Given integers $m,n$, find integers $a,b,c$ such that $a^3+b^3+c^3-3abc=m n$

Question

With a³+b³+c³-3abc=m (m-random integer) And a³+b³+c³-3abc=n(n-another integer)
How to find a³+b³+c³-3abc=mn(m and n are Co prime) I came across this in an online math contest. Hint was given as properties of determinants. We have to find integer values of a, b, c which satisfy the equation a³+b³+c³-3abc=mn (All three cases have different values for a,b,c) (in online question numbers where used instead of m and n) Pls help

Answer 1

There is an easy way to look at this. Take a three by three matrix $$ T = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right) $$ Note that $$ T^2 = \left( \begin{array}{rrr} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right) $$ and $$ T^3 = I$$

$$ aI + bT + cT^2 = \left( \begin{array}{rrr} a & b & c \\ c & a & b \\ b & c & a \end{array} \right) $$

Using $T^3 = I, \; T^4 = T,$ given $(a,b,c)$ and $(d,e,f),$ it is easy to calculate $(j,k,l)$ in $$ (aI + b T + c T^2)(dI + eT + fT^2) = jI + kT + lT^2 $$ Meanwhile, we are using determinants, in that $$ \det(aI + b T + c T^2) = a^3 + b^3 + c^3 - 3abc $$