Write this just as $ a^3+b^3+(-1)^3-3.a.b.(-1)=0 $, And now factorize.

rigor wrote: Write this just as $ a^3+b^3+(-1)^3-3.a.b.(-1)=0 $, And now factorize. And what do you do then? We get $(a+b-1)(a^2 + b^2 + 1-ab-a-b) = 0$

wowmariahaha wrote: rigor wrote: Write this just as $ a^3+b^3+(-1)^3-3.a.b.(-1)=0 $, And now factorize. And what do you do then? We get $(a+b-1)(a^2 + b^2 + 1-ab-a-b) = 0$ Need to explain a bit more, If such a pair exists, then either $ (a+b)=1 $ For the first case $ \exists $ infinitely many such pair $ (a,b) $, And also on the contrary possibly we have, $ (a-b)^2+(b+1)^2+(c+1)^2=0 $ which leads us to find $ (a,b)=(-1,-1) $

Thanks

wowmariahaha wrote: rigor wrote: Write this just as $ a^3+b^3+(-1)^3-3.a.b.(-1)=0 $, And now factorize. And what do you do then? We get $(a+b-1)(a^2 + b^2 + 1-ab-a-b) = 0$ There is a mistake in your factorization u it should be $(a+b-1)(a^2+b^2+1-ab+a+b)$ and so than we have $a=1-b$ and $(a-b)^2+(a+1)^2+(b+1)^2=0$ which is $a=b=-1$ so we have $a,b=(a,1-a),(-1,-1) $

We must use the following lemma: if $x$, $y$ and $z$ are real numbers such that $x^3+y^3+z^3=3xyz$, then $x+y+z=0$ or $x=y=z$. The given equation is equivalent to: $$a^3+b^3+(-1)^3=3ab(-1).$$Then, $a+b-1=0$ or $a=b=-1$. Therefore the only pairs that satisfy are $(-1, -1)$ and $(a, 1-a)$ for each integer number $a$.