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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?

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If $\pmod {10}$ doesn't work you could try other moduli. Maybe the $3$ in the equation makes you think of $3^n$. I would try $\pmod 9$ The fact that $2008 \equiv 0 \pmod 8$ seems interesting as well. No guarantees. – Ross Millikan Mar 29 '14 at 3:41
Why do you think the workload would be so massive? You don't actually need to do every pair - for a given prime $p$ just compute a table of all the fifth powers, mod $p$, and then a table of three times all the fifth powers; checking to see whether the shift of any given number shows up on the other is basically an $O(1)$ operation, and so you should be able to check modulo any given prime $p$ in $O(p)$ steps pretty easily. This isn't great, but it's not bad at all. – Steven Stadnicki Mar 29 '14 at 3:42
@RossMillikan Unfortunately neither of those suggestions works; every odd number is a fifth power mod $8$, and $2008$ is $1$ mod $9$. – Mike Miller Mar 29 '14 at 3:49
@Mike: Playing a bit more, also every number is a fifth power mod $9$. It seemed like a good shot to me. To me, it would be the next try. Looks like Dan Brumleve had a better thought than mine. – Ross Millikan Mar 29 '14 at 4:04
up vote 9 down vote accepted

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 $1$).

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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.

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I'm not sure how you get from $a^{10}\equiv1\pmod{11}$ to $a^{5}\equiv -1,0,1\pmod{11}$. Fermat's Little Thm only gives $a^{p-1}\equiv1\pmod{p}$ if $a,p$ are relatively prime. – Marconius Nov 20 '15 at 0:10
@Marconius Firstly Fermat's little theorem doesn't state that a and b must be relatively prime. It only says that p must be prime. Secondly that jump I have done from $a^{10}$ to $a^{5}$ it's just taking the square root of 1 from $a^{10}$ which is $-1$ or $1$, also adding the exclusion case where $a^5=0(mod\space\ 11)$ intuitively. – Sota Antonino Nov 20 '15 at 0:23

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