n begins with the digit 1?
NOT MY TITLE
How many positive integers n ≤ 1000
are such that 2n begins with the digit 1?
301
NOT MY TITLE
How many positive integers n ≤ 1000
are such that 2n begins with the digit 1?
301
log10 2 = 0.30102999566398 . . .
x #{n ≤ x : 2n begins with 1}10 3
x #{n ≤ x : 2n begins with 1}10 3
100 30
1000 301
x #{n ≤ x : 2n begins with 1}10 3
100 30
1000 301
10000 3010
100000 30102
1000000 301029
10000000 3010299
x #{n ≤ x : 2n begins with 1}10 3
100 30
1000 301
10000 3010
100000 30102
1000000 301029
10000000 3010299
log10 2 = 0.30102999566398 . . .
x #{n ≤ x : 2n begins with 1}10 3
100 30
1000 301
10000 3010
100000 30102
1000000 301029
10000000 3010299
To Ponder: Prove the pattern continues.
APPLICATIONS OF PADE APPROXIMATIONS
OF (1 − z)k TO NUMBER THEORY
by Michael Filaseta
University of South Carolina
General Areas of Applications:
General Areas of Applications:
• irrationality measures
General Areas of Applications:
• irrationality measures• diophantine equations
General Areas of Applications:
• irrationality measures• diophantine equations• Waring’s problem
General Areas of Applications:
• irrationality measures• diophantine equations• Waring’s problem• the factorization of n(n + 1)
General Areas of Applications:
• irrationality measures• diophantine equations• Waring’s problem• the factorization of n(n + 1)
• Galois groups of classical polynomials
General Areas of Applications:
• irrationality measures• diophantine equations• Waring’s problem• the factorization of n(n + 1)
• Galois groups of classical polynomials• the Ramanujan-Nagell equation
General Areas of Applications:
• irrationality measures• diophantine equations• Waring’s problem• the factorization of n(n + 1)
• Galois groups of classical polynomials• the Ramanujan-Nagell equation• k-free numbers in short intervals
General Areas of Applications:
• irrationality measures• diophantine equations• Waring’s problem• the factorization of n(n + 1)
• Galois groups of classical polynomials• the Ramanujan-Nagell equation• k-free numbers in short intervals• k-free values of polynomials and binary forms
General Areas of Applications:
• irrationality measures• diophantine equations• Waring’s problem• the factorization of n(n + 1)
• Galois groups of classical polynomials• the Ramanujan-Nagell equation• k-free numbers in short intervals• k-free values of polynomials and binary forms• the abc-conjecture
What are the Pad e approximations of (1− z)k?
What are the Pad e approximations of (1− z)k?
What are the Pad e approximations of (1− z)k?
What are the Pad e approximations of (1− z)k?
Answer: Rational functions that give good approxi-mations to (1−z)k near the origin.
What are the Pad e approximations of (1− z)k?
Answer: Rational functions that give good approxi-mations to (1−z)k near the origin.
What are the Pad e approximations of ez?
Answer: Rational functions that give good approxi-mations to ez near the origin.
What are the Pad e approximations of (1− z)k?
Answer: Rational functions that give good approxi-mations to (1−z)k near the origin.
What are the Pad e approximations of (1− z)k?
Answer: Rational functions that give good approxi-mations to (1−z)k near the origin.
What are the Pad e approximations of (1− z)k?
Answer: Rational functions that give good approxi-mations to (1−z)k near the origin.
Important Equation:
What are the Pad e approximations of (1− z)k?
Answer: Rational functions that give good approxi-mations to (1−z)k near the origin.
Important Equation:
(1 − z)k ≈P (z)
Q(z)
What are the Pad e approximations of (1− z)k?
Answer: Rational functions that give good approxi-mations to (1−z)k near the origin.
Important Equation:
(1 − z)k =P (z)
Q(z)− zmR(z)
What are the Pad e approximations of (1− z)k?
Answer: Rational functions that give good approxi-mations to (1−z)k near the origin.
Important Equation:
(1 − z)k =P (z)
Q(z)− zmR(z)
degree < k (usually)
What are the Pad e approximations of (1− z)k?
Answer: Rational functions that give good approxi-mations to (1−z)k near the origin.
Important Equation:
P − (1 − z)k Q = zmE
What are the Pad e approximations of (1− z)k?
Answer: Rational functions that give good approxi-mations to (1−z)k near the origin.
Important Equation:
Pr− (1 − z)kQr = zmEr
What are the Pad e approximations of (1− z)k?
Answer: Rational functions that give good approxi-mations to (1−z)k near the origin.
Important Equation:
Pr− (1 − z)kQr = z2r+1Er
What are the Pad e approximations of (1− z)k?
Answer: Rational functions that give good approxi-mations to (1−z)k near the origin.
Important Equation:
Pr− (1 − z)kQr = z2r+1Er
What are the Pad e approximations of (1− z)k?
Answer: Rational functions that give good approxi-mations to (1−z)k near the origin.
Important Equation:
Pr− (1 − z)kQr = z2r+1Er
What are the Pad e approximations of (1− z)k?
Answer: Rational functions that give good approxi-mations to (1−z)k near the origin.
Important Equation:
Pr− (1 − z)kQr = z2r+1Er
deg Pr = deg Qr = r < k, deg Er = k − r − 1
Some Properties of the Polynomials:
(i) Pr(z), (−z)kQr(z), and z2r+1Er(z) satisfy
z(z−1)y′′ +(2r(1−z)−(k−1)z
)y′ + r(k+r)y = 0.
(ii) Qr(z) =r∑
j=0
(2r − j
r
)(k − r + j − 1
j
)zj
(iii) Qr(z) =(k+r)!
(k−r−1)! r! r!
∫ 1
0
(1−t)rtk−r−1(1−t+zt)r dt
(iv) Pr(z)Qr+1(z) − Qr(z)Pr+1(z) = cz2r+1
Pr − (1 − z)kQr = z2r+1Er
Pr − (1 − z)kQr = z2r+1Er
WARNING: In the applications you areabout to see, the true identies used havebeen changed.
Pr − (1 − z)kQr = z2r+1Er
WARNING : In the applications you areabout to see, the true identies used havebeen changed. They have been changedto conform to the identity above. Theidentity above gives a result of the typewanted. Typically, a closer analysis ofthese polynomials or even a variant of thepolynomials is used to obtain the currentlybest known results in the applications.
Irrationality measures:
Irrationality measures:
Theorem (Liouville): Fix α ∈ R − Q with α algebraicand of degree n. Then there is a constant C =
C(α) > 0 such that∣∣∣α −a
b
∣∣∣ >C
bn
where a and b with b > 0 are arbitrary integers.
Irrationality measures:
Theorem (Liouville): Fix α ∈ R − Q with α algebraicand of degree n. Then there is a constant C =
C(α) > 0 such that∣∣∣α −a
b
∣∣∣ >C
bn
where a and b with b > 0 are arbitrary integers.
Irrationality measures:
Theorem (Roth): Fix ε > 0 and α ∈ R − Q with α al-gebraic. Then there is a constant C = C(α, ε) >
0 such that ∣∣∣α −a
b
∣∣∣ >C
b2+ε
where a and b with b > 0 are arbitrary integers.
Irrationality measures:
Theorem (Roth): Fix ε > 0 and α ∈ R − Q with α al-gebraic. Then there is a constant C = C(α, ε) >
0 such that ∣∣∣α −a
b
∣∣∣ >C
b2+ε
where a and b with b > 0 are arbitrary integers.
Comment: Liouville’s result is effective; Roth’s is not.
Irrationality measures:
Theorem( Baker ): For a and b integers with b > 0,∣∣∣ 3√2 −
a
b
∣∣∣ >C
b2.955
where C = 10−6.
Irrationality measures:
Theorem( Baker ): For a & b integers with b > 0,∣∣∣ 3√2 −
a
b
∣∣∣ >1
106b2.955.
Irrationality measures:
Theorem(Chudnovsky ): For a & b integers with b > 0,∣∣∣ 3√2 −
a
b
∣∣∣ >1
c · b2.43.
Irrationality measures:
Theorem( Bennett ): For a & b integers with b > 0,∣∣∣ 3√2 −
a
b
∣∣∣ >1
c · b2.47.
Irrationality measures:
Theorem( Bennett ): For a & b integers with b > 0,∣∣∣ 3√2 −
a
b
∣∣∣ >1
4 · b2.47.
Irrationality measures:
Theorem( Bennett ): For a & b integers with b > 0,∣∣∣ 3√2 −
a
b
∣∣∣ >1
4 · b2.47.
Comment: Similar explicit estimates have also beenmade for certain other cube roots.
The Basic Approach:
The Basic Approach:
Pr − ( 1 − z ) k Qr = z2r+1Er
The Basic Approach:
Pr − ( 1 − z )1/3Qr = z2r+1Er
The Basic Approach:
Pr − ( 1 − z↑
3/128
)1/3Qr = z2r+1Er
The Basic Approach:
Pr − (125/128)1/3Qr = z2r+1Er
The Basic Approach:
Pr − (125/128)1/3Qr = z2r+1Er
Rearrange and Normalize to Integers
The Basic Approach:
Pr − (125/128)1/3Qr = z2r+1Er
Rearrange and Normalize to Integers
3√2 br − ar = small
The Basic Approach:
Pr − (125/128)1/3Qr = z2r+1Er
Rearrange and Normalize to Integers
3√2 br − ar = smallr
The Basic Approach:
Pr − (125/128)1/3Qr = z2r+1Er
Rearrange and Normalize to Integers
3√2 br − ar = smallr∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
The Basic Approach:
Pr − (125/128)1/3Qr = z2r+1Er
Rearrange and Normalize to Integers
3√2 br − ar = smallr∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
The Basic Approach:
Pr − (125/128)1/3Qr = z2r+1Er
Rearrange and Normalize to Integers
3√2 br − ar = smallr∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
Wait!!
The Basic Approach:
Pr − (125/128)1/3Qr = z2r+1Er
Rearrange and Normalize to Integers
3√2 br − ar = smallr∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
Wait!! I thought we wanted that LARGE!!
The Basic Approach:∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
The Basic Approach:∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
What’s small r?
The Basic Approach:∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
What’s small r? Let b be a positive integer.
The Basic Approach:∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
What’s small r? Let b be a positive integer. Choosingr right, one can obtain
smallr <1
2b brand br < cb1.47.
The Basic Approach:∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
What’s small r? Let b be a positive integer. Choosingr right, one can obtain
smallr <1
2b brand br < cb1.47.
∣∣∣ 3√2 −
a
b
∣∣∣ ≥
The Basic Approach:∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
What’s small r? Let b be a positive integer. Choosingr right, one can obtain
smallr <1
2b brand br < cb1.47.
∣∣∣ 3√2 −
a
b
∣∣∣ ≥∣∣∣ar
br−
a
b
∣∣∣ −∣∣∣ 3√
2 −ar
br
∣∣∣ >
The Basic Approach:∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
What’s small r? Let b be a positive integer. Choosingr right, one can obtain
smallr <1
2b brand br < cb1.47.
∣∣∣ 3√2 −
a
b
∣∣∣ ≥∣∣∣ar
br−
a
b
∣∣∣ −∣∣∣ 3√
2 −ar
br
∣∣∣ >
The Basic Approach:∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
What’s small r? Let b be a positive integer. Choosingr right, one can obtain
smallr <1
2b brand br < cb1.47.
∣∣∣ 3√2 −
a
b
∣∣∣ ≥∣∣∣ar
br−
a
b
∣∣∣ −∣∣∣ 3√
2 −ar
br
∣∣∣ >1
b br
The Basic Approach:∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
What’s small r? Let b be a positive integer. Choosingr right, one can obtain
smallr <1
2b brand br < cb1.47.
∣∣∣ 3√2 −
a
b
∣∣∣ ≥∣∣∣ar
br−
a
b
∣∣∣ −∣∣∣ 3√
2 −ar
br
∣∣∣ >1
b br
The Basic Approach:∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
What’s small r? Let b be a positive integer. Choosingr right, one can obtain
smallr <1
2b brand br < cb1.47.
∣∣∣ 3√2 −
a
b
∣∣∣ ≥∣∣∣ar
br−
a
b
∣∣∣ −∣∣∣ 3√
2 −ar
br
∣∣∣ >1
b br
The Basic Approach:∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
What’s small r? Let b be a positive integer. Choosingr right, one can obtain
smallr <1
2b brand br < cb1.47.
∣∣∣ 3√2 −
a
b
∣∣∣ ≥∣∣∣ar
br−
a
b
∣∣∣ −∣∣∣ 3√
2 −ar
br
∣∣∣ >1
b br
The Basic Approach:∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
What’s small r? Let b be a positive integer. Choosingr right, one can obtain
smallr <1
2b brand br < cb1.47.
∣∣∣ 3√2 −
a
b
∣∣∣ ≥∣∣∣ar
br−
a
b
∣∣∣ −∣∣∣ 3√
2 −ar
br
∣∣∣ >1
b br−
The Basic Approach:∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
What’s small r? Let b be a positive integer. Choosingr right, one can obtain
smallr <1
2b brand br < cb1.47.
∣∣∣ 3√2 −
a
b
∣∣∣ ≥∣∣∣ar
br−
a
b
∣∣∣ −∣∣∣ 3√
2 −ar
br
∣∣∣ >1
b br−
1
2b br
The Basic Approach:∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
What’s small r? Let b be a positive integer. Choosingr right, one can obtain
smallr <1
2b brand br < cb1.47.
∣∣∣ 3√2 −
a
b
∣∣∣ ≥∣∣∣ar
br−
a
b
∣∣∣ −∣∣∣ 3√
2 −ar
br
∣∣∣ >1
2b br
The Basic Approach:∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
What’s small r? Let b be a positive integer. Choosingr right, one can obtain
smallr <1
2b brand br < cb1.47.
∣∣∣ 3√2 −
a
b
∣∣∣ ≥∣∣∣ar
br−
a
b
∣∣∣ −∣∣∣ 3√
2 −ar
br
∣∣∣ >1
2b br
The Basic Approach:∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
What’s small r? Let b be a positive integer. Choosingr right, one can obtain
smallr <1
2b brand br < cb1.47.
∣∣∣ 3√2 −
a
b
∣∣∣ ≥∣∣∣ar
br−
a
b
∣∣∣ −∣∣∣ 3√
2 −ar
br
∣∣∣ >1
2cb2.47
The Basic Approach:∣∣∣ 3√2 −
ar
br
∣∣∣ = smallr
What’s small r? Let b be a positive integer. Choosingr right, one can obtain
smallr <1
2b brand br < cb1.47.
∣∣∣ 3√2 −
a
b
∣∣∣ ≥∣∣∣ar
br−
a
b
∣∣∣ −∣∣∣ 3√
2 −ar
br
∣∣∣ >1
4 ·b2.47
Diophantine equations:
Diophantine equations:
Theorem (Bennett):For a and b integers with b > 0,∣∣∣ 3√2 −
a
b
∣∣∣ >1
4 · b2.47.
Diophantine equations:
Theorem (Bennett):For a and b integers with b 6= 0,∣∣∣ 3√2 −
a
b
∣∣∣ >1
4 · b2.47.
Diophantine equations:
Theorem (Bennett):For a and b integers with b 6= 0,∣∣∣ 3√2 −
a
b
∣∣∣ >1
4 · |b|2.47.
Diophantine equations:
Theorem (Bennett):For a and b integers with b 6= 0,∣∣∣ 3√2 −
a
b
∣∣∣ >1
4 · |b|2.5.
Diophantine equations:
Theorem (Bennett):For a and b integers with b 6= 0,∣∣∣ 3√2 −
a
b
∣∣∣ >1
4 · |b|2.5.
x3 − 2y3 = n
Diophantine equations:
Theorem (Bennett):For a and b integers with b 6= 0,∣∣∣ 3√2 −
a
b
∣∣∣ >1
4 · |b|2.5.
x3 − 2y3 = n, y 6= 0
Diophantine equations:
Theorem (Bennett):For a and b integers with b 6= 0,∣∣∣ 3√2 −
a
b
∣∣∣ >1
4 · |b|2.5.
x3 − 2y3 = n, y 6= 0∣∣∣ 3√2 −
x
y
∣∣∣ ∣∣∣ 3√2e2πi/3 −
x
y
∣∣∣ ∣∣∣ 3√2e4πi/3 −
x
y
∣∣∣ =|n||y|3
Diophantine equations:
Theorem (Bennett):For a and b integers with b 6= 0,∣∣∣ 3√2 −
a
b
∣∣∣ >1
4 · |b|2.5.
x3 − 2y3 = n, y 6= 0∣∣∣ 3√2 −
x
y
∣∣∣ ∣∣∣ 3√2e2πi/3 −
x
y
∣∣∣ ∣∣∣ 3√2e4πi/3 −
x
y
∣∣∣ =|n||y|3
Diophantine equations:
Theorem (Bennett):For a and b integers with b 6= 0,∣∣∣ 3√2 −
a
b
∣∣∣ >1
4 · |b|2.5.
x3 − 2y3 = n, y 6= 0∣∣∣ 3√2 −
x
y
∣∣∣ ∣∣∣ 3√2e2πi/3 −
x
y
∣∣∣ ∣∣∣ 3√2e4πi/3 −
x
y
∣∣∣ =|n||y|3∣∣∣ 3√
2 −x
y
∣∣∣ <|n||y|3
Diophantine equations:
Theorem (Bennett):For a and b integers with b 6= 0,∣∣∣ 3√2 −
a
b
∣∣∣ >1
4 · |b|2.5.
x3 − 2y3 = n, y 6= 0∣∣∣ 3√2 −
x
y
∣∣∣ ∣∣∣ 3√2e2πi/3 −
x
y
∣∣∣ ∣∣∣ 3√2e4πi/3 −
x
y
∣∣∣ =|n||y|3
1
4 |y|2.5<
∣∣∣ 3√2 −
x
y
∣∣∣ <|n||y|3
Diophantine equations:
x3 − 2y3 = n, y 6= 0∣∣∣ 3√2 −
x
y
∣∣∣ ∣∣∣ 3√2e2πi/3 −
x
y
∣∣∣ ∣∣∣ 3√2e4πi/3 −
x
y
∣∣∣ =|n||y|3
1
4 |y|2.5<
∣∣∣ 3√2 −
x
y
∣∣∣ <|n||y|3
Diophantine equations:
x3 − 2y3 = n, y 6= 0∣∣∣ 3√2 −
x
y
∣∣∣ ∣∣∣ 3√2e2πi/3 −
x
y
∣∣∣ ∣∣∣ 3√2e4πi/3 −
x
y
∣∣∣ =|n||y|3
1
4 |y|2.5<
∣∣∣ 3√2 −
x
y
∣∣∣ <|n||y|3
|y|1/2 < 4|n|
Diophantine equations:
x3 − 2y3 = n, y 6= 0∣∣∣ 3√2 −
x
y
∣∣∣ ∣∣∣ 3√2e2πi/3 −
x
y
∣∣∣ ∣∣∣ 3√2e4πi/3 −
x
y
∣∣∣ =|n||y|3
1
4 |y|2.5<
∣∣∣ 3√2 −
x
y
∣∣∣ <|n||y|3
|y|1/2 < 4|n| =⇒ |y| < 16n2
Diophantine equations:
x3 − 2y3 = n, y 6= 0∣∣∣ 3√2 −
x
y
∣∣∣ ∣∣∣ 3√2e2πi/3 −
x
y
∣∣∣ ∣∣∣ 3√2e4πi/3 −
x
y
∣∣∣ =|n||y|3
1
4 |y|2.5<
∣∣∣ 3√2 −
x
y
∣∣∣ <|n||y|3
|y|1/2 < 4|n| =⇒ |y| < 16n2
Diophantine equations:
Theorem: Let n be a non-zero integer. If x and y areintegers satisfying x3−2y3 = n, then |y| < 16n2.
Diophantine equations:
Theorem (Bennett):If a, b, and n are integers withab 6= 0 and n ≥ 3, then the equation
|axn + byn| = 1
has at most 1 solution in positive integers x and y.
Waring’s Problem:
Waring’s Problem:
Waring’s Problem: Let k be an integer ≥ 2. Thenthere exists a number s such that every natural num-ber is a sum of s kth powers.
Waring’s Problem:
Waring’s Problem: Let k be an integer ≥ 2. Thenthere exists a number s such that every natural num-ber is a sum of s kth powers. If g(k) is the least suchs, what is g(k)?
Waring’s Problem:
Waring’s Problem: Let k be an integer ≥ 2. Thenthere exists a number s such that every natural num-ber is a sum of s kth powers. If g(k) is the least suchs, what is g(k)?
Known: (i) g(k) = 2k +
[(3
2
)k]
− 2
Waring’s Problem:
Waring’s Problem: Let k be an integer ≥ 2. Thenthere exists a number s such that every natural num-ber is a sum of s kth powers. If g(k) is the least suchs, what is g(k)?
Known: (i) g(k) = 2k +
[(3
2
)k]
− 2
(ii) no one knows how to prove (i)
Waring’s Problem:
Waring’s Problem: Let k be an integer ≥ 2. Thenthere exists a number s such that every natural num-ber is a sum of s kth powers. If g(k) is the least suchs, what is g(k)?
Known: (i) g(k) = 2k +
[(3
2
)k]
− 2
(ii) no one knows how to prove (i)
(iii) (i) holds if∥∥∥(3
2
)k∥∥∥ > 0.75k
Waring’s Problem:
Waring’s Problem: Let k be an integer ≥ 2. Thenthere exists a number s such that every natural num-ber is a sum of s kth powers. If g(k) is the least suchs, what is g(k)?
Known: (i) g(k) = 2k +
[(3
2
)k]
− 2
(ii) no one knows how to prove (i)
(iii) (i) holds if∥∥∥(3
2
)k∥∥∥ > 0.75k
(iv) (iii) holds if k > 8
Waring’s Problem:
Waring’s Problem: Let k be an integer ≥ 2. Thenthere exists a number s such that every natural num-ber is a sum of s kth powers. If g(k) is the least suchs, what is g(k)?
Known: (i) g(k) = 2k +
[(3
2
)k]
− 2
(ii) no one knows how to prove (i)
(iii) (i) holds if∥∥∥(3
2
)k∥∥∥ > 0.75k
(iv) (iii) holds if k > 8
(v) no one knows how to prove (iv)
Waring’s Problem:
Known: (i) g(k) = 2k +
[(3
2
)k]
− 2
(ii) No one knows how to prove (i).
(iii) (i) holds if∥∥∥(3
2
)k∥∥∥ > 0.75k
Waring’s Problem:
Known: (i) g(k) = 2k +
[(3
2
)k]
− 2
(ii) No one knows how to prove (i).
(iii) (i) holds if∥∥∥(3
2
)k∥∥∥ > 0.75k
Theorem (Beukers): If k > 4, then∥∥∥(3
2
)k∥∥∥ > 0.5358k.
Waring’s Problem:
Known: (i) g(k) = 2k +
[(3
2
)k]
− 2
(ii) No one knows how to prove (i).
(iii) (i) holds if∥∥∥(3
2
)k∥∥∥ > 0.75k
Theorem(Dubitskas): If k > 4, then∥∥∥(3
2
)k∥∥∥ > 0.5767k.
The factorization of n(n + 1):
The factorization of n(n + 1):
Well-Known: The largest prime factor of n(n + 1)
tends to infinity with n.
The factorization of n(n + 1):
Well-Known: The largest prime factor of n(n + 1)
tends to infinity with n.
Let p1, p2, . . . , pr be primes. There is an N suchthat if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > 1.
Let p1, p2, . . . , pr be primes. There is an N suchthat if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > 1.
Let p1, p2, . . . , pr be primes. There is an N suchthat if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > 1.
Lehmer: Gave some explicit estimates:
Let p1, p2, . . . , pr be primes. There is an N suchthat if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > 1.
Lehmer: Gave some explicit estimates:
n(n+1) divisible only by primes ≤ 11 =⇒ n ≤
Let p1, p2, . . . , pr be primes. There is an N suchthat if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > 1.
Lehmer: Gave some explicit estimates:
n(n+1) divisible only by primes ≤ 11 =⇒ n ≤... only by primes ≤ 41 =⇒ n ≤
Let p1, p2, . . . , pr be primes. There is an N suchthat if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > 1.
Lehmer: Gave some explicit estimates:
n(n+1) divisible only by primes ≤ 11 =⇒ n ≤ 9800
... only by primes ≤ 41 =⇒ n ≤
Let p1, p2, . . . , pr be primes. There is an N suchthat if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > 1.
Lehmer: Gave some explicit estimates:
n(n+1) divisible only by primes ≤ 11 =⇒ n ≤ 9800
... only by primes ≤ 41 =⇒ n ≤ 63927525375
Let p1, p2, . . . , pr be primes. There is an N suchthat if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > 1.
Let p1, p2, . . . , pr be primes. There is an N suchthat if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > 1.
Let p1, p2, . . . , pr be primes. There is an N suchthat if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m >1 .
Let p1, p2, . . . , pr be primes. There is an N suchthat if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > nθ.
Want: Let p1, p2, . . . , pr be primes. There is anN = N(θ, p1, . . . , pr) such that if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > nθ.
Want: Let p1, p2, . . . , pr be primes. There is anN = N(θ, p1, . . . , pr) such that if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > nθ.
abc-conjecture =⇒ θ =
Want: Let p1, p2, . . . , pr be primes. There is anN = N(θ, p1, . . . , pr) such that if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > nθ.
abc-conjecture =⇒ θ = 1 − ε
Want: Let p1, p2, . . . , pr be primes. There is anN = N(θ, p1, . . . , pr) such that if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > nθ.
abc-conjecture =⇒ θ = 1 − ε
unconditionally one can obtain θ =
Want: Let p1, p2, . . . , pr be primes. There is anN = N(θ, p1, . . . , pr) such that if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > nθ.
abc-conjecture =⇒ θ = 1 − ε
unconditionally one can obtain θ = 1 − ε
Want: Let p1, p2, . . . , pr be primes. There is anN = N(θ, p1, . . . , pr) such that if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > nθ.
abc-conjecture =⇒ θ = 1 − ε
unconditionally one can obtain
(ineffective)
θ = 1 − ε
Want: Let p1, p2, . . . , pr be primes. There is anN = N(θ, p1, . . . , pr) such that if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > nθ.
Effective Approach:
Want: Let p1, p2, . . . , pr be primes. There is anN = N(θ, p1, . . . , pr) such that if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > nθ.
Effective Approach: (Linear Forms of Logarithms)
Want: Let p1, p2, . . . , pr be primes. There is anN = N(θ, p1, . . . , pr) such that if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > nθ.
Effective Approach: (Linear Forms of Logarithms)
θ =c
log log n
Want: Let p1, p2, . . . , pr be primes. There is anN = N(θ, p1, . . . , pr) such that if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > nθ.
Effective Approach: (Linear Forms of Logarithms)
θ =c
log log n
Problem: Can we narrow the gap betweenthese ineffective and effective results?
Want: Let p1, p2, . . . , pr be primes. There is anN = N(θ, p1, . . . , pr) such that if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > nθ.
Want: Let p1, p2, . . . , pr be primes. There is anN = N(θ, p1, . . . , pr) such that if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > nθ.
Theorem (Bennett, F., Trifonov): If n ≥ 9 and
n(n + 1) = 2k3`m,
thenm ≥
Want: Let p1, p2, . . . , pr be primes. There is anN = N(θ, p1, . . . , pr) such that if n ≥ N and
n(n + 1) = pe11 p
e22 · · · per
r m
for some integer m, then m > nθ.
Theorem (Bennett, F., Trifonov): If n ≥ 9 and
n(n + 1) = 2k3`m,
thenm ≥ n1/4.
Theorem (Bennett, F., Trifonov): The set of 4-tuples(k, `, M1, M2) of positive integers with
0 <∣∣3kM1 − 2`M2
∣∣ ≤ 100, gcd(6, M1M2) = 1,
M1M2 ≤ min{3kM1, 2
`M2}0.25
consists of 28 tuples of which 26 satisfy k ≤ 5,` ≤ 8 and M1M2 = 1 and the remaining two are(6, 7, 1, 5) and (8, 15, 5, 1).
Conjecture: For n > 512,
n(n + 1) = 2u3vm =⇒ m >√
n.
Conjecture: For n > 512,
n(n + 1) = 2u3vm =⇒ m >√
n.
Comment: The conjecture has been verified for
512 < n ≤
Conjecture: For n > 512,
n(n + 1) = 2u3vm =⇒ m >√
n.
Comment: The conjecture has been verified for
512 < n ≤ 101000.
The Method:
The Method:
n(n + 1) = 3k2`m
The Method:
n(n + 1) = 3k2`m
3km1 − 2`m2 = ±1
The Method:
n(n + 1) = 3k2`m
3km1 − 2`m2 = ±1
Main Idea: Find “small” integers P , Q, and E suchthat
3kP − 2`Q = E.
The Method:
n(n + 1) = 3k2`m
3km1 − 2`m2 = ±1
Main Idea: Find “small” integers P , Q, and E suchthat
3kP − 2`Q = E.
Then
3k (Qm1 − Pm2) = ±Q − Em2.
The Method:
n(n + 1) = 3k2`m
3km1 − 2`m2 = ±1
Main Idea: Find “small” integers P , Q, and E suchthat
3kP − 2`Q = E.
Then
3k (Qm1 − Pm2) = ±Q − Em2.
Main Idea: Find “small” integers P , Q, and E suchthat
3kP − 2`Q = E
and
Qm1 − Pm2 6= 0.
Then
3k (Qm1 − Pm2) = ±Q − Em2.
Main Idea: Find “small” integers P , Q, and E suchthat
3kP − 2`Q = E
and
Qm1 − Pm2 6= 0.
Then
3k (Qm1 − Pm2) = ±Q − Em2.
Obtain an upper bound on 3k.
Main Idea: Find “small” integers P , Q, and E suchthat
3kP − 2`Q = E
and
Qm1 − Pm2 6= 0.
Then
3k (Qm1 − Pm2) = ±Q − Em2.
Obtain an upper bound on 3k. Since 3km1 ≥ n, itfollows that m1 and m = m1m2 are not small.
The “small” integers P , Q, and E are obtained throughthe use of Pade approximations for (1 − z)k.
The “small” integers P , Q, and E are obtained throughthe use of Pade approximations for (1 − z)k.
More precisely, one takes z = 1/9 in the equation
Pr(z) − (1 − z)kQr(z) = z2r+1Er(z).
What’s Needed for the Method to Work:
What’s Needed for the Method to Work:
One largely needs to be dealing with two primes (like2 and 3) with a difference of powers of these primesbeing small (like 32 − 23 = 1).
Galois groups of classical polynomials:
Galois groups of classical polynomials:
• D. Hilbert (1892) used his now classical Hilbert’sIrreducibility Theorem to show that for each integern ≥ 1, there is polynomial f(x) ∈ Z[x] such thatthe Galois group associated with f(x) is the sym-metric group Sn.
Galois groups of classical polynomials:
• D. Hilbert (1892) used his now classical Hilbert’sIrreducibility Theorem to show that for each integern ≥ 1, there is polynomial f(x) ∈ Z[x] such thatthe Galois group associated with f(x) is the sym-metric group Sn. He also showed the analogousresult in the case of the alternating group An.
Galois groups of classical polynomials:
• D. Hilbert (1892) used his now classical Hilbert’sIrreducibility Theorem to show that for each integern ≥ 1, there is polynomial f(x) ∈ Z[x] such thatthe Galois group associated with f(x) is the sym-metric group Sn. He also showed the analogousresult in the case of the alternating group An.
• Hilbert’s work and work of E. Noether (1918) beganwhat is now called Inverse Galois Theory.
Galois groups of classical polynomials:
• D. Hilbert (1892) used his now classical Hilbert’sIrreducibility Theorem to show that for each integern ≥ 1, there is polynomial f(x) ∈ Z[x] such thatthe Galois group associated with f(x) is the sym-metric group Sn. He also showed the analogousresult in the case of the alternating group An.
• Hilbert’s work and work of E. Noether (1918) beganwhat is now called Inverse Galois Theory.
• Van der Waerden showed that for “almost all” poly-nomials f(x) ∈ Z[x], the Galois group associatedwith f(x) is the symmetric group Sn.
Galois groups of classical polynomials:
• Schur showed L(0)n (x) has Galois group Sn.
Galois groups of classical polynomials:
• Schur showed L(0)n (x) has Galois group Sn.
• Schur showed L(1)n (x) has Galois group An (the
alternating group) if n is odd.
Galois groups of classical polynomials:
• Schur showed L(0)n (x) has Galois group Sn.
• Schur showed L(1)n (x) has Galois group An (the
alternating group) if n is odd.
• Schur showedn∑
j=0
xj
j!has Galois group An if 4|n.
Galois groups of classical polynomials:
• Schur showed L(0)n (x) has Galois group Sn.
• Schur showed L(1)n (x) has Galois group An (the
alternating group) if n is odd.
• Schur showedn∑
j=0
xj
j!has Galois group An if 4|n.
• Schur did not find an explicit sequence of polyno-mials having Galois group An with n ≡ 2 (mod 4).
Galois groups of classical polynomials:
Theorem (R. Gow, 1989): If n > 2 is even and
L(n)n (x) =
n∑j=0
(2n
n − j
)(−x)j
j!
is irreducible, then the Galois group of L(n)n (x) is An.
Galois groups of classical polynomials:
Theorem (R. Gow, 1989): If n > 2 is even and
L(n)n (x) =
n∑j=0
(2n
n − j
)(−x)j
j!
is irreducible, then the Galois group of L(n)n (x) is An.
Theorem (joint work with R. Williams): For almostall positive integers n the polynomial L
(n)n (x) is irre-
ducible (and, hence, has Galois group An for almostall even n).
Galois groups of classical polynomials:
Theorem (joint work with R. Williams): For almostall positive integers n the polynomial L
(n)n (x) is irre-
ducible.
Galois groups of classical polynomials:
Theorem (joint work with R. Williams): For almostall positive integers n the polynomial L
(n)n (x) is irre-
ducible.
Comment: The method had an ineffective compo-nent to it. We could show that if n is sufficiently large
and L(n)n (x) is reducible, then L
(n)n (x) has a linear
factor. But we didn’t know what sufficiently large was.
Galois groups of classical polynomials:
Theorem (joint work with R. Williams): For almostall positive integers n the polynomial L
(n)n (x) is irre-
ducible.
Comment: The method had an ineffective compo-nent to it. We could show that if n is sufficiently large
and L(n)n (x) is reducible, then L
(n)n (x) has a linear
factor. But we didn’t know what sufficiently large was.
Theorem (Kidd, Trifonov, F.): If n > 2 and n ≡ 2
(mod 4), then L(n)n (x) is irreducible.
Galois groups of classical polynomials:
Theorem (joint work with R. Williams): For almostall positive integers n the polynomial L
(n)n (x) is irre-
ducible.
Comment: The method had an ineffective compo-nent to it. We could show that if n is sufficiently large
and L(n)n (x) is reducible, then L
(n)n (x) has a linear
factor. But we didn’t know what sufficiently large was.
Theorem (Kidd, Trifonov, F.): If n > 2 and n ≡ 2
(mod 4), then L(n)n (x) has Galois group An.
The Ramanujan-Nagell equation:
The Ramanujan-Nagell equation:
Classical Ramanujan-Nagell Theorem: If x and n
are positive integers satisfying
x2 + 7 = 2n,
then
The Ramanujan-Nagell equation:
Classical Ramanujan-Nagell Theorem: If x and n
are positive integers satisfying
x2 + 7 = 2n,
thenx ∈ {1, 3, 5, 11, 181}.
The Ramanujan-Nagell equation:
Some Background: Beukers used a method “simi-lar” to the approach for finding irrationality measuresto show that
√2 cannot be approximated too well by
rationals a/b with b a power of 2. This implies boundsfor solutions to the Diophantine equation x2 + D =
2n with D fixed. He showed that if D 6= 7, then theequation has ≤ 4 solutions. Related work by Apery,Beukers, and Bennett establishes that for odd primesp not dividing D, the equation x2 + D = pn has atmost 3 solutions. All of these are in some sense bestpossible (though more can and has been said).
The Ramanujan-Nagell equation:
Classical Ramanujan-Nagell Theorem: If x and n
are positive integers satisfying
x2 + 7 = 2n,
thenx ∈ {1, 3, 5, 11, 181}.
The Ramanujan-Nagell equation:
Classical Ramanujan-Nagell Theorem: If x and n
are positive integers satisfying
x2 + 7 = 2n,
thenx ∈ {1, 3, 5, 11, 181}.
Problem: If x2 + 7 = 2nm and x is not in the setabove, then can we say that m must be large?
Problem: If x2 + 7 = 2nm and x is not in the setabove, then can we say that m must be large?
Connection with n(n + 1) problem:
Problem: If x2 + 7 = 2nm and x is not in the setabove, then can we say that m must be large?
Connection with n(n + 1) problem:
x2 + 7 = 2nm
Problem: If x2 + 7 = 2nm and x is not in the setabove, then can we say that m must be large?
Connection with n(n + 1) problem:
x2 + 7 = 2nm
(x+
√−7
2
)(x−
√−7
2
)=
(1+
√−7
2
)n−2(1−
√−7
2
)n−2
m
Problem: If x2 + 7 = 2nm and x is not in the setabove, then can we say that m must be large?
Connection with n(n + 1) problem:
x2 + 7 = 2nm
(x+
√−7
2
)↑
linear
(x−
√−7
2
)↑
linear
=
(1+
√−7
2
)n−2(1−
√−7
2
)n−2
m
Problem: If x2 + 7 = 2nm and x is not in the setabove, then can we say that m must be large?
Connection with n(n + 1) problem:
x2 + 7 = 2nm
(x+
√−7
2
)↑
linear
(x−
√−7
2
)↑
linear
=
(1+
√−7
2
)↑
prime
n−2(1−
√−7
2
)↑
prime
n−2
m
Theorem (Bennett, F., Trifonov): If x, n and m are pos-itive integers satisfying
x2 + 7 = 2nm and x 6∈ {1, 3, 5, 11, 181},
thenm ≥ ???
Theorem (Bennett, F., Trifonov): If x, n and m are pos-itive integers satisfying
x2 + 7 = 2nm and x 6∈ {1, 3, 5, 11, 181},
thenm ≥ x1/2.
Theorem (Bennett, F., Trifonov): If x, n and m are pos-itive integers satisfying
x2 + 7 = 2nm and x 6∈ {1, 3, 5, 11, 181},
thenm ≥ x1/2.
Comment: In the case of x2 + 7 = 2nm, the differ-ence of the primes (1 +
√−7)/2 and (1 −
√−7)/2
each raised to the 13th power has absolute value≈ 2.65 and the powers themselves have absolutevalue ≈ 90.51.
k-free numbers in short intervals:
k-free numbers in short intervals:
k-free numbers in short intervals:
Problem: Find θ = θ(k) as small as possible suchthat, for x sufficiently large, the interval (x, x + xθ]
contains a k-free number.
k-free numbers in short intervals:
Problem: Find θ = θ(k) as small as possible suchthat, for x sufficiently large, the interval (x, x + xθ]
contains a k-free number.
Main Idea: Show there are integers in (x, x + xθ]
not divisible by the kth power of a prime. Considerprimes in different size ranges. Deal with small primesand large primes separately.
Problem: Find θ = θ(k) as small as possible suchthat, for x sufficiently large, the interval (x, x + xθ]
contains a k-free number.
Problem: Find θ = θ(k) as small as possible suchthat, for x sufficiently large, the interval (x, x + xθ]
contains a k-free number.
Small Primes: p ≤ z
Problem: Find θ = θ(k) as small as possible suchthat, for x sufficiently large, the interval (x, x + xθ]
contains a k-free number.
Small Primes: p ≤ z where z = xθ√log x
Problem: Find θ = θ(k) as small as possible suchthat, for x sufficiently large, the interval (x, x + xθ]
contains a k-free number.
Small Primes: p ≤ z where z = xθ√log x
The number of integers n ∈ (x, x + xθ] divisible bysuch a pk is bounded by (2/3)xθ.
Large Primes: p ∈ (N, 2N ], N ≥ z = xθ√log x
Large Primes: p ∈ (N, 2N ], N ≥ z = xθ√log x
x < pkm ≤ x + xθ
Large Primes: p ∈ (N, 2N ], N ≥ z = xθ√log x
x < pkm ≤ x + xθ =⇒x
pk< m ≤
x
pk+
xθ
pk
Large Primes: p ∈ (N, 2N ], N ≥ z = xθ√log x
x < pkm ≤ x + xθ =⇒x
pk< m ≤
x
pk+
xθ
pk
=⇒∥∥∥ x
pk
∥∥∥ <xθ
Nk
Large Primes: p ∈ (N, 2N ], N ≥ z = xθ√log x
x < pkm ≤ x + xθ =⇒x
pk< m ≤
x
pk+
xθ
pk
=⇒∥∥∥ x
pk
∥∥∥ <xθ
Nk
where ‖t‖ = min{|t − `| : ` ∈ Z}
Large Primes: p ∈ (N, 2N ], N ≥ z = xθ√log x
x < pkm ≤ x + xθ =⇒x
pk< m ≤
x
pk+
xθ
pk
=⇒∥∥∥ x
pk
∥∥∥ <xθ
Nk
where ‖t‖ = min{|t − `| : ` ∈ Z}
Idea: Show there are few primes p∈ (N, 2N ] withx/pk that close to an integer.
Large Primes: p ∈ (N, 2N ], N ≥ z = xθ√log x
x < pkm ≤ x + xθ =⇒x
pk< m ≤
x
pk+
xθ
pk
=⇒∥∥∥ x
pk
∥∥∥ <xθ
Nk
where ‖t‖ = min{|t − `| : ` ∈ Z}
Idea: Show there are few integers p∈ (N, 2N ] withx/pk that close to an integer.
Large Primes: p ∈ (N, 2N ], N ≥ z = xθ√log x
x < pkm ≤ x + xθ =⇒x
pk< m ≤
x
pk+
xθ
pk
=⇒∥∥∥ x
pk
∥∥∥ <xθ
Nk
where ‖t‖ = min{|t − `| : ` ∈ Z}
Idea: Show there are few integers u∈ (N, 2N ] withx/uk that close to an integer.
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
Differences:
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
Differences:∥∥∥ x
uk
∥∥∥ <xθ
Nk,
∥∥∥ x
(u + a)k
∥∥∥ <xθ
Nk
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
Differences:∥∥∥ x
uk
∥∥∥ <xθ
Nk,
∥∥∥ x
(u + a)k
∥∥∥ <xθ
Nk
x
uk−
x
(u + a)k
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
Differences:∥∥∥ x
uk
∥∥∥ <xθ
Nk,
∥∥∥ x
(u + a)k
∥∥∥ <xθ
Nk
x
uk−
x
(u + a)k�
ax
uk+1
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
Differences:∥∥∥ x
uk
∥∥∥ <xθ
Nk,
∥∥∥ x
(u + a)k
∥∥∥ <xθ
Nk
x
uk−
x
(u + a)k�
ax
uk+1�
ax
Nk+1
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
Differences:∥∥∥ x
uk
∥∥∥ <xθ
Nk,
∥∥∥ x
(u + a)k
∥∥∥ <xθ
Nk
x
uk−
x
(u + a)k�
ax
uk+1�
ax
Nk+1
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
Differences:∥∥∥ x
uk
∥∥∥ <xθ
Nk,
∥∥∥ x
(u + a)k
∥∥∥ <xθ
Nk
x
uk−
x
(u + a)k�
ax
uk+1�
ax
Nk+1
consider N = x1/k
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
Differences:∥∥∥ x
uk
∥∥∥ <xθ
Nk,
∥∥∥ x
(u + a)k
∥∥∥ <xθ
Nk
x
uk−
x
(u + a)k�
ax
uk+1�
a
x1/k
consider N = x1/k
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
Differences:∥∥∥ x
uk
∥∥∥ <xθ
Nk,
∥∥∥ x
(u + a)k
∥∥∥ <xθ
Nk
x
uk−
x
(u + a)k�
ax
uk+1�
a
x1/k
consider N = x1/k, a < x1/(2k)
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
Differences:∥∥∥ x
uk
∥∥∥ <xθ
Nk,
∥∥∥ x
(u + a)k
∥∥∥ <xθ
Nk
x
uk−
x
(u + a)k�
ax
uk+1�
a
x1/k
consider N = x1/k, a < x1/(2k), θ ≈ 1/k
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
Differences:∥∥∥ x
uk
∥∥∥ <xθ
Nk,
∥∥∥ x
(u + a)k
∥∥∥ <xθ
Nk
x
uk−
x
(u + a)k�
ax
uk+1�
a
x1/k
consider N = x1/k, a < x1/(2k), θ ≈ 1/k
LHS small compared to RHS
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
“Modified” Differences:
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
“Modified” Differences:∥∥∥ x
uk
∥∥∥ <xθ
Nk,
∥∥∥ x
(u + a)k
∥∥∥ <xθ
Nk
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
“Modified” Differences:∥∥∥ x
uk
∥∥∥ <xθ
Nk,
∥∥∥ x
(u + a)k
∥∥∥ <xθ
Nk
x
ukP −
x
(u + a)kQ small
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
“Modified” Differences:∥∥∥ x
uk
∥∥∥ <xθ
Nk,
∥∥∥ x
(u + a)k
∥∥∥ <xθ
Nk
x
ukP −
x
(u + a)kQ small (but not too small)
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
“Modified” Differences:∥∥∥ x
uk
∥∥∥ <xθ
Nk,
∥∥∥ x
(u + a)k
∥∥∥ <xθ
Nk
x
ukP −
x
(u + a)kQ small (but not too small)
(u + a)kP − ukQ small (but not too small)
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
“Modified” Differences:∥∥∥ x
uk
∥∥∥ <xθ
Nk,
∥∥∥ x
(u + a)k
∥∥∥ <xθ
Nk
x
ukP −
x
(u + a)kQ small (but not too small)
(u + a)kP − ukQ small (but not too small)
consider Pr(z) − (1 − z)kQr(z) with z =a
u + a
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
“Modified” Differences:
Theorem (Halberstam & Roth):
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
“Modified” Differences:
Theorem (Halberstam & Roth & Nair):
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
“Modified” Differences:
Theorem (Halberstam & Roth & Nair): For x large,there is a k-free number in (x, x + x1/(2k)].
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
Modified Differences plus Divided Differences:
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
Modified Differences plus Divided Differences:
Theorem (F. & Trifonov): For x sufficiently large, thereis a squarefree number in (x, x + cx1/5 log x].
∥∥∥ x
uk
∥∥∥ <xθ
Nk, u ∈ (N, 2N ], N ≥ xθ
√log x
Modified Differences plus Divided Differences:
Theorem (F. & Trifonov): For x sufficiently large, thereis a squarefree number in (x, x + cx1/5 log x].
Theorem (Trifonov): For x sufficiently large, there isa k-free number in (x, x + cx1/(2k+1) log x].
k-free values of polynomials and binary forms:
k-free values of polynomials and binary forms:
The method for obtaining results about gaps betweenk-free numbers generalizes to k-free values of poly-nomials.
k-free values of polynomials and binary forms:
The method for obtaining results about gaps betweenk-free numbers generalizes to k-free values of poly-nomials. Suppose f(x) ∈ Z[x] is irreducible anddeg f = n.
k-free values of polynomials and binary forms:
The method for obtaining results about gaps betweenk-free numbers generalizes to k-free values of poly-nomials. Suppose f(x) ∈ Z[x] is irreducible anddeg f = n. In what follows, we suppose further thatf has no fixed kth power divisors.
k-free values of polynomials and binary forms:
The method for obtaining results about gaps betweenk-free numbers generalizes to k-free values of poly-nomials. Suppose f(x) ∈ Z[x] is irreducible anddeg f = n. In what follows, we suppose further thatf has no fixed kth power divisors.
Theorem (Nair): Let k ≥ n + 1. For x sufficientlylarge, there is an integer m such that f(m) is k-free with
x < m ≤ x + cxn
2k−n+1.
Theorem (Nair): Let k ≥ n + 1. For x sufficientlylarge, there is an integer m such that f(m) is k-free with
x < m ≤ x + cxn
2k−n+1.
Theorem (Nair): Let k ≥ n + 1. For x sufficientlylarge, there is an integer m such that f(m) is k-free with
x < m ≤ x + cxn
2k−n+1.
Theorem: Let k ≥ n + 1. For x sufficiently large,there is an integer m such that f(m) is k-free with
x < m ≤ x + cxn
2k−n+r,
where r =
Theorem (Nair): Let k ≥ n + 1. For x sufficientlylarge, there is an integer m such that f(m) is k-free with
x < m ≤ x + cxn
2k−n+1.
Theorem: Let k ≥ n + 1. For x sufficiently large,there is an integer m such that f(m) is k-free with
x < m ≤ x + cxn
2k−n+r,
where r =√
2n − 12.
Basic Idea: One works in a number field where f(x)
has a linear factor. As in the case f(x) = x, onewants to show certain u (in the ring of algebraic inte-gers in the field) are not close by considering
(u + a)kP − ukQ
arising from Pade approximations. One uses that thisexpression is an integer and, hence, either 0 or ≥ 1.
Basic Idea: One works in a number field where f(x)
has a linear factor. As in the case f(x) = x, onewants to show certain u (in the ring of algebraic inte-gers in the field) are not close by considering
(u + a)kP − ukQ
arising from Pade approximations. One uses that thisexpression is an integer and, hence, either 0 or ≥ 1.
Difficulty: An “integer” in this context can be smallwithout being 0.
Basic Idea: One works in a number field where f(x)
has a linear factor. As in the case f(x) = x, onewants to show certain u (in the ring of algebraic inte-gers in the field) are not close by considering
(u + a)kP − ukQ
arising from Pade approximations. One uses that thisexpression is an integer and, hence, either 0 or ≥ 1.
Difficulty: An “integer” in this context can be smallwithout being 0.
Solution: If it’s small, work with a conjugate instead.
Comment: In the case that k ≤ n, one can try thesame methods. The gap size becomes “bad” in thesense that one obtains m ∈ (x, x + h] where f(m)
is k-free but h increases as k decreases. There is apoint where h exceeds x itself and the method fails(the size of f(m) is no longer of order xn). Nairtook the limit of what can be done with k ≤ n andobtained
Comment: In the case that k ≤ n, one can try thesame methods. The gap size becomes “bad” in thesense that one obtains m ∈ (x, x + h] where f(m)
is k-free but h increases as k decreases. There is apoint where h exceeds x itself and the method fails(the size of f(m) is no longer of order xn). Nairtook the limit of what can be done with k ≤ n andobtained
Theorem (Nair): If f(x) is irreducible of degree n andk ≥
(2√
2 − 1)n/2, then there are infinitely many
integers m for which f(m) is k-free.
Theorem (Nair): If f(x) is irreducible of degree n andk ≥
(2√
2 − 1)n/2, then there are infinitely many
integers m for which f(m) is k-free.
Theorem (Nair): If f(x) is irreducible of degree n andk ≥
(2√
2 − 1)n/2, then there are infinitely many
integers m for which f(m) is k-free.
Theorem: If f(x, y) is an irreducible binary form of
degree n and k ≥(2√
2 − 1)n/4, then there are
infinitely many integer pairs (a, b) for which f(a, b)
is k-free.
The abc-conjecture:
The abc-conjecture:
Notation: Q(n) =∏p|n
p
The abc-conjecture:
Notation: Q(n) =∏p|n
p
The abc-Conjecture: For a and b in Z+, define
La,b =log(a + b)
log Q(ab(a + b)
)and
L = {La,b : a ≥ 1, b ≥ 1, gcd(a, b) = 1}.
The abc-conjecture:
Notation: Q(n) =∏p|n
p
The abc-Conjecture: For a and b in Z+, define
La,b =log(a + b)
log Q(ab(a + b)
)and
L = {La,b : a ≥ 1, b ≥ 1, gcd(a, b) = 1}.
The set of limit points of L is the interval [1/3, 1].
La,b =log(a + b)
log Q(ab(a + b)
)L = {La,b : a ≥ 1, b ≥ 1, gcd(a, b) = 1}
La,b =log(a + b)
log Q(ab(a + b)
)L = {La,b : a ≥ 1, b ≥ 1, gcd(a, b) = 1}
Theorem: The set of limit points of L includes theinterval [1/3, 36/37].
La,b =log(a + b)
log Q(ab(a + b)
)L = {La,b : a ≥ 1, b ≥ 1, gcd(a, b) = 1}
Theorem: The set of limit points of L includes theinterval [1/3, 36/37].
(work of Browkin, Greaves, F., Nitaj, Schinzel)
Approach: Makes use of a preliminary result aboutsquarefree values of binary forms.
Approach: Makes use of a preliminary result aboutsquarefree values of binary forms. In particular, for
f(x, y) = xy(x + y)(x − y)(x2 + y2)(2x2 + y2)(x2 + 2y2)
× (x4 − x2y2 + y4)(3x4 + 3x2y2 + y4)(x4 + 3x2y2 + 3y4)
the number f(x, y)/6 takes on the right proportionof squarefree values for
X < x ≤ 2X, Y < y ≤ 2Y, X = Y α,
where α ∈ (1, 3).
Polynomial Identity:
Polynomial Identity:
P3(z) − (1 − z)7Q3(z) = z7E3(z)
where
P3(z) = (2z − 1)(3z2 − 3z + 1),
Q3(z) = −(z + 1)(z2 + z + 1),
and
E3(z) = −(z − 2)(z2 − 3z + 3)
Polynomial Identity:
P3(z) − (1 − z)7Q3(z) = z7E3(z)
Polynomial Identity:
P3(z) − (1 − z)7Q3(z) = z7E3(z)
z =x
x + y=⇒
Polynomial Identity:
P3(z) − (1 − z)7Q3(z) = z7E3(z)
z =x
x + y=⇒
{(x + y)7(x − y)(x2 − xy + y2)
+ y7(2x + y)(3x2 + 3xy + y2)
= x7(x + 2y)(x2 + 3xy + 3y2)
(x + y)7(x − y)(x2 − xy + y2)
+ y7(2x + y)(3x2 + 3xy + y2)
= x7(x + 2y)(x2 + 3xy + 3y2)
(x + y)7(x − y)(x2 − xy + y2)
+ y7(2x + y)(3x2 + 3xy + y2)
= x7(x + 2y)(x2 + 3xy + 3y2)
(x2 + y2)7(x2 − y2)(x4 − x2y2 + y4)
+ y14(2x2 + y2)(3x4 + 3x2y2 + y4)
= x14(x2 + 2y2)(x4 + 3x2y2 + 3y4)
(x + y)7(x − y)(x2 − xy + y2)
+ y7(2x + y)(3x2 + 3xy + y2)
= x7(x + 2y)(x2 + 3xy + 3y2)
(x2 + y2)7(x2 − y2)(x4 − x2y2 + y4)
+ y14(2x2 + y2)(3x4 + 3x2y2 + y4)
= x14(x2 + 2y2)(x4 + 3x2y2 + 3y4)
f(x, y) = xy(x + y)(x − y)(x2 + y2)(2x2 + y2)(x2 + 2y2)
× (x4 − x2y2 + y4)(3x4 + 3x2y2 + y4)(x4 + 3x2y2 + 3y4)
(x + y)7(x − y)(x2 − xy + y2)
+ y7(2x + y)(3x2 + 3xy + y2)
= x7(x + 2y)(x2 + 3xy + 3y2)
(x2 + y2)7(x2 − y2)(x4 − x2y2 + y4)
+ y14(2x2 + y2)(3x4 + 3x2y2 + y4)
= x14(x2 + 2y2)(x4 + 3x2y2 + 3y4)
f(x, y) = xy(x + y)(x − y)(x2 + y2)(2x2 + y2)(x2 + 2y2)
× (x4 − x2y2 + y4)(3x4 + 3x2y2 + y4)(x4 + 3x2y2 + 3y4)
(x + y)7(x − y)(x2 − xy + y2)
+ y7(2x + y)(3x2 + 3xy + y2)
= x7(x + 2y)(x2 + 3xy + 3y2)
(x2 + y2)7(x2 − y2)(x4 − x2y2 + y4)
+ y14(2x2 + y2)(3x4 + 3x2y2 + y4)
= x14(x2 + 2y2)(x4 + 3x2y2 + 3y4)
f(x, y) = xy(x + y)(x − y)(x2 + y2)(2x2 + y2)(x2 + 2y2)
× (x4 − x2y2 + y4)(3x4 + 3x2y2 + y4)(x4 + 3x2y2 + 3y4)
a = (x2 + y2)7(x2 − y2)(x4 − x2y2 + y4)
b = y14(2x2 + y2)(3x4 + 3x2y2 + y4)
X = Y α, 1 < α < 3
a + b = x14(x2 + 2y2)(x4 + 3x2y2 + 3y4)
a = (x2 + y2)7(x2 − y2)(x4 − x2y2 + y4)
b = y14(2x2 + y2)(3x4 + 3x2y2 + y4)
X = Y α, 1 < α < 3
a + b = x14(x2 + 2y2)(x4 + 3x2y2 + 3y4)
f(x, y) = xy(x + y)(x − y)(x2 + y2)(2x2 + y2)(x2 + 2y2)
× (x4 − x2y2 + y4)(3x4 + 3x2y2 + y4)(x4 + 3x2y2 + 3y4)
a = (x2 + y2)7(x2 − y2)(x4 − x2y2 + y4)
b = y14(2x2 + y2)(3x4 + 3x2y2 + y4)
X = Y α, 1 < α < 3
a + b = x14(x2 + 2y2)(x4 + 3x2y2 + 3y4)
f(x, y) = xy(x + y)(x − y)(x2 + y2)(2x2 + y2)(x2 + 2y2)
× (x4 − x2y2 + y4)(3x4 + 3x2y2 + y4)(x4 + 3x2y2 + 3y4)
La,b =log(a + b)
log Q(ab(a + b)
)
a = (x2 + y2)7(x2 − y2)(x4 − x2y2 + y4)
b = y14(2x2 + y2)(3x4 + 3x2y2 + y4)
X = Y α, 1 < α < 3
a + b = x14(x2 + 2y2)(x4 + 3x2y2 + 3y4)
f(x, y) = xy(x + y)(x − y)(x2 + y2)(2x2 + y2)(x2 + 2y2)
× (x4 − x2y2 + y4)(3x4 + 3x2y2 + y4)(x4 + 3x2y2 + 3y4)
La,b =log(a + b)
log Q(ab(a + b)
)
a = (x2 + y2)7(x2 − y2)(x4 − x2y2 + y4)
b = y14(2x2 + y2)(3x4 + 3x2y2 + y4)
X = Y α, 1 < α < 3
a + b = x14(x2 + 2y2)(x4 + 3x2y2 + 3y4)
f(x, y) = xy(x + y)(x − y)(x2 + y2)(2x2 + y2)(x2 + 2y2)
× (x4 − x2y2 + y4)(3x4 + 3x2y2 + y4)(x4 + 3x2y2 + 3y4)
La,b =log(a + b)
log Q(ab(a + b)
) ≈20α log Y
a = (x2 + y2)7(x2 − y2)(x4 − x2y2 + y4)
b = y14(2x2 + y2)(3x4 + 3x2y2 + y4)
X = Y α, 1 < α < 3
a + b = x14(x2 + 2y2)(x4 + 3x2y2 + 3y4)
f(x, y) = xy(x + y)(x − y)(x2 + y2)(2x2 + y2)(x2 + 2y2)
× (x4 − x2y2 + y4)(3x4 + 3x2y2 + y4)(x4 + 3x2y2 + 3y4)
La,b =log(a + b)
log Q(ab(a + b)
) ≈20α log Y
a = (x2 + y2)7(x2 − y2)(x4 − x2y2 + y4)
b = y14(2x2 + y2)(3x4 + 3x2y2 + y4)
X = Y α, 1 < α < 3
a + b = x14(x2 + 2y2)(x4 + 3x2y2 + 3y4)
f(x, y) = xy(x + y)(x − y)(x2 + y2)(2x2 + y2)(x2 + 2y2)
× (x4 − x2y2 + y4)(3x4 + 3x2y2 + y4)(x4 + 3x2y2 + 3y4)
La,b =log(a + b)
log Q(ab(a + b)
) ≈20α log Y
(21α + 1) log Y
a = (x2 + y2)7(x2 − y2)(x4 − x2y2 + y4)
b = y14(2x2 + y2)(3x4 + 3x2y2 + y4)
X = Y α, 1 < α < 3
a + b = x14(x2 + 2y2)(x4 + 3x2y2 + 3y4)
f(x, y) = xy(x + y)(x − y)(x2 + y2)(2x2 + y2)(x2 + 2y2)
× (x4 − x2y2 + y4)(3x4 + 3x2y2 + y4)(x4 + 3x2y2 + 3y4)
La,b =log(a + b)
log Q(ab(a + b)
) ≈20α
21α + 1
La,b =log(a + b)
log Q(ab(a + b)
) ≈20α
21α + 1
La,b =log(a + b)
log Q(ab(a + b)
) ≈20α
21α + 1
1 < α < 3 =⇒
La,b =log(a + b)
log Q(ab(a + b)
) ≈20α
21α + 1
1 < α < 3 =⇒ ?? < La,b < ??
La,b =log(a + b)
log Q(ab(a + b)
) ≈20α
21α + 1
1 < α < 3 =⇒10
11< La,b <
15
16
La,b =log(a + b)
log Q(ab(a + b)
) ≈20α
21α + 1
1 < α < 3 =⇒10
11< La,b <
15
16
Comment: This shows [10/11, 15/16] is containedin the set of limit points of La,b. A similar argumentis given for other subintervals of [1/3, 36/37] (not allinvolving Pade approximations).
The End
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