No integer solutions for $x^5 - 3y^5 = 2008$

Question

I have a tutorial question (not homework) that asks to prove that there exist no integer solution for $$x^5 - 3y^5 = 2008$$

I originally thought I would solve it by taking all cases $\bmod 10$, but realised that would be a massive workload, and also realised it fell through instantly as $x = 1 \bmod 1, y = 1 \bmod 1$, gave $x^5 - 3y^5 = 8 \bmod 10$ instantly. I only thought that method might work because we were doing mod proofs directly prior to the question.

I have absolutely no idea how to solve this problem without using computation, any tips or methods?

Answer 1

Hint: Try mod $11$ instead, noting that both $x^5$ and $y^5$ are in $\{-1,0,1\}$. The intuition behind choosing $11$ is Fermat's little theorem (all tenth powers mod $11$ are either $0$ or $1$).

Answer 2

Fermat Theorem : For any a$\in$Z and prime p, it's well known that $a^p=a (mod\space\ p)$. $\Longrightarrow$ $a^{11}=a(mod\space\ 11)$. So it follows $a^{10}=1(mod\space\ 11)$ $\Longleftrightarrow (a^{5})^{2}=1(mod \space\ 11)$. So from here $a^5=\{-1,0,1\} (mod\space\ 11)$. So having established this and also knowing that $2008=6(mod\space\ 11)$ the rest is easy considering the possible residues of $x^5-3y^5 (mod\space\ 11)$.

$\longrightarrow$ And in conclusion the equation has no integer solutions.