The equation can be rewritten as $(a-b)[-ba^{2} + (49-b^{2})a-b(b-7)^{2}] = 0$. If $a\neq b$, then $-ba^{2} + (49-b^{2})a-b(b-7)^{2} =0$ ... (*) would be a quadratic equation whose discriminant $\Delta$ would have to be a perfect square. Since $\Delta = (b-7)^{2}(7-b)(7+3b),$ it follows that $b$ could be any member of the set $\{7,6,3,0,-2\}$. Trying out these possibilities in (*), we get that the only valid possibilities for $b$ are $7, 3,$ and $-2$. Therefore, the set of solutions of the original equation is $\{(n,n) \colon n \in \mathbb{Z}\} \cup \{(0,7), (12,3), (-18,-2)\}.$

hint

indeed ,This problem is acquired from Belorussian Math olympiads-2014, The Author of the Problem was I.Gorodnin