Belarusian Mathematical Olympiad

$\{(x,y,z)\}=\{(0,1,1),(1,1,2),(0,3,2),(2,5,4)\}$. We rewrite the equation in the form $7^x+2^y=3^z$. Note that for $y=0$ the left-hand side of the equation is an even integer while the right-hand side is an odd integer. So, $y>0$. Consider three cases: $y=1$, $y=2$ and $y\ge 3$. 1) $y=1$. In this case we have $$7^x+2=3^z.(1)$$ If $x=0$, then $z=1$; if $x=1$, then $z=2$. Thus, it remains to consider the case $x>2$, $z>2$. In this case $(7^x+2)\ge 3^3$. Consider the residues of $7^n$ modulo 27: $$7\to -5\to -8\to -2\to 13\to 10\to -11\to 4\to 1.$$ It follows that $x\equiv 4(\mathrm{mod}9)$. Note that $7^6+7^3+1=117993=3\cdot 37\cdot 1063$, so $(7^9-1)\equiv (7^6+7^3+1)\equiv 37$. Thus, $7^9\equiv 1(\mathrm{mod}37)$ and then $$7^x\equiv 7^4\equiv 49^2\equiv 12^2\equiv 144\equiv 33(\mathrm{mod}37).$$ Hence $7^x+2\equiv 35(\mathrm{mod}37)$. On the other hand, considering the residues of $3^n$ modulo 37, we have $$3\to 9\to 27\to 7\to 21\to 26\to 4\to 12\to 36\to 34\to 28\to 10\to 30\to 16\to 11\to 33\to 25\to 1.$$ We see that $3^z\not\equiv 35(\mathrm{mod}35)$ for all $z$. Therefore, there are no solutions of (1) for $x>2$, $z>2$. 2) $y=2$. We have $3^z=7^x+4\equiv 2(\mathrm{mod}3)$, a contradiction. 3) $y\ge 3$. Then $7^x\equiv 3^z(\mathrm{mod}8)$, whence $x\equiv 2(\mathrm{mod}2)$ and $z\equiv 2(\mathrm{mod}2)$. Let $x=2x_1$, $z=2z_1$. Now the initial equation can be rewritten in the form $$(3^{z_1}-7^{x_1})(3^{z_1}+7^{x_1})=2^y.$$ So, it follows that $$3^{z_1}-7^{x_1}=2^a,(2)$$ $$3^{z_1}+7^{x_1}=2^b,(3)$$ and, since $(3^{z_1}-7^{x_1})\equiv 2$, we have $b>a\ge 1$. Summing (2) and (3), we obtain $2^a+2^b=2\cdot 3^{z_1}$. Hence $a=1$. Then (2) can be rewritten as $3^{z_1}-7^{x_1}=2$, or $3^{z_1}=7^{x_1}+2$, which coincides with (1). Therefore, we have $\{(x_1,z_1)\}=\{(0,1),(1,2)\}$, and $b$ is equal to 2 and 4 respectively. Finally, in this case we have $\{(x,y,z)\}=\{(0,3,2),(2,5,4)\}$.