How to solve the equation in integer?

Question

$$x^3+y^3=x^2+42xy+y^2$$

How to solve in integer numbers?

Answer 1

Here is an outline of an approach.

Let $d = \gcd(x,y)$, with $x = ad, y = bd$. We now have $d (a^3 + b^3) = a^2 + 42ab + b^2$. We rewrite this as $d(a+b)(a^2-ab+b^2) = (a^2-ab+b^2) + 43 ab$, and rearrange to obtain $$(d(a+b)-1)(a^2-ab+b^2) = 43ab.$$ Note that since $a$ and $b$ are coprime, we have $\gcd(a, a^2 + b^2 - ab) = \gcd(a, b^2) = 1$, and a similar result for $b$, so we have $a^2-ab+b^2 \mid 43$.

If $a^2-ab+b^2 = 43$, WLOG that $a \geq b$, and notice that the LHS is at least $b^2$. So we test $b$ from $1$ to $6$, and find all $(a,b)$, and then sub back into the equation to find $d$. We obtain solutions $(x,y) = (1,7), (7,1)$ in this way.

If $a^2-ab+b^2 = 1$, then since the LHS is at least $ab$ by basic inequalities, we necessarily have $a=b=1$. This then yields the solution $(x,y) = (22,22)$.