33rd Hellenic Mathematical Olympiad

We distinguish the cases:

If $z=0$, then the equation becomes $x^2+y^2=80$. Then $x,y$ must be multiples of $4$, that is $x=4a$, $y=4b$, $0\le a\le b$, and the equation becomes $a^2+b^2=5$, hence $(a,b)=(1,2)$, that is $(x,y)=(4,8)$.

If $z>0$, then $7\mid 2016^z$ (since $7\mid 2016$) and $7\mid 77$, and therefore $7$ must divide the left part, that is $7\mid x^2+y^2$. The possible remainders of the division of a square by $7$ are the same as the remainders of the numbers $0^2,1^2,2^2,3^2,4^2,5^2,6^2$, that is $0,1,2,4$. Therefore, in order to have $7\mid x^2+y^2$ we need $7\mid x$, $7\mid y$. We write $x=7x_1$ and $y=7y_1$, with $0\le x_1\le y_1$.

By substitution in (1) we get $49(x_1^2+y_1^2)=3\cdot 2016^z+77$.

If $z\ge 2$, then $49\mid 2016^z$, and then we conclude that $49\mid 77$, absurd.

If $z=1$, then $49(x_1^2+y_1^2)=3\cdot 7\cdot 288+77=7(3\cdot 288+11)=7\cdot 7\cdot 125$. Hence $x_1^2+y_1^2=125$. Since $x_1\le y_1$, we have $2y_1^2\ge 125\Rightarrow y_1\ge 8$. If $y_1=8,9,10,11$, we find the solutions $(x_1,y_1)\in \{(5,10),(2,11)\}$. Therefore we have the solutions: $(x,y)\in \{(4,8),(35,70),(14,77)\}$.