Sums of Consecutive Perfect Powers is Seldom a Perfect Power

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Sums of Consecutive Perfect Powers is Seldom a Perfect Power Journées Algophantiennes Bordelaises 2017, Université de Bordeaux June 7, 2017

A Diophantine Equation Question x k + (x + 1) k + + (x + d 1) k = y n. Fix k 2 and d 2. Determine all of the integer solutions (x, y, n), n 2.

A Brief History Theorem (Zhang and Bai, 2013) Let q be a prime such that q 5, 7 (mod 12). Suppose q d. Then the equation x 2 + (x + 1) 2 + + (x + d 1) 2 = y n has no integer solutions. Corollary (Use Dirichlet s Theorem) Let A 2 be the set of integers d 2 such that the equation x 2 + (x + 1) 2 + + (x + d 1) 2 = y n has a solution (x, y, n). Then A 2 has natural density zero.

A Brief History Theorem (Zhang and Bai, 2013) Let q be a prime such that q 5, 7 (mod 12). Suppose q d. Then the equation x 2 + (x + 1) 2 + + (x + d 1) 2 = y n has no integer solutions. Corollary (Use Dirichlet s Theorem) Let A 2 be the set of integers d 2 such that the equation x 2 + (x + 1) 2 + + (x + d 1) 2 = y n has a solution (x, y, n). Then A 2 has natural density zero.

The Result Theorem (V. Patel, S. Siksek) Let k 2 be an even integer. Let A k be the set of integers d 2 such that the equation x k + (x + 1) k + (x + d 1) k = y n, x, y, n Z, n 2 has a solution (x, y, n). Then A k has natural density zero. In other words we have #{d A k : d X} lim = 0. X X

The Result Theorem (V. Patel, S. Siksek) Let k 2 be an even integer and let r be a non-zero integer. Let A k,r be the set of integers d 2 such that the equation x k + (x + r) k + (x + r(d 1)) k = y n, x, y, n Z, n 2 has a solution (x, y, n). Then A k,r has natural density zero. In other words we have #{d A k,r : d X} lim = 0. X X

Bernoulli polynomials and relation to sums of consecutive powers Definition (Bernoulli Numbers, b k ) x e x 1 = x k b k k!. k=0 b 0 = 1, b 1 = 1/2, b 2 = 1/6, b 3 = 0, b 4 = 1/30, b 5 = 0, b 6 = 1/42. Lemma b 2k+1 = 0 for k 1.

Bernoulli polynomials and relation to sums of consecutive powers Definition (Bernoulli Polynomial, B k ) B k (x) := k m=0 ( ) k b m x k m. m Lemma x k + (x + 1) k + + (x + d 1) k = 1 k + 1 (B k+1(x + d) B k (x)).

Bernoulli polynomials and relation to sums of consecutive powers Definition (Bernoulli Polynomial, B k ) B k (x) := k m=0 ( ) k b m x k m. m Lemma x k + (x + 1) k + + (x + d 1) k = 1 k + 1 (B k+1(x + d) B k (x)).

Bernoulli polynomials and relation to sums of consecutive powers Lemma x k + (x + 1) k + + (x + d 1) k = 1 k + 1 (B k+1(x + d) B k (x)). Apply Taylor s Theorem and use B k+1 (x) = (k + 1) B k(x). Lemma Let q k + 3 be a prime. Let d 2. Suppose that q d. Then x k + (x + 1) k + + (x + d 1) k d B k (x) (mod q 2 ).

Bernoulli polynomials and relation to sums of consecutive powers Lemma x k + (x + 1) k + + (x + d 1) k = 1 k + 1 (B k+1(x + d) B k (x)). Apply Taylor s Theorem and use B k+1 (x) = (k + 1) B k(x). Lemma Let q k + 3 be a prime. Let d 2. Suppose that q d. Then x k + (x + 1) k + + (x + d 1) k d B k (x) (mod q 2 ).

Bernoulli polynomials and relation to sums of consecutive powers x k + (x + 1) k + + (x + d 1) k = y n. Proposition (Criterion) Let k 2. Let q k + 3 be a prime such that the congruence B k (x) 0 (mod q) has no solutions. Let d be a positive integer such that ord q (d) = 1. Then the equation has no solutions. (i.e. d / A k ). Remark: Computationally we checked k 75,000 and we could always find such a q.

Bernoulli polynomials and relation to sums of consecutive powers x k + (x + 1) k + + (x + d 1) k = y n. Proposition (Criterion) Let k 2. Let q k + 3 be a prime such that the congruence B k (x) 0 (mod q) has no solutions. Let d be a positive integer such that ord q (d) = 1. Then the equation has no solutions. (i.e. d / A k ). Remark: Computationally we checked k 75,000 and we could always find such a q.

Relation to Densities? We need to use Chebotarev s density theorem, which can be seen as a generalisation of Dirichlet s theorem on primes in arithmetic progression. Proposition Let k 2 be even and let G be the Galois group of B k (x). Then there is an element µ G that acts freely on the roots of B k (x). Assuming the proposition, we may then use Chebotarev s density theorem to find a set of primes q i with positive Dirichlet density such that Frob qi G is conjugate to µ. Then we can apply Niven s results to deduce our Theorem.

Relation to Densities? We need to use Chebotarev s density theorem, which can be seen as a generalisation of Dirichlet s theorem on primes in arithmetic progression. Proposition Let k 2 be even and let G be the Galois group of B k (x). Then there is an element µ G that acts freely on the roots of B k (x). Assuming the proposition, we may then use Chebotarev s density theorem to find a set of primes q i with positive Dirichlet density such that Frob qi G is conjugate to µ. Then we can apply Niven s results to deduce our Theorem.

Niven s Results (Flash!) The setup: 1 Let A be a set of positive integers. 2 Define: A(X) = #{d A : d X} for positive X. 3 Natural Density: δ(a) = lim X A(X)/X. 4 Given a prime q, define: A (q) = {d A : ord q (d) = 1}. Theorem (Niven) Let {q i } be a set of primes such that δ(a (q i) ) = 0 and q 1 i =. Then δ(a) = 0.

A Legendre Symbol analogue Proposition Let k 2 be even and let G be the Galois group B k (x). Then there is an element µ G that acts freely on the roots of B k (x). Conjecture For any even integer k, B k (x) is irreducible over Q. Remark: The conjecture implies the Proposition. This then proves our Theorem.

A Legendre Symbol analogue Proposition Let k 2 be even and let G be the Galois group B k (x). Then there is an element µ G that acts freely on the roots of B k (x). Conjecture For any even integer k, B k (x) is irreducible over Q. Remark: The conjecture implies the Proposition. This then proves our Theorem.

Tough Stuff A sketch of an unconditional proof! Proposition Let k 2 be even and let G be the Galois group B k (x). Then there is an element µ G that acts freely on the roots of B k (x). Theorem (von Staudt-Clausen) Let n 2 be even. Then b n + (p 1) n 1 p Z.

2 is the Oddest Prime The Newton Polygon of B k (x) for k = 2 s t, s 1. ν 2 (a i ) B k (x) = k i=0 ( ) k b k i x i = k i k a i x i i=0 (0, 0) (0, 1) (k 2 s, 0) (k 2 s, 1) slope = 1/2 s (k, 0) i

Another nice result 1 Sloping part corresponds to irreducible factor over Q 2. 2 Root in Q 2 must have valuation zero. 3 Root belongs to Z 2 and is odd. 4 Symmetry ( 1) k B k (x) = B k (1 x) gives a contradiction. Theorem (V. Patel, S. Siksek) Let k 2 be an even integer. Then B k (x) has no roots in Q 2. Theorem (K. Inkeri, 1959) Let k 2 be an even integer. Then B k (x) has no roots in Q.

Another nice result 1 Sloping part corresponds to irreducible factor over Q 2. 2 Root in Q 2 must have valuation zero. 3 Root belongs to Z 2 and is odd. 4 Symmetry ( 1) k B k (x) = B k (1 x) gives a contradiction. Theorem (V. Patel, S. Siksek) Let k 2 be an even integer. Then B k (x) has no roots in Q 2. Theorem (K. Inkeri, 1959) Let k 2 be an even integer. Then B k (x) has no roots in Q.

Another nice result 1 Sloping part corresponds to irreducible factor over Q 2. 2 Root in Q 2 must have valuation zero. 3 Root belongs to Z 2 and is odd. 4 Symmetry ( 1) k B k (x) = B k (1 x) gives a contradiction. Theorem (V. Patel, S. Siksek) Let k 2 be an even integer. Then B k (x) has no roots in Q 2. Theorem (K. Inkeri, 1959) Let k 2 be an even integer. Then B k (x) has no roots in Q.

What is Going On? L = Splitting Field of B k (x) L P F P G = Galois Group H G C = Cyclic Q Q 2 F 2 = Residue Field

What is Going On? L = Splitting Field of B k (x) L P F P G = Galois Group H G C = Cyclic Q Q 2 F 2 = Residue Field µ lives here!

A sketch proof of the Proposition The Setup: k 2 is even. L is the splitting field of B k (x). G is the Galois group of B k (x). P be a prime above 2. ν 2 on Q 2 which we extend uniquely to L P (also call it ν 2 ). H = Gal(L P /Q 2 ) G be the decomposition subgroup corresponding to P.

A sketch proof of the Proposition B k (x) = g(x)h(x) where g(x) has degree k 2 s. Label the roots {α 1,..., α k 2 s}, and h(x) has degree 2 s. Label the roots {β 1,..., β 2 s}. All roots L β. h(x) is irreducible. Therefore H acts transitively on β j. Pick µ H such that µ acts freely on the roots of h(x). Check it doesn t end up fixing a root of g(x).

Bad Prime = Extremely Useful Prime! The Newton Polygon of B k (x) for k = 2 s t, s 1. ν 2 (a i ) (0, 0) (k 2 s, 0) slope = 1/2 s (k, 0) i (0, 1) (k 2 s, 1)

Finding µ Lemma Let H be a finite group acting transitively on a finite set {β 1,..., β n }. Let H i H be the stabiliser of β i and suppose H 1 = H 2. Let π : H C be a surjective homomorphism from H onto a cyclic group C. Then there exists some µ H acting freely on {β 1,..., β n } such that π(µ) is a generator of C. 1 Let F P be the residue field of P. 2 Let C = Gal (F P /F 2 ). 3 C is cyclic generated by the Frobenius map: γ γ 2. 4 Let π : H C be the induced surjection. 5 Finally use the Lemma.

Finding µ Lemma Let H be a finite group acting transitively on a finite set {β 1,..., β n }. Let H i H be the stabiliser of β i and suppose H 1 = H 2. Let π : H C be a surjective homomorphism from H onto a cyclic group C. Then there exists some µ H acting freely on {β 1,..., β n } such that π(µ) is a generator of C. 1 Let F P be the residue field of P. 2 Let C = Gal (F P /F 2 ). 3 C is cyclic generated by the Frobenius map: γ γ 2. 4 Let π : H C be the induced surjection. 5 Finally use the Lemma.

Check g(x) B k (x) = g(x)h(x) where g(x) has degree k 2 s. Label the roots {α 1,..., α k 2 s}, and h(x) has degree 2 s. Label the roots {β 1,..., β 2 s}. Lemma µ acts freely on the α i. 1 Suppose not. Let α be a root that is fixed by µ. 2 ν 2 (α) = 0 so let ᾱ = α (mod P), ᾱ F P. 3 α fixed by µ hence ᾱ fixed by π(µ) = C. 4 Hence ᾱ F 2. f(x) = 2B k (x) Z 2 [x]. 5 f( 1) = f( 0) = 1. A contradiction!

Thank you for Listening!

Solving the equations for k = 2 ( ( d x + d + 1 ) 2 + 2 ) (d 1)(d + 1) = y p. 12 X 2 + C 1 p = (1/d)y p

Solving the equations for k = 2 d Equation Level Dimension 6 2y p 5 7 = 3(2x + 7) 2 2 7 3 2 5 7 480 11 11 p 1 y p 2 5 = (x + 6) 2 2 7 5 11 160 13 13 p 1 y p 2 7 = (x + 7) 2 2 7 7 13 288 22 2 11 p 1 y p 7 23 = (2x + 23) 2 2 7 7 11 23 5,280 23 23 p 1 y p 2 2 11 = (x + 12) 2 2 3 11 23 54 26 2 13 p 1 y p 3 2 5 2 = (2x + 27) 2 2 7 3 5 13 384 33 11 p 1 y p 2 4 17 = 3(x + 17) 2 2 3 3 2 11 17 200 37 37 p 1 y p 2 3 19 = (x + 19) 2 2 7 3 19 37 5,184 39 13 p 1 y p 2 2 5 19 = 3(x + 20) 2 2 3 3 2 5 13 19 1,080 46 2 23 p 1 y p 3 2 5 47 = (2x + 47) 2 2 7 3 5 23 47 32,384 47 47 p 1 y p 2 3 23 = (x + 24) 2 2 5 23 47 1,012 59 59 p 1 y p 2 5 29 = (x + 30) 2 2 7 5 29 59 25,984

Solving the equations for k = 4 d Equation Level Dimension 5 y p + 2 73 = 5(X) 2 2 7 5 2 73 5,472 6 y p + 7 53 = 6(X) 2 2 8 3 2 7 53 12,480 7 7 p 1 y p + 2 2 29 = (X) 2 2 3 7 29 42 10 y p + 3 11 149 = 10(X) 2 2 8 5 2 3 11 149 449,920 13 13 p 1 y p + 2 7 101 = (X) 2 2 7 7 13 101 28,800 14 7 p 1 y p + 13 293 = 2(X) 2 2 8 7 13 293 168,192 15 y p + 2 3 7 673 = 15(X) 2 2 5 3 2 5 2 7 673 383,040 17 17 p 1 y p + 2 3 3 173 = (X) 2 2 5 3 17 173 5,504 19 19 p 1 y p + 2 3 23 47 = (X) 2 2 7 3 19 23 47 145,728 21 7 p 1 y p + 2 11 1321 = 3(X) 2 2 7 3 2 7 11 1321 1,584,000 26 13 p 1 y p + 3 2 5 1013 = 2(X) 2 2 8 3 5 13 1013 777,216 29 29 p 1 y p + 2 7 2521 = (X) 2 2 7 7 29 2521 1,693,440 30 y p + 19 29 31 71 = 30(X) 2 2 8 3 2 5 2 19 29 31 71 804,384,000 Where X is a quadratic in the original variable x.

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