Show that the equation $x^{13} +12x + 13y^6 = 1$ doesn't have integer solutions

So I'm asked to show that the equation $x^{13} + 12x + 13y^6 = 1$ doesn't have integer solutions.

I'm not quite sure how to approach the problem as this doesn't seem to look like anything I had in my number theory course so far (or perhaps I missed this particular lecture).

In similar problems where the exponents are not that high, I generally can look for solutions mod n for some small n natural, so that's where I decided to start.

I started looking for solutions for each component of this equation in mod 2. Basically, 12x is always congruent to 0 mod 2.

$x^{13}$ is congruent to 0 mod 2 when x is even, and congruent to 1 otherwise.

$13y^6$ is congruent to 0 mod 2 when y is even, and congruent to 1 otherwise.

However, all I could conclude is that there can be no integer solutions when both x and y are even, or when both x and y are odd. It looks like when x is even and y is odd (or vice-versa) there could be a solution and I can't show otherwise...

Perhaps I started it all wrong, but that's all I could come up with so far. I would appreciate if someone could point a different direction or help me finish with this. Thanks in advance!

Hint: Consider modulo 13. Use Fermat's Little Theorem to conclude that if there are any solutions, then $0 \equiv 1 \pmod{13}$.
Hint: You have a good idea looking for solutions $\pmod n$, but $2$ is not the most effective $n$. There are lots of things going on near $12$ and $13$, so maybe you should look there. Note that $6=\frac {12}2$
• So here's what I got. Supposing the equation has integer solutions, and considering mod 13: From F.L.T, $x^{12} \equiv 1 mod 13$ $x^{12} + 12 \equiv 0 mod 13$ $x^{13} + 12x \equiv 0 mod 13$ Again, $y^{12} \equiv 1 mod 13$ $(y^{12})^{1/2} \equiv 1^{1/2} \equiv \pm 1 mod 13$ $13y^6 \equiv 0 mod 13$ Thus, $x^{13} + 12x + 13y^6 \equiv 0 mod 13.$ Since $x^{13} + 12x + 13y^6 = 1$, then $1 \equiv 0 mod 13$ Which is absurd. If the equation has no solutions modulus 13, then it has no integer solutions. – Bronski Apr 7 '13 at 19:38
• @Bronski: If you are working $\pmod {13}$ you don't have to worry about the $y^{12}$ (though what you said is right) because it is multiplied by zero. Good answer – Ross Millikan Apr 8 '13 at 2:58
Consider the equation modulo $$13$$. The left side turns to be $$x^{13}+(-1)x+(0)y^6$$ which equals $$x^{13}-x$$. Now this has to be congruent mod $$13$$ with the right side, which is $$1$$, so $$x^{13}-x \equiv 1\pmod{13}$$. But by Fermat´s Little Theorem $$x^{13}\equiv x \pmod{13}$$ so we can rewrite the expression as $$x-x \equiv 1\pmod{13}$$ and finally $$0 \equiv 1\pmod{13}$$ which is a clear contradiction. So we can conclude that the equation has no integer solutions. Q.E.D