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

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?

• 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 $r=0,1$? – tkf Sep 3 at 15:36
• Maybe , but how ? – Anonymous Sep 3 at 15:39
• 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. – tkf Sep 3 at 15:42
• taking case when x=2m+1 and similarly may help – Albus Dumbledore Sep 3 at 15:45
• Ok the expansion of $(2m + 1)^5$ is $32m^5 + 80m^4 + 80m^3 + 40m^2 + 10m + 1$ (I hope I am not wrong) . – 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 $$x=1$$.
This is Mihailescu's theorem in the case $$a=5, b=2$$. The case $$b=2$$ was solved by Victor-Amédée Lebesgue in 1850.