Find $m,n$ such that $(m^2+n)(n^2+m)=(m-n)^3$

Question

Find all $m,n$ such that

$(m^2+n)(n^2+m)=(m-n)^3$ where $m,n$ are nonzero integers.

I expand this and found some equation.By checking ,I found two solutions $(-1,-1)$&$(8,-10)$. Please help me some this. My expression failed to find some finite relation so that I can find these.

I had gotten my answer . So don't post any command, answer .

Answer 1

From $(m^2 + n)(n^2+m) = (m-n)^3$ we get $$m^2 n^2 + 3 m^2 n - 3 m n^2 + m n + 2 n^3 = 0.$$ Since $n,m$ are non-zero, we can safely divide the whole thing by $n$ to get $$2n^2 + (m^2-3m)n + (3m^2 + m) = 0.$$ We can solve this quadratic equation in $n$: $$n = \dfrac{3m-m^2 \pm \sqrt{m(m-8)(m+1)^2}}{4}.$$

This means that $m(m-8)$ is a perfect square, say $p^2$. Let $q = m+4$ then $(q+p)(q-p) = q^2- p^2 = 16$.

There are up to $10$ integral solutions to the system of equations

$$\begin{cases}q+p = \pm 2^u \\ q-p = \pm2^v\end{cases} \qquad u,v \geq 0, u+v = 4.$$

Then you'd have to test if that solution for $m$ yields a valid solution for $n$ (we want the numerator to be a non-zero multiple of $4$). This yields all solutions.

Answer 2

$(m,0)$ is a solution.

When $n\ne0$, the equation reduces to a quadratic equation in $n$ whose discriminant is $m(m-8)(m+1)^2$. So, we want to find $m$ such that $m(m-8)$ is a square: $m(m-8)=y^2$.

Let $x=m-4$. Then $(x+y)(x-y)=x^2-y^2=16$. There are only finitely many possibilities, corresponding the factorizations of $16=ab$, with $a,b$ of the same parity:

\begin{array}{brrrr} a & b & x & y & m \\ -8 & -2 & -5 & -3 & -1 \\ -4 & -4 & -4 & 0 & 0 \\ -2 & -8 & -5 & 3 & -1 \\ 2 & 8 & 5 & -3 & 9 \\ 4 & 4 & 4 & 0 & 8 \\ 8 & 2 & 5 & 3 & 9 \\ \end{array}

Answer 3

Hint: rewriting the equation as $$ m^2n(n + 3) - mn(3n - 1) + 2n^3=0, $$ we have a quadratic equation in $m$, so the term under the square root for the formula needs to be a square, i.e., $1-8n$ needs to be a square.