First we consider m>=3. m^4+1=n^3(m-2)+n. If n<=m so it is easy to see the contradition. So n>=m+1. So we have:m^4+1>=(m+1)^3(m-2)+(m+1) Or m^3-3m^2-4m-1<=0.(1) If m>=5 so (1) is not true. The rest is easy

What happened when $m <3$ ?

If m<0 we consider n. If n>0 so again contradition. If n<0 so we do again like my above post.

This equation has 2 root:(m;n)=(0;-1);(2;-17).

anhtaitran wrote: This equation has 2 root:(m;n)=(0;-1);(2;-17). I think $(m,n)=(-1,-1)$ works as well.

Write the equation as : $mn^3 - m^4 = 2n^3-n+1$ $m(n^3-m^3) = 2n^3-n+1$ $m(n-m)(n^2+m^2+mn) = 2n^3-n+1$ This means that, $m | 2n^3-n+1$ and $n-m | 2n^3-n+1$ $ \Rightarrow n | 2(2n^3-n+1)$ which means that $n|2$ and there are four possible values of n, viz. $1,2,-1,-2$ Substituting for each value of n in the equation and solving for integral m, we get that the only solutions are $(m,n)=(0,-1),(-1,-1)$, both of which satisfy the original equation