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

NOT MY TITLE

How many positive integers n ≤ 1000

are such that 2n begins with the digit 1?

NOT MY TITLE



301

NOT MY TITLE



301

log10 2 = 0.30102999566398 . . .

x #{n ≤ x : 2n begins with 1}10 3


100 30

1000 301


100 30

1000 301

10000 3010

100000 30102

1000000 301029

10000000 3010299


100 30

1000 301

10000 3010

100000 30102

1000000 301029

10000000 3010299

log10 2 = 0.30102999566398 . . .


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:


• irrationality measures


• irrationality measures• diophantine equations


• irrationality measures• diophantine equations• Waring’s problem


• irrationality measures• diophantine equations• Waring’s problem• the factorization of n(n + 1)



• Galois groups of classical polynomials



• Galois groups of classical polynomials• the Ramanujan-Nagell equation



• Galois groups of classical polynomials• the Ramanujan-Nagell equation• k-free numbers in short intervals



• Galois groups of classical polynomials• the Ramanujan-Nagell equation• k-free numbers in short intervals• k-free values of polynomials and binary forms



• 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?




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.







Important Equation:



Important Equation:

(1 − z)k ≈P (z)

Q(z)



Important Equation:

(1 − z)k =P (z)

Q(z)− zmR(z)



Important Equation:

(1 − z)k =P (z)

Q(z)− zmR(z)

degree < k (usually)

Michael Filaseta

P(z)

Michael Filaseta

P(z)

Michael Filaseta

Michael Filaseta



Important Equation:

P − (1 − z)k Q = zmE



Important Equation:

Pr− (1 − z)kQr = zmEr



Important Equation:

Pr− (1 − z)kQr = z2r+1Er



Important Equation:




Important Equation:




Important Equation:


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


WARNING: In the applications you areabout to see, the true identies used havebeen changed.


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:


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.


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



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



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


Comment: Liouville’s result is effective; Roth’s is not.


Theorem( Baker ): For a and b integers with b > 0,∣∣∣ 3√2 −

a

b

∣∣∣ >C

b2.955

where C = 10−6.


Theorem( Baker ): For a & b integers with b > 0,∣∣∣ 3√2 −

a

b

∣∣∣ >1

106b2.955.


Theorem(Chudnovsky ): For a & b integers with b > 0,∣∣∣ 3√2 −

a

b

∣∣∣ >1

c · b2.43.


Theorem( Bennett ): For a & b integers with b > 0,∣∣∣ 3√2 −

a

b

∣∣∣ >1

c · b2.47.



a

b

∣∣∣ >1

4 · b2.47.



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


3√2 br − ar = small

The Basic Approach:

Pr − (125/128)1/3Qr = z2r+1Er


3√2 br − ar = smallr

The Basic Approach:

Pr − (125/128)1/3Qr = z2r+1Er


3√2 br − ar = smallr∣∣∣ 3√2 −

ar

br

∣∣∣ = smallr

The Basic Approach:

Pr − (125/128)1/3Qr = z2r+1Er


3√2 br − ar = smallr∣∣∣ 3√2 −

ar

br

∣∣∣ = smallr

The Basic Approach:

Pr − (125/128)1/3Qr = z2r+1Er


3√2 br − ar = smallr∣∣∣ 3√2 −

ar

br

∣∣∣ = smallr

Wait!!

The Basic Approach:

Pr − (125/128)1/3Qr = z2r+1Er


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


ar

br

∣∣∣ = smallr

What’s small r?


ar

br

∣∣∣ = smallr

What’s small r? Let b be a positive integer.


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.


ar

br

∣∣∣ = smallr


smallr <1


∣∣∣ 3√2 −

a

b

∣∣∣ ≥


ar

br

∣∣∣ = smallr


smallr <1


∣∣∣ 3√2 −

a

b

∣∣∣ ≥∣∣∣ar

br−

a

b

∣∣∣ −∣∣∣ 3√

2 −ar

br

∣∣∣ >


ar

br

∣∣∣ = smallr


smallr <1


∣∣∣ 3√2 −

a

b

∣∣∣ ≥∣∣∣ar

br−

a

b

∣∣∣ −∣∣∣ 3√

2 −ar

br

∣∣∣ >


ar

br

∣∣∣ = smallr


smallr <1


∣∣∣ 3√2 −

a

b

∣∣∣ ≥∣∣∣ar

br−

a

b

∣∣∣ −∣∣∣ 3√

2 −ar

br

∣∣∣ >1

b br


ar

br

∣∣∣ = smallr


smallr <1


∣∣∣ 3√2 −

a

b

∣∣∣ ≥∣∣∣ar

br−

a

b

∣∣∣ −∣∣∣ 3√

2 −ar

br

∣∣∣ >1

b br


ar

br

∣∣∣ = smallr


smallr <1


∣∣∣ 3√2 −

a

b

∣∣∣ ≥∣∣∣ar

br−

a

b

∣∣∣ −∣∣∣ 3√

2 −ar

br

∣∣∣ >1

b br


ar

br

∣∣∣ = smallr


smallr <1


∣∣∣ 3√2 −

a

b

∣∣∣ ≥∣∣∣ar

br−

a

b

∣∣∣ −∣∣∣ 3√

2 −ar

br

∣∣∣ >1

b br


ar

br

∣∣∣ = smallr


smallr <1


∣∣∣ 3√2 −

a

b

∣∣∣ ≥∣∣∣ar

br−

a

b

∣∣∣ −∣∣∣ 3√

2 −ar

br

∣∣∣ >1

b br−


ar

br

∣∣∣ = smallr


smallr <1


∣∣∣ 3√2 −

a

b

∣∣∣ ≥∣∣∣ar

br−

a

b

∣∣∣ −∣∣∣ 3√

2 −ar

br

∣∣∣ >1

b br−

1

2b br


ar

br

∣∣∣ = smallr


smallr <1


∣∣∣ 3√2 −

a

b

∣∣∣ ≥∣∣∣ar

br−

a

b

∣∣∣ −∣∣∣ 3√

2 −ar

br

∣∣∣ >1

2b br


ar

br

∣∣∣ = smallr


smallr <1


∣∣∣ 3√2 −

a

b

∣∣∣ ≥∣∣∣ar

br−

a

b

∣∣∣ −∣∣∣ 3√

2 −ar

br

∣∣∣ >1

2b br


ar

br

∣∣∣ = smallr


smallr <1


∣∣∣ 3√2 −

a

b

∣∣∣ ≥∣∣∣ar

br−

a

b

∣∣∣ −∣∣∣ 3√

2 −ar

br

∣∣∣ >1

2cb2.47


ar

br

∣∣∣ = smallr


smallr <1


∣∣∣ 3√2 −

a

b

∣∣∣ ≥∣∣∣ar

br−

a

b

∣∣∣ −∣∣∣ 3√

2 −ar

br

∣∣∣ >1

4 ·b2.47

Diophantine equations:


Theorem (Bennett):For a and b integers with b > 0,∣∣∣ 3√2 −

a

b

∣∣∣ >1

4 · b2.47.


Theorem (Bennett):For a and b integers with b 6= 0,∣∣∣ 3√2 −

a

b

∣∣∣ >1

4 · b2.47.



a

b

∣∣∣ >1

4 · |b|2.47.



a

b

∣∣∣ >1

4 · |b|2.5.



a

b

∣∣∣ >1

4 · |b|2.5.

x3 − 2y3 = n



a

b

∣∣∣ >1

4 · |b|2.5.

x3 − 2y3 = n, y 6= 0



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



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



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



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


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


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|


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


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


Theorem: Let n be a non-zero integer. If x and y areintegers satisfying x3−2y3 = n, then |y| < 16n2.


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:


Known: (i) g(k) = 2k +

[(3

2

)k]

− 2

Waring’s Problem:



[(3

2

)k]

− 2

(ii) no one knows how to prove (i)

Waring’s Problem:



[(3

2

)k]

− 2


(iii) (i) holds if∥∥∥(3

2

)k∥∥∥ > 0.75k

Waring’s Problem:



[(3

2

)k]

− 2



2

)k∥∥∥ > 0.75k

(iv) (iii) holds if k > 8

Waring’s Problem:



[(3

2

)k]

− 2



2

)k∥∥∥ > 0.75k

(iv) (iii) holds if k > 8

(v) no one knows how to prove (iv)

Waring’s Problem:


[(3

2

)k]

− 2

(ii) No one knows how to prove (i).


2

)k∥∥∥ > 0.75k

Waring’s Problem:


[(3

2

)k]

− 2



2

)k∥∥∥ > 0.75k

Theorem (Beukers): If k > 4, then∥∥∥(3

2

)k∥∥∥ > 0.5358k.

Waring’s Problem:


[(3

2

)k]

− 2



2

)k∥∥∥ > 0.75k

Theorem(Dubitskas): If k > 4, then∥∥∥(3

2

)k∥∥∥ > 0.5767k.

The factorization of n(n + 1):


Well-Known: The largest prime factor of n(n + 1)

tends to infinity with n.


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.


n(n + 1) = pe11 p

e22 · · · per

r m



n(n + 1) = pe11 p

e22 · · · per

r m


Lehmer: Gave some explicit estimates:


n(n + 1) = pe11 p

e22 · · · per

r m



n(n+1) divisible only by primes ≤ 11 =⇒ n ≤


n(n + 1) = pe11 p

e22 · · · per

r m



n(n+1) divisible only by primes ≤ 11 =⇒ n ≤... only by primes ≤ 41 =⇒ n ≤


n(n + 1) = pe11 p

e22 · · · per

r m



n(n+1) divisible only by primes ≤ 11 =⇒ n ≤ 9800

... only by primes ≤ 41 =⇒ n ≤


n(n + 1) = pe11 p

e22 · · · per

r m



n(n+1) divisible only by primes ≤ 11 =⇒ n ≤ 9800

... only by primes ≤ 41 =⇒ n ≤ 63927525375


n(n + 1) = pe11 p

e22 · · · per

r m



n(n + 1) = pe11 p

e22 · · · per

r m



n(n + 1) = pe11 p

e22 · · · per

r m

for some integer m, then m >1 .


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



n(n + 1) = pe11 p

e22 · · · per

r m


abc-conjecture =⇒ θ =


n(n + 1) = pe11 p

e22 · · · per

r m


abc-conjecture =⇒ θ = 1 − ε


n(n + 1) = pe11 p

e22 · · · per

r m



unconditionally one can obtain θ =


n(n + 1) = pe11 p

e22 · · · per

r m



unconditionally one can obtain θ = 1 − ε


n(n + 1) = pe11 p

e22 · · · per

r m



unconditionally one can obtain

(ineffective)

θ = 1 − ε


n(n + 1) = pe11 p

e22 · · · per

r m


Effective Approach:


n(n + 1) = pe11 p

e22 · · · per

r m


Effective Approach: (Linear Forms of Logarithms)


n(n + 1) = pe11 p

e22 · · · per

r m



θ =c

log log n


n(n + 1) = pe11 p

e22 · · · per

r m



θ =c

log log n

Problem: Can we narrow the gap betweenthese ineffective and effective results?


n(n + 1) = pe11 p

e22 · · · per

r m



n(n + 1) = pe11 p

e22 · · · per

r m


Theorem (Bennett, F., Trifonov): If n ≥ 9 and

n(n + 1) = 2k3`m,

thenm ≥


n(n + 1) = pe11 p

e22 · · · per

r m


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.


n(n + 1) = 2u3vm =⇒ m >√

n.

Comment: The conjecture has been verified for

512 < n ≤


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


3kP − 2`Q = E.

Then

3k (Qm1 − Pm2) = ±Q − Em2.

The Method:

n(n + 1) = 3k2`m

3km1 − 2`m2 = ±1


3kP − 2`Q = E.

Then

3k (Qm1 − Pm2) = ±Q − Em2.


3kP − 2`Q = E

and

Qm1 − Pm2 6= 0.

Then

3k (Qm1 − Pm2) = ±Q − Em2.


3kP − 2`Q = E

and

Qm1 − Pm2 6= 0.

Then

3k (Qm1 − Pm2) = ±Q − Em2.

Obtain an upper bound on 3k.


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:


• 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.


• 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.



• 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.


• 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 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).


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 (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).



(n)n (x) is irre-

ducible.



(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.



(n)n (x) is irre-

ducible.





Theorem (Kidd, Trifonov, F.): If n > 2 and n ≡ 2

(mod 4), then L(n)n (x) is irreducible.



(n)n (x) is irre-

ducible.





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:


Classical Ramanujan-Nagell Theorem: If x and n

are positive integers satisfying

x2 + 7 = 2n,

then




x2 + 7 = 2n,

thenx ∈ {1, 3, 5, 11, 181}.


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).




x2 + 7 = 2n,

thenx ∈ {1, 3, 5, 11, 181}.




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?


Connection with n(n + 1) problem:



x2 + 7 = 2nm



x2 + 7 = 2nm

(x+

√−7

2

)(x−

√−7

2

)=

(1+

√−7

2

)n−2(1−

√−7

2

)n−2

m



x2 + 7 = 2nm

(x+

√−7

2

)↑

linear

(x−

√−7

2

)↑

linear

=

(1+

√−7

2

)n−2(1−

√−7

2

)n−2

m



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 ≥ ???


x2 + 7 = 2nm and x 6∈ {1, 3, 5, 11, 181},

thenm ≥ x1/2.


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:



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.





Small Primes: p ≤ z



Small Primes: p ≤ z where z = xθ√log x



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


x < pkm ≤ x + xθ


x < pkm ≤ x + xθ =⇒x

pk< m ≤

x

pk+

xθ

pk



pk< m ≤

x

pk+

xθ

pk

=⇒∥∥∥ x

pk

∥∥∥ <xθ

Nk



pk< m ≤

x

pk+

xθ

pk

=⇒∥∥∥ x

pk

∥∥∥ <xθ

Nk

where ‖t‖ = min{|t − `| : ` ∈ Z}



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.



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.



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


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


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


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


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


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


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


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


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


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


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


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


uk

∥∥∥ <xθ

Nk,

∥∥∥ x

(u + a)k

∥∥∥ <xθ

Nk

x

ukP −

x


(u + a)kP − ukQ small (but not too small)

∥∥∥ x

uk

∥∥∥ <xθ

Nk, u ∈ (N, 2N ], N ≥ xθ

√log x


uk

∥∥∥ <xθ

Nk,

∥∥∥ x

(u + a)k

∥∥∥ <xθ

Nk

x

ukP −

x


(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


Theorem (Halberstam & Roth):

∥∥∥ x

uk

∥∥∥ <xθ

Nk, u ∈ (N, 2N ], N ≥ xθ

√log x


Theorem (Halberstam & Roth & Nair):

∥∥∥ x

uk

∥∥∥ <xθ

Nk, u ∈ (N, 2N ], N ≥ xθ

√log x


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


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


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:


The method for obtaining results about gaps betweenk-free numbers generalizes to k-free values of poly-nomials.


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.


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.


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.


x < m ≤ x + cxn

2k−n+1.


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 =


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.



(u + a)kP − ukQ


Difficulty: An “integer” in this context can be smallwithout being 0.



(u + a)kP − ukQ


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.


(2√




(2√



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:


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:


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:


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)


P3(z) − (1 − z)7Q3(z) = z7E3(z)


P3(z) − (1 − z)7Q3(z) = z7E3(z)

z =x

x + y=⇒


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)


× (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)


× (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)


× (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)


× (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)


× (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)


× (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)


× (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)


× (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)


× (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)


× (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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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