Finding the number of positive integer pairs $(a,b)$ satisfying $a^2 + b^2 < 2013$ and $a^2b\mid b^3 - a^3$

Question

Find the number of positive integer pairs $(a,b)$ satisfying $a^2 + b^2 < 2013$ and $a^2b \mid b^3 - a^3$.

Let $\dfrac{b^3-a^3}{a^2b} = \dfrac{b^2}{a^2} - \dfrac{a}{b} = k$, where $k$ is an integer. Rearranging a bit, we can get the cubic equation $\dfrac{a^3}{b^3} + k\dfrac{a^2}{b^2} - 1 = 0$, or, substituting $\dfrac{a}{b} = x$, we have $x^3 + kx^2 - 1 = 0$. Since $a$ and $b$ are integers, the roots need to be rational. However, I am not sure how to do that.

Answer 1

A more direct way: Denote $d = (a, b)$, $a = a_0 d$, $b = b_0 d$, then $a^2 b \mid b^3 - a^3 \Leftrightarrow a_0^2 b_0 \mid b_0^3 - a_0^3$. If $a_0 > 1$, for any prime $p \mid a_0$, because $(a_0, b_0) = 1$, then$$ p \mid b_0^3 - a_0^3,\ p \mid a_0^3 \Longrightarrow p \mid b_0^3 \Longrightarrow p \mid 1, $$ a contradiction. Thus $a_0 = 1$, and analogously, $b_0 = 1$. Therefore, $a = b$.

Answer 2

Depending on the tools allowed, I would just calculate in a spreadsheet. We know $b \lt \sqrt {2013}$ so $b \le 44$ I made a spreadsheet with $a$ ranging from $1$ to $44$ in a column and $b$ ranging from $1$ to $44$ in a row. Using mixed references I was able to put a formula for $(b^3-a^3)/(a^2b)$ in the $1,1$ cell, then copy right and down to fill it all. A quick visual scan shows the only cases that give an integer are $a=b$, so we need $a=b\lt \sqrt{\frac{2013}2}=31+$ and there are $31$ pairs.

Answer 3

We will show that if $a\ne b$, there is no solution.

For $a^2b \mid b^3 - a^3$, both $a$ and $b$ must be even. Hence: $a=2m, b=2n$. Then $m^2n\mid n^3-m^3$. By the method of infinite descent, there is no solution.