Tessellate S.T.E.M.S. (2019) - Mathematics - Category A (School) - Set 2 - Objective Problem 2

There are two solutions: $(2,-3)$ and $(6,5).$

It's easiest to split into three cases: $y \le -11,$ $-10 \le y \le 1,$ and $y \ge 2.$

For the second case, there are only finitely many possibilities for $x$ given $y,$ so we can check manually for $-10 \le y \le 1$ that the only solution is $(2,-3).$

For the third case, note that the left side is a strictly increasing function of $x,$ and note that for $y \ge 2,$ $y^3+10y-1 \le y^3+6y^2,$ so if $(x,y)$ is a solution then $x > y.$ But also note that $(y+2)^3 + 10(y+2)-1 = y^3+6y^2+22y+27$ is greater than $y^3+6y^2,$ so $x < y+2.$ The only possibility left is $x=y+1.$ Plugging in and simplifying gives $\begin{aligned} 3y^2-13y-10 &= 0 \\ (3y+2)(y-5) &= 0 \end{aligned}$ which gives the one integer solution $x=6,y=5.$

For the first case, note that the right side is $\le -605,$ so the left side is, so $x$ is negative as well. Now let $a = -x, b = -y$ and rewrite $a^3+10a+1 = b^3-6b^2, \quad b \ge 11.$ Again the left side is a strictly increasing function of $a,$ so it's clear that $a < b.$ But now consider plugging in $a=b-3$ : the right side minus the left side would be $b^3-6b^2-((b-3)^3+10(b-3)+1) = 3b^2-37b+56 > 0 \text{ if } b \ge 11.$ So $a$ must be larger than $b-3,$ because plugging in $a=b-3$ gives a left side that is too small.

Since $b-3 < a < b,$ the only possibilities are $a=b-2$ and $a=b-1.$ The former gives $b=27/22,$ not an integer, and the latter leads to the same $x=y+1$ equation whose only integer solution was $y=5.$ Hence the first case gives no new solutions.

So we are done.