The Diophantine equation $x^5-2y^2=1$ I'm trying to solve the Diophantine equation $x^5-2y^2=1$.
Here's my progress so far. We can write the Diophantine equation as
$$\frac{x-1}{2}\cdot(x^4+x^3+x^2+x+1)=y^2.$$
If $x\not\equiv1\pmod{5}$, then $\gcd(\frac{x-1}{2},x^4+x^3+x^2+x+1)=1$, so both $\frac{x-1}{2}$ and $x^4+x^3+x^2+x+1$ must be perfect squares (note: $x^4+x^3+x^2+x+1>0$).
In particular, $4(x^4+x^3+x^2+x+1)$ is a perfect square.
Comparison with $(2x^2+x)^2$ and $(2x^2+x+1)^2$ forces $-1\leq x\leq3$.
This results in the solutions $(3,\pm11)$.
If $x\equiv1\pmod{5}$, then we can write the Diophantine equation as
$$\frac{x-1}{10}\cdot\frac{x^4+x^3+x^2+x+1}{5}=\left(\frac{y}{5}\right)^2,$$
where $\gcd(\frac{x-1}{10},\frac{x^4+x^3+x^2+x+1}{5})=1$, so both $\frac{x-1}{10}$ and $\frac{x^4+x^3+x^2+x+1}{5}$ must be perfect squares.
Thus,
$$x=10a^2+1,$$
$$x^4+x^3+x^2+x+1=5b^2.$$
Unfortunately, this is where I get stuck.
I can substitute the first equation into the second, giving
$$10000a^8+5000a^6+1000a^4+100a^2+5=5b^2,$$
$$2000a^8+1000a^6+200a^4+20a^2+1=b^2,$$
$$2000a^8+1000a^6+200a^4+20a^2=(b-1)(b+1),$$
$$5a^2(100a^6+50a^4+10a^2+1)=\frac{b-1}{2}\cdot\frac{b+1}{2},$$
but this doesn't seem to be making progress, even with modular arithmetic considerations.
 A: Also not a super elementary argument, but the only non-trivial result I use here is that $K=\Bbb Q(\sqrt{-2})$ has class number $1$, i.e. $\Bbb Z[\sqrt{-2}]$ is a PID.
Let $x^5-2y^2=1$ with $x,y\in\Bbb Z$. Note that $x$ is odd. We can factor the equation as $$x^5=(1-\sqrt{-2}y)(1+\sqrt{-2}y).$$ Let $d$ be a gcd of $(1-\sqrt{-2}y),(1+\sqrt{-2}y)$ in $\Bbb Z[\sqrt{-2}]$. Note that $d\mid 2$ and $d\mid x^5$. As $x^5$ is odd this implies that $d\mid 1$, i.e. the elements are coprime, hence (as $\Bbb Z[\sqrt{-2}]$ is a UFD) there is a unit $\varepsilon\in \Bbb Z[\sqrt{-2}]^\times=\{-1,1\}$ and $z=a+b\sqrt{-2}\in\Bbb Z[\sqrt{-2}]$ such that $1+\sqrt{-2}y=\varepsilon z^5$. As $(-1)^5=-1$ we may assume that $\varepsilon=1$. Then expanding the fifth power and comparing coefficients we get:
\begin{align*}
1&=a^5-20a^3b^2+20ab^4\\
y&=5a^4b-20a^2b^3+4b^5
\end{align*}
The first equation implies $a=\pm1$ and for $a=-1$ we don't get any solutions and for $a=1$ we get $b^4-b^2=0$, i.e. $b=0$ or $b=\pm1$. These correspond to $y=0$ and $y=\pm11$. Hence the only possible solutions for the original equation are $(1,0),(3,-11),(3,11)$.
A: You don't really say if you just want to know the solutions, or if you want a nice elementary argument for why the solutions are only $(3, \pm 11)$, if you just want a proven answer and not an elementary argument the following works, its a bit overkill but its easier than thinking if you know these methods already:
The equation $x^5 - 2y^2 = 1$ considered over the rationals defines a hyperelliptic curve, of genus 2.
So there is a big hammer called Chabauty's method that often determines all rational points on such curves.
Our curve is isomorphic via change of variables ($y\mapsto 4y,x\mapsto 2x$) to the curve
$$y^2 = x^5 - \frac{1}{2^5}$$
or even to an integral model
$$y^2 = -2x^6 + 2x.$$
The computer algebra system Magma can determine the rank of the Mordell-Weil group of the Jacobian of this curve (using the integral model above) to be 1 (and hence Chabauty's method applies), using a generator Magma can also run Chabauty's method automatically in this case, and provably find all rational points:
> R<x>:=PolynomialRing(Rationals());
> H:=HyperellipticCurve(x^5-1/(2^5));       
> HH,ma:=MinimalWeierstrassModel(H);  
> a,b,P:=RankBounds(Jacobian(HH):ReturnGenerators);
> a,b,P;
1 1 [ (x^2 - 1/3*x, 22/9*x, 2) ]
> pts:=Chabauty(P[1]);
> pts;
{ (1 : 22 : 3), (1 : -22 : 3), (0 : 0 : 1), (1 : 0 : 1) }
> [P : P in pts][1]@(ma^(-1));  
(3/2 : -11/4 : 1)
> [P : P in pts][3]@(ma^(-1));
(1 : 0 : 0)

From this list we see that translating back to the original equation/curve the only interesting rational solutions are those you found already.
If you only wanted integral solutions to begin with there should be less high-tech methods to do this!
A: Here is an "elementary" proof. The given diophantine equation $x^5 = 1+2y^2$ admits the obvious solution $x=1, y=0$. Exclude this trivial solution and consider $a=x^5$ as an integral parameter which one wants to represent as the value of the quadratic form $t^2+2y^2$, with unknown integers $(t,y)$. Geometrically, the problem is equivalent to find the points of the sublattice $\mathbf Z^2$ of $\mathbf R^2$ which belong to the ellipse with equation $a= t^2+2y^2$. Since the lattice is discrete and the ellipse is compact, the set $S$ of wanted points is finite. If $S$ is not empty and $t=1$, symmetry w.r.t. the $t$-axis imposes that card $S=2$. If one follows this elementary approach, the only reason for the hypothesis $a=x^5$ seems to be the quick growth of the 5-th power, which allows to determine $S$ without too many trials.
NB : As usual, "elementary" methods often mask the power - and slickness - of more  elaborate methods. The general process at work here is, as suggested by the answer given by @leoli1, the representation of a positive integer by a binary quadratic positive definite form.
