How to solve $5^{2x}-3\cdot2^{2y}+5^x\cdot2^{y-1}-2^{y-1}-2\cdot5^x+1=0$ in $\mathbb{Z}$

Question

how to solve in $\Bbb Z$:

$$5^{2x}-3\cdot2^{2y}+5^x\cdot2^{y-1}-2^{y-1}-2\cdot5^x+1=0$$

Answer 1

Note that $5^{2x}-2\cdot5^x+1=(5^x-1)^2$, and $5^x\cdot2^{y-1}-3\cdot2^{2y}-2^{y-1}=(5^x-1)2^{y-1}-3\cdot2^{2y}$, so the original equation can be written

$$(5^x-1)(5^x-1+2^{y-1})-3\cdot2^{2y}=0\;.\tag{1}$$

The recurring element $5^x-1$ suggests making a substitution $u=5^x-1$, and since $3\cdot2^{2y}$ can be expressed simply in terms of $2^{y-1}$, we might also try substituting $v=2^{y-1}$ and rewriting $(1)$ as

$$u(u+v)-12v^2=u^2+uv-12v^2=(u-3v)(u+4v)=0\;.$$

Thus, either $u=3v$, or $u=-4v$. That is, $5^x-1=3\cdot2^{y-1}$, or $5^x-1=-4\cdot2^{y-1}=-2^{y+1}$. It seems preferable to write these as $5^x=3\cdot2^{y-1}+1$ and $5^x=1-2^{y+1}$. Clearly the latter produces no solutions, so we need only find the solutions to $5^x=3\cdot2^{y-1}+1$. By inspection $x=2,y=4$ is a solution; it’s not immediately clear to me whether there are any others.

Added: As Ivan Loh points out in the comments, this is the unique solution. First, it’s clear that $x$ and $y$ must be positive. since $5^x\equiv(-1)^x\pmod3$, we must have $x$ even; say $x=2a$, with $a\ge 1$. Then we have $5^{2a}-1=3\cdot2^{y-1}$, and we can factor the lefthand side to get

$$(5^a-1)(5^a+1)=3\cdot2^{y-1}\;.$$

Clearly $5^a+1\equiv2\pmod 4$. On the other hand, $5^a+1$ must be either $2^k$ or $3\cdot2^k$ for some $k\le y-1$, so we must have $k=1$ and $5^a$ either $2$ or $3\cdot2=6$. Then $a=0$ or $a=1$. If $a=0$, then $5^a-1=0$, which is impossible, since $3\cdot2^{y-1}\ne0$., so $a=1$, $5^1-1=4$, and we have the solution $x=2,y=4$.

Answer 2

Hint: Denote $z=5^x, t=2^{y-1}$ and obtain the equation $$ (z-3t)(z+4t)-2z-t+1=0 $$