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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!

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3 Answers 3

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

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    $\begingroup$ I felt discouraged from trying mods much higher than 2 due to the exponents and coefficients, but I wasn't considering Fermat's Little Theorem. $\endgroup$
    – Bronski
    Apr 7, 2013 at 19:39
  • $\begingroup$ I posted my solution below. Thank you both very much for the tips :) $\endgroup$
    – Bronski
    Apr 7, 2013 at 19:40
  • $\begingroup$ @Bronski At times, it can be hard to decide what mod to take. Sometimes, you have to take several mods before you can reach the conclusion. As Ross said, trying 12 or 13 makes a lot of sense in this problem. $\endgroup$
    – Calvin Lin
    Apr 8, 2013 at 19:36
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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 $6=\frac {12}2$

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    $\begingroup$ 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. $\endgroup$
    – Bronski
    Apr 7, 2013 at 19:38
  • $\begingroup$ @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 $\endgroup$
    – Ross Millikan
    Apr 8, 2013 at 2:58
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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

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