Solve in $\mathbb N^{2}$ the following equation : $5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$ Question : 
Solve for natural number the equation : 
$5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$
My try : 
Let : $X=5^{x}$  and $Y=2^{y}$ so above equation 
equivalent : 
$2X^{2}+(Y-4)X-6Y^{2}-Y+2=0$
We solve this equation for $X$ 
$\Delta =(7Y)^{2}$ mean : $X_{1}=\frac{3}{2}Y+1$ and $X_{2}=1-2Y$
From here how I can find $X$ and $Y$ , this is 
all my effort ? 
Thanks!

$5^{2x}-3\cdot 2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$
$(5^{2x}-2\cdot 5^x + 1)+ (5^{x}2^{y-1}-2^{y-1})-3\cdot 2^{2y}=0$
$(5^x-1)^2 + 2^{y-1}(5^x - 1)-3\cdot 2^{2y}=0$
If we solve for $5^x-1$ we get
$5^x-1 =\frac {-2^{y-1} \pm\sqrt{2^{2y-2} + 12*2^{2y}}}{2}=$
$\frac {-2^{y-1} \pm\sqrt{2^{2y-2} + 48*2^{2y-2}}}{2}=\frac {-2^{y-1} \pm\sqrt{49*2^{2y-2}}}{2}=$
$\frac {-2^{y-1} \pm 7*2^{y-1}}{2}=3*2^{y-1}$
Or $5^x = 3*2^{y-1} + 1$.
$5^x = 2^y + 2^{y-1} + 1$
But $5^x = (4+1)^x = \sum_{j=0}^x C_{x,j} 4^j$
We have a solution for $x=2$ and $y=4$ ($5^2 = 3*2^3+1$) which seems to work as ${2\choose 1}4 + 4^2 = 2*2^2 + 2^4= 2^3+2^4$.  But in general it doesn't seem like a likely occurrence.
For this to occur we need $\sum_{j=1}^x C_{x,j} 4^j = 2^{y-1} + 2^y$.
There's probably a really easy way to prove this is impossible if $x > 2$. If we divide the LHS by $4$ we get a remainder of $x\pmod 4$ and the RHS is $0\pmod 4$ if $y> 4$.... I think we have some chasing involved. $2^{y-2}|x$ but then $5^x$ is huge compared to $3*2^{y-1}+1$....
