Find all positive integers a, b, n and prime numbers p that satisfy a2013+b2013=pn. Proposed by Matija Bucić.
Problem
Source: European Mathematical Cup 2012, Senior Division, Problem 1
Tags: number theory, prime numbers, number theory proposed
27.07.2013 06:25
27.07.2013 06:48
Matematika wrote: Find all positive integers a, b, n and prime numbers p that satisfy a^{2013} + b^{2013} = p^n\text{.} Author: Matija Bucić Let a=p^x \cdot y,b= p^z \cdot t with x,y,z,t \in \mathbb{N}, \; t,y \ge 1 and \gcd(y,p)=1, \; \gcd(t,p)=1. Without loss of generality, assume that x \ge z. From the equation we obtain t^{2013}+p^{2013x-2013z} \cdot y^{2013}= p^{n-2013z} If x>z then p \nmid LHS. Therefore p \nmid p^{n-2013z} implies n=2013z. Thus, t^{2013}+p^{2013x-2013z} \cdot y^{2013}=1, a contradiction. Thus, x=z. It follows that t^{2013}+y^{2013}=p^{n-2013x}=p^k \; (k=n-2013z \in \mathbb{N}^*) By applying LTE theorem we have \begin{array}{l} v_p \left( t^{2013}+y^{2013} \right) = v_p(2013)+v_p(t+y) \\ k=v_p(2013)+v_p(t+y) \end{array} Case 1. If \gcd (p,2013)=1 then v_p(t+y)=k. Let t+y=p^k =y^{2013}+t^{2013}. Therefore t+y=t^{2013}+y^{2013}. We obtain t=y=1 implies p=2,a=b=2^k,n=2013k+1. Case 2. If \gcd (p,2013) \ne 1 then p \in \{ 3;11;61 \}. Therefore v_p(2013)=1. It follows that k-1=v_p(t+y). Let t+y=p^{k-1}. Therefore A= t^{2012}-t^{2011}y+ \cdots +y^{2012}=\frac{t^{2013}+y^{2013}}{y+t}=p. That means A < t+y or t^{2013}+y^{2013} < (t+y)^2. It follows that t=y=1. Thus, p=2, a contradiction. In conclusion, thw solution is \boxed{ (a,b,n,p)= \left( 2^k,2^k,2013k+1,2 \right)} forall k \in \mathbb{N}.
27.07.2013 07:02
admin25 wrote:
It must be \boxed{(a, b, p,n)=(2^k, 2^k, 2, 2013k+1)} not \boxed{(a, b, n, p)=(2^k, 2^k, 2, 2013k+1)} .
27.07.2013 21:15
This can be generalized to: a^n+b^n=p^k has solutions: (a,b,p,n)=(2^{\frac{k-1}n},2^{\frac{k-1}n},2,k),(2\cdot3^r,3^k,3,3r+2).
13.09.2015 01:41
Zsigmondy's Theorem and boom. The exceptional cases yield the solutions.
18.11.2017 14:09
mathmdmb wrote: This can be generalized to: a^n+b^n=p^k has solutions: (a,b,p,n)=(2^{\frac{k-1}n},2^{\frac{k-1}n},2,k),(2\cdot3^r,3^k,3,3r+2). The condition of n s.t. n\geq 3 and n is odd is necessary