Find all prime solutions of equation $5x^2-7x+1=y^2.$ Find all prime solutions of the equation $5x^2-7x+1=y^2.$
It is easy to see that 
$y^2+2x^2=1 \mod 7.$ Since $\mod 7$-residues are $1,2,4$ it follows that $y^2=4 \mod 7$, $x^2=2 \mod 7$  or $y=2,5 \mod 7$ and $x=3,4 \mod 7.$ 
In the same way from $y^2+2x=1 \mod 5$ we have that $y^2=1 \mod 5$ and $x=0 \mod 5$ or $y^2=-1 \mod 5$ and $x=4 \mod 5.$
How put together the two cases?
Computer find two prime solutions $(3,5)$ and $(11,23).$

Can we just charge straight at this?
$y$ is odd. $x=2 \ (\Rightarrow y^2=7)$ is not a solution, so $x$ is an odd prime.
$x(5x-7) = (y-1)(y+1)$, so $x \mid (y-1) $ or $x \mid (y+1)$  ($x$ is prime)
 so $kx=y\pm1$, $k$ even
$k\ge4$ is too large: $(kx\pm1)^2\ge (4x-1)^2 $ $= 16x^2-8x+1$ $>5x^2-7x+1$.  So only $k=2$, that is $x=\frac 12(y\pm1)$, makes the equality feasible.
Considering the two cases:  


*

*(1) $x=\frac 12(y+1)$, $y=2x-1$:
$x(5x-7) = 4x(x-1) \implies x = 3, y=5$

*(2) $x=\frac 12(y-1)$, $y=2x+1$:
$x(5x-7) = 4x(x+1) \implies x = 11, y=23$
Note that I didn't constrain $y$ at any point - the two solutions just happened to have $y$ prime.
