# $x^5=y^2+10$ has no solutions

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.

• See also this post, for $x^5=y^2+1$. Apr 16 at 15:59
• 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.
– Sil
Apr 16 at 16:09
• 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. Apr 16 at 16:29
• Do you have any reason to expect that there would be an alternative elementary method to show this equation has no solutions? May 3 at 10:44

## 2 Answers

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.

• 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. Apr 18 at 12:17

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 $$\boxed{(x,y)=(10x\pm1,10y\pm1), (10x+9,10y\pm3)}$$

• 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. Apr 17 at 0:25
• @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 $(x,y)=(10x\pm1,10y\pm1)$ Apr 17 at 22:07
• 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 $\mathbb{Z}$. Apr 17 at 23:59
• 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. Apr 18 at 12:38