BMO 2017

Let $d=(x,y)$ be the greatest common divisor of positive integers $x$ and $y$. So, $x=ad$, $y=bd$, where $d\in \mathbb{N}$, $(a,b)=1$, $a,b\in \mathbb{N}$. We have $$\begin{array}{rl}{x}^3+y^3& =x^2+42xy+y^2\\ & \iff d^3(a^3+b^3)=d^2(a^2+42ab+b^2)\\ & \iff d(a+b)(a^2-ab+b^2)=a^2+42ab+b^2\\ & \iff (da+db-1)(a^2-ab+b^2)=43ab.\end{array}$$ If we denote $c=da+db-1\in \mathbb{N}$, then the equality $a^{2}c-abc+b^{2}c=43ab$ implies the relations $$\begin{gathered} \{\begin{array}{l}b|c{a}^2\Rightarrow b|c\\ a|c{b}^2\Rightarrow a|c\end{array}\Rightarrow (ab)|c \\ \iff c=mab,m\in {\mathbb{N}}^{+} \\ \Rightarrow m(a^2-ab+b^2)=43 \\ \Rightarrow (a^2-ab+b^2)|43 \\ \iff a^2-ab+b^2=1\text{or}{a}^2-ab+b^2=43. \end{gathered}$$ If $a^2-ab+b^2=1$, then $(a-b{)}^2=1-ab\ge 0\Rightarrow a=b=1$, $2d=44$, $(x,y)=(22,22)$. If $a^2-ab+b^2=43$, then, by virtue of symmetry, we suppose that $x\ge y\Rightarrow a\ge b$. We obtain that $43=a^2-ab+b^2\ge ab\ge b^2\Rightarrow b\in \{1,2,3,4,5,6\}$. If $b=1$, then $a=7$, $d=1$, $(x,y)=(7,1)$ or $(x,y)=(1,7)$. If $b=6$, then $a=7$, $d=\frac{43}{13}\notin \mathbb{N}$. For $b\in \{2,3,4,5\}$ there no positive integer solutions for $a$. Finally, we have $(x,y)\in \{(1,7),(7,1),(22,22)\}$.