Solve $x^5 - 1 = y^2$ for integer values of $(x,y)$ .

What I Tried :- I see that $x^5 = y^2 + 1$ , from here I can conclude that $x$ has to be positive , because if $x \leq 0$ , then $x^5 \leq 0$ , but $y^2 + 1 > 0$.

Also I thought that maybe $(x^\frac{5}{2} + 1)(x^\frac{5}{2} - 1) = 1$ would do the trick [which would have implied that both the terms are $(1,1)$ or $(-1,-1)$], but that dosen't necessarily mean that $x^\frac{5}{2}$ is an integer .

Then I see that :-

$x^5 - 1 = y^2$

$\rightarrow (x-1)(x^4 + x^3 + x^2 + x + 1) = y^2$

From here the only idea is that $x \neq 2$ , because if it is then it can't be a perfect square (it's a foolish idea though) .

But I tried playing this problem with many ways , but I am stuck at the same place . Can anyone help ?

Note :- It's given that answer is only $(1,0)$ , but how is it coming?

  • $\begingroup$ Does it help to factorise the LHS over $\mathbb{Z}[\zeta]$ (where $\zeta=e^{2\pi i/5}$) and use the fact that this is a UFD, so $x-\zeta=z^2(1-\zeta)^ru$, for unit $u$ and $r=0,1$? $\endgroup$ – tkf Sep 3 at 15:36
  • $\begingroup$ Maybe , but how ? $\endgroup$ – Anonymous Sep 3 at 15:39
  • $\begingroup$ This type of argument often leads to a reduction proof (given one integer solution produce a strictly smaller one). I haven't done one of these in ages - just thought it might be worth playing with if you haven't already tried it. $\endgroup$ – tkf Sep 3 at 15:42
  • $\begingroup$ taking case when x=2m+1 and similarly may help $\endgroup$ – Albus Dumbledore Sep 3 at 15:45
  • $\begingroup$ Ok the expansion of $(2m + 1)^5$ is $32m^5 + 80m^4 + 80m^3 + 40m^2 + 10m + 1$ (I hope I am not wrong) . $\endgroup$ – Anonymous Sep 3 at 15:46

So $$x^5=y^2+1=(y+i)(y-i).$$ Considering this modulo $4$ gives $y$ even and $x$ odd. The gcd of $y\pm i$ in the Gaussian integers divides $2$ and $y^2+1$ (which is odd) so it is $1$. As the Gaussian integers has unique factorisation and has four units, both $y\pm i$ are fifth powers, so $$y+i=(a+bi)^5=(a^5-10a^3b^2+5ab^4)+(5a^4b-10a^2b^3+b^5)i.$$ So $b\mid 1$ and $b=\pm1 $ and $$5a^4-10a^2+1=\pm1.$$ The only integer solution of this is $a=0$ leading to $y=0$ and $x=1$.

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  • $\begingroup$ Ok now that seems understandable to me . Thank you for that solution $\endgroup$ – Anonymous Sep 3 at 16:05

This is Mihailescu's theorem in the case $a=5, b=2$. The case $b=2$ was solved by Victor-Amédée Lebesgue in 1850.

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    $\begingroup$ Ok that's what I want as my solution , how was it solved ? I tried a lot of things but couldn't get to a solution. $\endgroup$ – Anonymous Sep 3 at 15:51
  • $\begingroup$ See Chapter 3 here. $\endgroup$ – Robert Israel Sep 3 at 15:59

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