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 $$2^x\cdot 3^y+25^d=49^d\iff 2^x\cdot 3^y=(7^d-5^c)(7^d+5^c).$$ As $\gcd (7^d-5^c,7^d+5^c)=2$ and $7^d+5^c>2$, there exist exactly three possibilities: $$(1)\{\begin{array}{l}{7}^d-5^d=2^{x-1}\\ 7^d+5^d=2\cdot 3^y\end{array}$$ $$(2)\{\begin{array}{l}{7}^d-5^d=2\cdot 3^y\\ 7^d+5^d=2^{x-1}\end{array}$$ $$(3)\{\begin{array}{l}{7}^d-5^d=2\\ 7^d+5^d=2^{x-1}\cdot 3^y\end{array}.$$

### Case 1. We have $7^d=2^{x-2}+3^y$ and reducing modulo $3$, we get $2^{x-2}\equiv 1(\mathrm{mod}3)$, 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 $$7^d-5^d\equiv 2\cdot 4^a(\mathrm{mod}4)\Rightarrow 7^d\equiv 1(\mathrm{mod}4)\Rightarrow d=2e,e\in \mathbb{N}.$$ Then we have $$49^c-5^d\equiv 2\cdot 4^a(\mathrm{mod}8)\Rightarrow 5^c\equiv 1(\mathrm{mod}8)\Rightarrow c=2f,f\in \mathbb{N}.$$ We obtain $49^c-25^f=2\cdot 4^a\Rightarrow 0\equiv 2(\mathrm{mod}3)$, 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 $7^d+5^c\equiv 0(\mathrm{mod}4)$, i.e. $3^d+1\equiv 0(\mathrm{mod}4)$, 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 $7\equiv 2(\mathrm{mod}2)$ which is false. For $c=1$ we get $d=1$ and the solution $x=3,y=1,z=t=2$.