# How may pairs of $(m,n)$ are there satisfying $m^{3}-n^{3}=21$? [closed]

This is from the entrance exam for Indian Statistical Institute. $m$ and $n$ are supposed to be positive integers.

• Are $m$ and $n$ supposed to be integers? – Airdish Mar 15 '16 at 12:46
• Yes.m and n are supposed to be positive integers. Sorry I did not mention that. – aknn83 Mar 15 '16 at 13:33

We must obviously have $m>n$, which means $m^3\geq (n+1)^3 = n^3+3n^2+3n+1$. This yields $$21=m^3-n^3\geq (n+1)^3-n^3= 3n^2+3n+1$$ which means that $n$ is at most equal to $2$ (and if we allow negative numbers, $n$ also cannot be smaller than $-3$). From there you can just check every case.
Hint: Use $a^3 - b^3$ formula, then assume the two terms to be factors of $21$. Then solve simultaneous equations for each pair of factors. This should give you integer pairs.