XVI-th Junior Balkan Mathematical Olympiad

Reducing modulo $3$ we get $5^z \equiv 1$, therefore $z$ is even, $z = 2c$, $c \in \mathbb{N}$.
Next we prove that $t$ is even.
Obviously, $t \ge 2$. Let us suppose that $t$ is odd, say $t = 2d+1$, $d \in \mathbb{N}$. The equation becomes $2^x \cdot 3^y + 25^c = 7 \cdot 49^d$. If $x \ge 2$, reducing modulo $4$ we get $1 \equiv 3$, a contradiction. And if $x=1$, we have $2 \cdot 3^y + 25^c = 7 \cdot 49^d$ and reducing modulo $24$ we obtain
$$2 \cdot 3^y + 1 \equiv 7 \Rightarrow 24 \mid 2(3^y - 3), \text{ i.e. } 4 \mid 3^{y-1} - 1$$
which means that $y-1$ is even. Then $y = 2b+1$, $b \in \mathbb{N}$. We obtain $6 \cdot 9^b + 25^c = 7 \cdot 49^d$, and reducing modulo $5$ we get $(-1)^b = 2(-1)^d$ which is false for all $b, d \in \mathbb{N}$. Hence $t$ is even, $t=2d, d \in \mathbb{N}$, as claimed.

Now the equation can be written as
$$2^x \cdot 3^y + 25^d = 49^d \Leftrightarrow 2^x \cdot 3^y = (7^d - 5^c)(7^d + 5^c).$$
As $\gcd(7^d - 5^c, 7^d + 5^c) = 2$ and $7^d + 5^c > 2$, there exist exactly three possibilities:
$$(1) \begin{cases} 7^d - 5^d = 2^{x-1} \\ 7^d + 5^d = 2 \cdot 3^y \end{cases}$$
$$(2) \begin{cases} 7^d - 5^d = 2 \cdot 3^y \\ 7^d + 5^d = 2^{x-1} \end{cases}$$
$$(3) \begin{cases} 7^d - 5^d = 2 \\ 7^d + 5^d = 2^{x-1} \cdot 3^y \end{cases}.$$

### Case 1.
We have $7^d = 2^{x-2} + 3^y$ and reducing modulo $3$, we get $2^{x-2} \equiv 1 \pmod{3}$, hence $x-2$ is even, i.e. $x=2a+2$, $a \in \mathbb{N}$, where $a>0$, since $a=0$ would mean $3^y+1=7^d$ which is impossible (even=odd).
We obtain
$$7^d - 5^d \equiv 2 \cdot 4^a \pmod{4} \Rightarrow 7^d \equiv 1 \pmod{4} \Rightarrow d=2e, e \in \mathbb{N}.$$
Then we have
$$49^c - 5^d \equiv 2 \cdot 4^a \pmod{8} \Rightarrow 5^c \equiv 1 \pmod{8} \Rightarrow c=2f, f \in \mathbb{N}.$$
We obtain $49^c - 25^f = 2 \cdot 4^a \Rightarrow 0 \equiv 2 \pmod 3$, false. In conclusion, in this case there are no solutions to the equation.

### Case 2.
From $2^{x-1} = 7^d + 5^c \ge 12$ we obtain $x \ge 5$. Then $7^d + 5^c \equiv 0 \pmod 4$, i.e. $3^d + 1 \equiv 0 \pmod 4$, hence $d$ is odd. As $7^d = 5^c + 2 \cdot 3^y \ge 11$, we get $d \ge 2$, hence $d=2e+1$, $e \in \mathbb{N}$.
As in the previous case, from $7^d = 2^2 + 3^y$ reducing modulo $3$ we obtain $x=2a+2$ (because $x \ge 5$). We get $7^d = 4^a + 3^y$ i.e. $7 \cdot 49^c = 4^a + 3^y$, hence, reducing modulo $8$ we obtain $7 \equiv 3^y$ which is false, because $3^y$ is congruent either to $1$ (if $y$ is even) or to $3$ (if $y$ is odd). In conclusion, in this case there are no solutions to the equation.

### Case 3.
From $7^d = 5^c + 2$ it follows that the last digit of $7^d$ is $7$, hence $d=4k+1$, $k \in \mathbb{N}$.
If $c \ge 2$, from $7^{4k+1} = 5^c + 2$ reducing modulo $25$ we obtain $7 \equiv 2 \pmod 2$ which is false.
For $c=1$ we get $d=1$ and the solution $x=3, y=1, z=t=2$.