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A common technique for proving that a diophantine equation does not have a solution is to prove that it does not have a solution mod $m$ for a suitable modulus $m$.

This technique works with $m=11$ for the equation $x^5 - y^2 = 4$ mentioned in this question.

Now, $11$ seems to be the only prime modulus that works. Moreover, the other moduli that work are all multiples of $11$. (I've tested all moduli up to $10^4$.)

(This is probably related to the rational points in the hyperelliptic curve $y^2=x^5-4$, but I don't know anything about this.)

Is it true that $11$ is the only prime modulus that works for that equation?

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lhf
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  • I wouldn't be surprised if it turns out that 11 is the only prime modulus that works. Note that $x^5$ takes on only 3 values modulo 11. – Gerry Myerson Apr 19 '16 at 11:46
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    Starting: For $p\ne5k+1$ then $x^5$ is a permutation polynomial. For $p=5k+1$ it is not. – student forever Apr 19 '16 at 11:51
  • Well, I don't know for a fact that 13 works for that equation, and the first prime I would try would be 7, since $x^6$ takes on only two values modulo 7. But anyway that equation is a lot easier – you are asking for two squares to differ by 4, and there aren't many of those. [This was in response to a comment that has since been deleted] – Gerry Myerson Apr 19 '16 at 11:51
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    Checking: $x=0$ can give a solution or not. This will be related with the quadratic residue of $-4$ and so $-1$. This also gives (for non-solution) $p$ need to be $4k+3$. Hence we need to search the class $p=20k+11$. We can also check eaisly the values $x=\pm1$. – student forever Apr 19 '16 at 12:41

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It is possible to use a geometric argument that goes as follows: every time you want to show that a curve $C$ over a finite field $\mathbb F_p$ has a rational point, what you can do is to show that there exists a place of $\mathbb P^1_{\mathbb F_p}$ of degree $1$ which has a place of degree $1$ of $C$ above it (simply because every projective curve comes with a morphism to $\mathbb P^1$ given by a corresponding embedding of function fields). In order to prove this, one can use a strong version of Chebotarev's theorem for function fields. I learned about it here: http://arxiv.org/abs/1404.6345. According to corollary 1.2, if you choose $\gamma$ to be the identity, so that the sum in the formula basically gives you the number of degree $1$ places of $C$, the lower bound that you get looks like $1/N\cdot (p+1-2g\sqrt{p})$, where $g$ is the genus of your curve $C$. Since that particular hyperelliptic curve has genus $2$, this lower bound is certainly positive for $p>7+4\sqrt{3}$, answering positively your question.

I know I did not fill in the details, since this can be pretty long, it was just to give you an idea of a possible method!

Ferra
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