Problem 1725

Solution:
If $x=0$ we have
$$z^{2} - 2^{2y} = 5^{t} \Longleftrightarrow (z + 2^{y})(z - 2^{y}) = 5^{t}$$
Putting $z + 2^{y} = 5^{a}$ and $z - 2^{y} = 5^{b}$ with $a + b = t$ we get $5^{a} - 5^{b} = 2^{y+1}$. This gives us $b = 0$ and now we have $5^{t} - 1 = 2^{y+1}$. If $y \geq 2$ then consideration by modulo 8 gives $2 \mid t$. Putting $t = 2s$ we get $(5^{s} - 1)(5^{s} + 1) = 2^{y+1}$. This means $5^{s} - 1 = 2^{c}$ and $5^{s} + 1 = 2^{d}$ with $c + d = y + 1$. Subtracting we get $2 = 2^{d} - 2^{c}$. Then we have $c = 1, d = 2$, but the equation $5^{s} - 1 = 2$ has no solutions over nonnegative integers. Therefore so $y \geq 2$ in this case gives us no solutions. If $y = 0$ we get again $5^{t} - 1 = 2$ which again has no solutions in nonnegative integers. If $y = 1$ we get $t = 1$ and $z = 3$ which gives us the solution $(t, x, y, z) = (1, 0, 1, 3)$.

Now if $x \geq 1$ then by modulo 3 we have $2 \mid t$. Putting $t = 2s$ we get
$$3^{x} 4^{y} = z^{2} - 5^{2s} \Longleftrightarrow 3^{x} 4^{y} = (z + 5^{s})(z - 5^{s})$$
Now we have $z + 5^{s} = 3^{m} 2^{k}$ and $z - 5^{s} = 3^{n} 2^{l}$, with $k + l = 2y$ and $m + n = x \geq 1$. Subtracting we get
$$2 \cdot 5^{s} = 3^{m} 2^{k} - 3^{n} 2^{l}$$
Here we get that $\min\{m, n\} = 0$. We now have a couple of cases.

Case 1. $k = l = 0$. Now we have $n = 0$ and we get the equation $2 \cdot 5^{s} = 3^{m} - 1$. From modulo 4 we get that $m$ is odd. If $s \geq 1$ we get modulo 5 that $4 \mid m$, a contradiction. So $s = 0$ and we get $m = 1$. This gives us $t = 0, x = 1, y = 0, z = 2$.

Case 2. $\min\{k, l\} = 1$. Now we deal with two subcases:

Case 2a. $l > k = 1$. We get $5^{s} = 3^{m} - 3^{n} 2^{l-1}$. Since $\min\{m, n\} = 0$, we get that $n = 0$. Now the equation becomes $5^{s} = 3^{m} - 2^{l-1}$. Note that $l-1 = 2y - 2$ is even. By modulo 3 we get that $s$ is odd and this means $s \geq 1$. Now by modulo 5 we get $3^{m} \equiv 2^{2y-2} \equiv 1, -1 \pmod{5}$. Here we get that $m$ is even as well, so we write $m = 2q$. Now we get $5^{s} = (3^{q} - 2^{y-1})(3^{q} + 2^{y-1})$.
Therefore $3^{q} - 2^{y-1} = 5^{v}$ and $3^{q} + 2^{y-1} = 5^{u}$ with $u + v = s$. Then $2^{y} = 5^{u} - 5^{v}$, whence $v = 0$ and we have $3^{q} - 2^{y-1} = 1$. Plugging in $y = 1, 2$ we get the solution $y = 2, q = 1$. This gives us $m = 2, s = 1, n = 0, x = 2, t = 2$ and therefore $z = 13$. Thus we have the solution $(t, x, y, z) = (2, 2, 2, 13)$. If $y \geq 3$ we get modulo 4 that $q, q = 2r$. Then $(3^{r} - 1)(3^{r} + 1) = 2^{y-1}$. Putting $3^{r} - 1 = 2^{e}$ and $3^{r} + 1 = 2^{f}$ with $e + f = y - 1$ and subtracting these two and dividing by 2 we get $2^{f-1} - 2^{e-1} = 1$, whence $e = 1, f = 2$. Therefore $r = 1, q = 2, y = 4$. Now since $2^{4} = 5^{u} - 1$ does not have a solution, it follows that there are no more solutions in this case.

Case 2b. $k > l = 1$. We now get $5^{s} = 3^{m} 2^{k-1} - 3^{n}$. By modulo 4 (which we can use since $0 < k-1 = 2y-2$) we get $3^{n} \equiv -1 \pmod{4}$ and therefore $n$ is odd. Now since $\min\{m, n\} = 0$ we get that $m = 0, 0 + n = m + n = x \geq 1$. The equation becomes $5^{s} = 2^{2y-2} - 3^{x}$. By modulo 3 we see that $s$ is even. We now put $s = 2g$ and obtain $(2^{y-1} - 5^{g})(5^{g} + 2^{y-1}) = 3^{x}$. Putting $2^{y-1} - 5^{g} = 3^{h}, 2^{y-1} + 5^{g} = 3^{i}$, where $i + h = x$, and subtracting the equations we get $3^{i} - 3^{h} = 2^{y}$. This gives us $h = 0$ and now we are solving the equation $3^{x} + 1 = 2^{y}$.
The solution $x = 0, y = 1$ gives $1 - 5^{g} = 1$ without solution. If $x \geq 1$ then by modulo 3 we get that $y$ is even. Putting $y = 2y_{1}$ we obtain $3^{x} = (2^{y_{1}} - 1)(2^{y_{1}} + 1)$. Putting $2^{y_{1}} - 1 = 3^{x_{1}}$ and $2^{y_{1}} + 1 = 3^{x_{2}}$ and subtracting we get $3^{x_{2}} - 3^{x_{1}} = 2$. This equation gives us $x_{1} = 0, x_{2} = 1$. Then $y_{1} = 1, x = 1, y = 2$ is the only solution to $3^{x} + 1 = 2^{y}$ with $x \geq 1$. Now from $2 - 5^{g} = 1$ we get $g = 0$. This gives us $t = 0$. Now this gives us the solution $1 + 3 \cdot 16 = 49$ and $(t, x, y, z) = (0, 1, 2, 7)$.

This completes all the cases and thus the solutions are $(t, x, y, z) = (1, 0, 1, 3), (0, 1, 0, 2), (2, 2, 2, 13),$ and $(0, 1, 2, 7)$.