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Prove that there does not exist an integer solution for the diophantine equation $x^5 - y^2 = 4$.

It's obvious that $x$ and $y$ are of the same parity. We can also claim that if $x$ is odd, then it is $1 \pmod 4$. Also, if $x$ and $y$ are even, then $y \equiv 2 \pmod 4\text{ since } x^5 \text{ is a multiple of } 32$ and if $y$ were a multiple of $4$ then it would be at a distance of at least $16$ from $x^5$ or be equal to it. These are my observations.

How should I proceed with the proof? Please give hints.

TheRandomGuy
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1 Answers1

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Hint: Consider working modulo $11$.

André Nicolas
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    $11$ is the only prime modulus less than $10000$ that works! Any explanation? – lhf Apr 18 '16 at 14:29
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    There are not many fifth powers modulo $11$, since $5=\frac{11-1}{2}$. – André Nicolas Apr 18 '16 at 14:30
  • Great Answer. But I did not realize the logic behind 'There are not many fifth powers modulo $11$, since $5=\frac{11-1}{2}$.' especially after the 'since'. – TheRandomGuy Apr 18 '16 at 14:37
  • @lhf Did you compute it? – TheRandomGuy Apr 18 '16 at 14:40
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    @Dhruv: By Fermat's Theorem, if $a$ is not divisible by $11$, then $(a^5)^2\equiv 1\pmod{11}$, so $a^5\equiv \pm 1\pmod{11}$. One has the same phenomenon for $a^k$ whenever $2k+1$ is prime. – André Nicolas Apr 18 '16 at 14:41
  • @Dhruv, yes I wrote a program to find a suitable modulus but André beat me to it... – lhf Apr 18 '16 at 14:43
  • I really don't understand how to find a suitable modulo. Is there any paper on this? – N.S.JOHN Apr 19 '16 at 02:28
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    For this problem, I gave a suitable modulus. Often modulus $2$, $4$, or $8$ is useful. Every so often other small numbers, like $3$, $5$, $7$, $9$ (which can be useful for cubics). But in many Diophantine problems, modular calculations do not settle the matter, or only give partial help. I don't know of any paper on how to find a suitable modulus. One picks up these things by doing problems. – André Nicolas Apr 19 '16 at 02:31
  • See also http://math.stackexchange.com/questions/1749467/x5-y2-4-has-no-solution-mod-m. – lhf Apr 19 '16 at 11:19
  • Great answer, and to teach the student to find a suitable modulus would make an invaluable chapter in a graduate level number theory text, even if it is just a collection of problems. – Adam Ledger Dec 15 '20 at 04:59
  • @lhf the program you or Andre wrote, can I please see the algorithm? – Adam Ledger Dec 15 '20 at 05:01