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 $k \in \mathbb{N}$.

admin25 wrote:

It must be $\boxed{(a, b, p,n)=(2^k, 2^k, 2, 2013k+1)}$ not $\boxed{(a, b, n, p)=(2^k, 2^k, 2, 2013k+1)}$.

This can be generalized to: $a^n+b^n=p^k$ has solutions: $(a,b,p,n)=(2^{\frac{k-1}n},2^{\frac{k-1}n},2,k),(2\cdot3^r,3^k,3,3r+2)$.

Zsigmondy's Theorem and boom. The exceptional cases yield the solutions.

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