Final Round

Answer: $a=b=t$, $t\in \mathbb{Z}$, $(a;b)=(-18;-2)$, $(a;b)=(0;7)$, $(a;b)=(12;3)$.

We rearrange the initial equation $(b^2+7(a-b){)}^2=a^{3}b$: $$\begin{array}{rl}(b^2+7(a-b){)}^2& =a^{3}b\iff b^4+14b^2(a-b)+49(a-b{)}^2=a^{3}b\iff \\ & a^{3}b-b^4-14b^2(a-b)-49(a-b{)}^2=0\iff \\ & b(a^3-b^3)-14b^2(a-b)-49(a-b{)}^2=0\iff \\ & (a-b)(b{a}^2+a{b}^2+b^3-14b^2-49(a-b))=0.\end{array}\text{\hspace{1em}(1)}$$ If $a=b$, then it is evident that (1) holds. Therefore we have the infinite set of the integer solutions: $a=b=t$, where $t\in \mathbb{Z}$.

Let $a\ne b$, then from (1) it follows that $$b{a}^2+a{b}^2+b^3-14b^2-49(a-b)=0.(2)$$ If $b=0$, then from (2) it follows that $a=0$, i.e. $a=b=0$, a contradiction. Thus $b\ne 0$. Consider (2) as a quadratic equation with respect to $a$. $$b{a}^2+(b^2-49)a+b^3-14b^2+49b=0.(3)$$ The discriminant $D(b)$ of this equation is $$\begin{array}{rl}D(b)=(b^2-49{)}^2-4b(b^3-14b^2+49b)& =((b-7)(b+7){)}^2-4b^2(b-7{)}^2=\\ & =(b-7{)}^2((b+7{)}^2-4b^2)=-(b-7{)}^3(3b+7).\end{array}$$ Equation (3) has real solutions iff $D(b)\ge 0$, i.e. for $b\in [-7/3;7]$. The segment $[-7/3;7]$ contains exactly ten integers: $-2,-1,0,1,2,3,4,5,6$ and $7$. Since $b\ne 0$ and $a_{1,2}=(49-b^2\pm \sqrt{D})/(2b)$, we have exactly four integers ($b=-2,3,6,7$) such that $D(b)$ is a perfect square (we need integer solutions of (3)).

1) If $b=-2$, then $D=27^2$ and either $a_1=-9/2$ or $a_2=-18$, so, in this case $(a;b)=(-18;-2)$ is a required solution.

2) If $b=3$, then $D=2^{10}$ and either $a_1=4/3$ or $a_2=12$, so $(a;b)=(12;3)$ is a required solution.

3) If $b=6$, then $D=5^2$ and either $a_1=13/12$ or $a_2=85/12$, so there are no solutions in this case.

4) If $b=7$, then $D=0$ and $a_{1,2}=0$, so $(a;b)=(0;7)$.