Introduction Descent The Modular Method and FLT Eliminating Newforms A Unique Perfect Power Decagonal Number Philippe Michaud-Rodgers University of Warwick Young Researchers in Algebraic Number Theory III 20.08.2021 Introduction Descent The Modular Method and FLT Eliminating Newforms Polygonal Numbers The nth s-gonal number is (s − 2)n2 − (s − 4)n P (n)= : s 2 n2+n Triangular numbers: P3(n) = 2 . 2 Square numbers: P4(n) = n . Decagonal numbers: P10(n) = D(n) = n(4n − 3). Introduction Descent The Modular Method and FLT Eliminating Newforms Perfect Powers in Sequences Perfect Power m N > 1 is a perfect power if N = y for some y 2 Z and m > 1. Perfect powers in Fibonacci and Lucas sequences. Perfect powers in arithmetic progressions. Perfect power polygonal numbers (Kim, Park, Pint´er,2013): m All solutions to Ps (n) = y when m > 2, Ps (n) > 1, and s 2 f3; 5; 6; 8; 20g are 3 9 3 P8(2) = 2 and P20(8) = 2 = 8 : A unique perfect power decagonal number (M. 2021) m The only solution to P10(n) = D(n) = y when m > 1 and D(n) > 1 is D(3) = 33. Introduction Descent The Modular Method and FLT Eliminating Newforms Descent We have D(n) = n(4n − 3) = y p, with p prime. Case 1: 3 - n. Then n and 4n − 3 are coprime. So n = ap and 4n − 3 = bp; with (a; b) = 1. So 4ap − bp = 3: (1) Case 2: 3 k n. Then n = 3tp and 4n − 3 = 3p−1up; with (t; u) = 1 and 3 - t. So 4tp − 3p−2up = 1: (2) Case 3: 32 j n. Introduction Descent The Modular Method and FLT Eliminating Newforms The Modular Method These are examples of binomial Thue equations. We can solve them for p = 2; 3; 5; 7 easily. From now on: p ≥ 11. We use the modular method: Suppose we have a non-zero solution. Associate an elliptic curve E (the Frey curve) to this solution. Show that E is linked to a newform f in some way. Obtain a contradiction. A newform at level N is a normalised cusp form in S2(Γ0(N)) that is `new' at its level. Eichler{Shimura: f −! Af , an abelian variety =Q. Introduction Descent The Modular Method and FLT Eliminating Newforms Level-Lowering Level-Lowering Theorem (Ribet) Let E be a (modular) elliptic curve over Q of conductor N and let p ≥ 5 be prime. Suppose ρE;p is irreducible. Then E arises mod p from a newform f , written E ∼p f , at level Np, where N Np = Y : q qkN;pjordq(∆min) Introduction Descent The Modular Method and FLT Eliminating Newforms FLT and Case 2 FLT: x p + y p = zp. Case 2: 4tp − 3p−2up = 1. E : Y 2 = X (X − x p)(X + y p). E : Y 2 + 3XY − 3p−2up Y = X 3. E ∼p f , f a newform at level 2. E ∼p f , f a newform at level 6. But! No newforms at level 2. But! No newforms at level 6. We were very lucky here! Usually there are newforms at the predicted level. Introduction Descent The Modular Method and FLT Eliminating Newforms Frey curves for Case 1 Case 1: 3 - n: 4ap − bp = 3: Frey curve depends on a (mod 4). 2 3 2 p If a ≡ 1 (mod 4) then E1 : Y = X − 3X + 3a X : E1 ∼p f1, the unique newform at level 36. 2 3 2 p If a ≡ 3 (mod 4) then E2 : Y = X + 3X + 3a X : E2 ∼p f2, the unique newform at level 72. 2 3 2 3ap If a is even, we set E3 : Y + XY = X − X + 16 X : E2 ∼p f at level 18, a contradiction. We want E1 6∼p f1 and E2 6∼p f2. Focus on E1 6∼p f1. Introduction Descent The Modular Method and FLT Eliminating Newforms The Trivial Solution Suppose E1 ∼p f1. Now 2 3 f1 −! W1 : Y = X + 1: Here, E1 ∼p f1 means ρE1;p ∼ ρW1;p. All methods of elimination will fail. Why? Because we have a non-zero trivial solution for all p: 4(1p) − 1(1p) = 3: Comes from D(1) = 1 = 1p. ∼ When a = 1, we have E1 = W1 for all p. Introduction Descent The Modular Method and FLT Eliminating Newforms Complex Multiplication p The curve W1 has complex multiplication by K = Q( −3). + So ρW1;p(GQ) ⊆ C ⊂ GL2(Fp). C is a Cartan subgroup. 0+ Since ρE1;p ∼ ρW1;p, we have ρE1;p(GQ) ⊆ C : + Either E1 Q 2 Xsplit(p)(Q) (when p splits in K); + or E1 Q 2 Xnonsplit(p)(Q) (when p is inert in K). 1 Also, E1 has a 2-torsion point =Q. Forces j(E1) 2 Z[ p ]. So a = b = 1. p Same idea works for s = 3; 6; 8; 20 to deal with Ps (1) = 1 . Does not work for any other s. For more details, see: M. A unique perfect power decagonal number, Bulletin of the Australian Mathematical Society, 1{5, 2021. Thank you!.