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.

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!
