Discrete math. Solve the equation in the set of natural numbers. I have to solve the equation $$m^4-n^4=5(m^3+n^3)$$ in the set of natural numbers. I wrote a simple code in java and i solved the equation. Only solution in the set of natural numbers is $m = 6$ and $n = 3$.
I have been trying to get this solution mathematically for at least 3-4 hours and i am stuck. I really hope that there is someone who could show me how to solve this equation? Thank you!
 A: Hint 1: We can assume that $n\le m$;
Hint 2: We can simplify the equation to $$(m-n)(m^2+n^2)=5(m^2+n^2-mn);$$
Hint 3: The new equation gives us $(m-n)<5$ (?);
Hint 4: If $(m-n)=k$, then $k=1,2,3$ or $4$;
Hint 5: From the new equation we have $$k(2n^2+2nk+k^2)=5(k^2+n^2+kn);$$
Hint 6: By $3^{rd}$ equation the only possible case is $k=3$.
A: Expanding $m^4-n^4$ , subtracting the right hand side, regrouping the terms by factoring $m^3+n^3$, and dividing through $m$$+n$, we finally arrive at $\underbrace{(m^2+n^2-mn)}_{(m-n)^2\ +\ mn\ \ge\ 0}\Big[m-(n+5)\Big]=mn\,(n-m)$, which would imply that $m-5\le n\le m\iff n=m-k$, where $0\le k\le5$. By replacing this in the former equation, we ultimately arrive at a quadratic equation in m, from where we deduce that $m=k\left[\dfrac12\pm\dfrac{\sqrt{4k\,(10-k)-75}}{4k-10}\right]$. Testing all six possible values, we have $k=3\to m=6\to$ $\to n=3$ , and $k=5\to m=5\to n=0$. QED.
A: Obviously   $ m>n.
m^4-n^4=(m-n)*(m^3+m^2*n+m*n^2+n^3)=5*(m^3+n^3).$
Now it is easy to locate that we can bound this thing.If $ m-n>=5 $ L.H.S>R.H.S.
If $m-n=1$ we have  $m^3+m^2*n+m*n^2+n^3<4m^3$
If $m-n=2$ we have $3*(m^3+n^3)=2mn*(n+m)$ , substituting $m^3+n^3=(m+n)^3-mn*(m+n)$ we get $3m^2+3n^2=5mn$, but by mean inequality we have $3m^2+3n^2>=6mn$
If $m-n=3$ doing the same we get $2m^2+2n^2=5mn$ , substituting $m=n+3$ yields $n^3+3n-18=0;(n-3)*(n+6)=0,$ By check (6,3) satisfiyes the equality
If $m-n=4$,we get $4mn*(m+n)=m^3+n^3=>m^2+n^2=5mn$ so $n|m$ , hence $n<=m/2$; so $ m<=8$ ; $n<=4$, by check 
no other solutions appear.
