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I am looking for an elementary way to show the equation $x^5=y^2+10$ has no integer solutions. I have checked the equation mod $n$ for $n<1000$ and it had solutions every time.

Here is my proof, using algebraic number theory: In the extension $\mathbb{Q}(\sqrt{-10})$, we have $(x)^5 = (y+\sqrt{-10})(y-\sqrt{-10})$. One can show the ideals on the RHS are coprime (assuming $(x,y)$ is a solution). Thus $(y+\sqrt{-10})=J^5$ for some ideal $J$. The class group of $\mathbb{Q}(\sqrt{-10})$ is order 2, so $J$ is principal, thus $y+\sqrt{-10}$ is a 5th power and one can show this leads to a contradiction.

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    $\begingroup$ See also this post, for $x^5=y^2+1$. $\endgroup$ Apr 16 at 15:59
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    $\begingroup$ I wonder if something similar to this one could be done, though I don't see numbers aligning (only that $x^5-1^5=y^2+3^2$). We know $x\equiv 3\pmod 4$. Maybe that could be of some use. $\endgroup$
    – Sil
    Apr 16 at 16:09
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    $\begingroup$ This equation is solvable in $\mathbb{F}_p$ for all $p>13$ by the Weil bounds, so no amount of wrangling with elementary congruences will work - it fails the Hasse local-global principle. $\endgroup$
    – K B Dave
    Apr 16 at 16:29
  • $\begingroup$ Do you have any reason to expect that there would be an alternative elementary method to show this equation has no solutions? $\endgroup$
    – Sahaj
    May 3 at 10:44

2 Answers 2

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Josef Blass proved in Math. Comp. 30 (1976) 638-640 the following

$\textbf{Theorem.}$ If $k$ is a square-free positive integer for which $k \not\equiv 7 \pmod 8 $ and $(h(-k),5)=1$, then the equation $x^5 = y^2 + k $ has only the solutions $(k, y, x) = (1,0,1), (19, \pm22434, 55)$ and $(341, \pm275964, 377),\,$ where $h(-k)$ is the class number of $\Bbb Q(\sqrt{-k})$.

For $k = 10$ we have $h(-10)=2$ is coprime to $5$ so the theorem implies that the equation $x^5=y^2+10$ has no integral solutions.

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    $\begingroup$ But is the underlying theorem elementary as defined by the OP? The question implies that algebraic number theory and class field theory, while technically correct, miss the OP's box. $\endgroup$ Apr 18 at 12:17
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A PARTIAL ANSWER.

$x,y$ should have the same parity; if $x$ is even then $x^5=y^2+10\implies 0\equiv y^2+10\pmod{16}$ which is not possible because $y$ is even and its square can be only $0$ or $4$ modulo $16$. Then $x,y$ are both odd. Besides

$x^5=y^2+10\implies x\equiv y^2\pmod{10}$ because $x^5\equiv x\pmod{10}$. It follows $$(x,y)=(10x\pm1,10y\pm1), (10x+9,10y\pm3), (10x+5,10y+5)$$ $\underline{(10x+5,10y+5)\space\text {it's impossible}}$

$(10x+5)^5=(10y+5)^2+10\implies 31250x+3125\equiv 35\pmod{100}$ so if $x$ is even one has $25\equiv35\pmod{100}$, impossible. Then $x=10(2x+1)+5=20x+15$ and $(20x+15)^5=100(y^2+y)+35\implies 75\equiv 35\pmod{100}$, impossible.

It remains to prove the impossibility for the cases $$\boxed{(x,y)=(10x\pm1,10y\pm1), (10x+9,10y\pm3)}$$

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    $\begingroup$ This style of thinking is a dead end: if $u$ is a unit and a square modulo $10$, then $u^5 - 10$ lifts to a square modulo $10^n$ for every $n$ by the Chinese Remainder Theorem and Hensel's Lemma. $\endgroup$
    – K B Dave
    Apr 17 at 0:25
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    $\begingroup$ @KBDave: of what unit are you spiking?. In this context the only units are $1$ and $-1$. I have proven that the last possibility is impossible so explain me, please, your objection of the method I use for leave this problem and no more pay attention to the case $(x,y)=(10x\pm1,10y\pm1)$ $\endgroup$
    – Piquito
    Apr 17 at 22:07
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    $\begingroup$ The basic issue is that, once we've found solutions modulo small powers of a number, we can find solutions modulo arbitrarily large powers of that number. For example, $11^5\equiv 5487849321^2 + 10\pmod{10^{10}}$, $19^5 \equiv 2519493717^2 + 10\pmod{10^{10}}$. In fact, this equation has solutions in $\mathbb{Z}/(n)$ for every $n$, but nevertheless has no solutions in $\mathbb{Z}$. $\endgroup$
    – K B Dave
    Apr 17 at 23:59
  • $\begingroup$ It is one thing to find a solution modulo a number $n$ and another thing absolutely different is to prove that there are no solutions modulo $n$. It is clear to me that you do not know how to appreciate an elementary proof where others have a sophisticated proof. I leave this I don't try it anymore. However I say the problem can be solved by elementary means except maybe for the cases $10x\pm1$ (I feel this last case can be solved too without something so high level as Class Field Theory. $\endgroup$
    – Piquito
    Apr 18 at 12:38

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