Number Theory

Problem : Let $a,b,c$ be three natural numbers which are greater or equal to $2$ and such that :

$$(abc{)}^2\mid a^3+b^3+c^3-3abc$$

Prove that $a=b=c$.

My solution :

The assumption of this problems can be writen as :

$$(abc{)}^2\mid a^3+b^3+c^3-3abc(\ast )$$

Take the any prime number $p$ and let $v_p(a)=A,v_p(b)=B,v_p(c)=C$.

Assume that $a=p^A.x,b=p^B.y=,c=p^C.z$ with $x,y,z$ are positive integers and coprime with $p$.

Wlog, we can suppose that $A\ge B\ge C$.

We have :

$$v_p((abc{)}^2)=2v_p(abc)=2\left(A+B+C\right)$$

And :

$v_p\left(a^3+b^3+c^3-3abc\right)$$=v_p\left(p^{3A}{x}^3+p^{3B}{y}^3+p^{3C}{z}^3-3p^{A+B+C}xyz\right)$

$$v_p\left[p^{3C}\left(p^{3A-3C}{x}^3+p^{3B-3C}{y}^3+z^3-3p^{A+B-2C}xyz\right)\right]$$

If $A=C$, we find it obvious to get $A=B=C$, which implies that $a=b=c$.

If $B=C$, we get $b=c$ and $y=z$ of course. From the previous results :

$$v_p((abc{)}^2)=2A+4B$$

$v_p(a^3+b^3+c^3-3abc)$$=v_p\left[p^{3C}\left(p^{3A-3C}{x}^3+2y^3+p^{A-C}x{y}^2\right)\right]$$=3C$$=3B$

Owing to $(\ast )$ we have :

$$2A+4B\le 3B\iff 2A+B\le 0$$

This is a contradiction !

If $B\ne C$, it’s obvious that $A\ne C$. We get :

$$v_p(a^3+b^3+c^3-3abc)=3C$$.

Owing to $(\ast )$ we have :

$$2(A+B+C)\le 3C\iff 2(A+B)\le C$$

This is also a contradiction !

To sum up, we have $a=b=c$ which is Q.E.D.