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!
Best Answer
Hint: Consider modulo 13. Use Fermat's Little Theorem to conclude that if there are any solutions, then $ 0 \equiv 1 \pmod{13} $.