<s>[INST] 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!
 [/INST] Now substitute back: you get that $5^x=3\cdot 2^{y-1}+1$ or $5^x=1-2^{y+1}$. The second one is impossible cause the RHS is $\leq 0$. For the first one to hold, looking mod 3 you see that $x$ needs to be even, so write $x=2x'$. Then $(5^{x'}-1)(5^{x'}+1)=3\cdot 2^{y-1}$. So the only chance is that $5^{x'}-1=2^{y_1}$ and $5^{x'}+1=3\cdot 2^{y_2}$ for some $y_1,y_2\in \mathbb N$, or the other way round. Substract term by term and you'll see immediately that the minimum of $y_1,y_2$ has to be either $0$ or $1$. Now you can check by hand case by case.
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