Prove that integer Let $a, b,$ and $c$ be real such that $ a ^ 2b + b ^ 2 c + c ^ 2a - ab ^ 2 - bc ^ 2 - ca ^ 2 = 6 $ and $ a ^ 2 + b ^ 2 + c ^ 2 -ab - ac -  bc = 7$. Prove that if a is integer, then $b$ and $c$ are also integers.
What I tried:
$ ab (a-b) + bc (b-c) + ca (c-a) = 6 $
$a (a-b) -c (c-a) + b (b-c) = 7 $
$ca (a-b) -c^2 (c-a) + bc (b-c) = 7c$
$ca (a-b) -c^2 (c-a) -ca (c-a) -ab (a-b) = k $
($k$ integer)
$a (c-b) (a-b) -c (c^2-a^2) = k$
$a (c-b) (a-b) -ca^2 = k '$
$a ((c-b) (a-b) -ca) = k '$
$a (-cb-ba + b^2) = k '$
What I tried, but I think something is missing

Put $x = a - b$, $y = b - c$, $z = c - a$. Then:


*

*$x + y + z = 0$;

*$x^2 + y^2 + z^2 = 2(a^2 + b^2 + c^2 - ab - bc - ca) = 14$;

*$xyz = -(a^2b + b^2c + c^2a - ab^2 - bc^2 - ca^2) = -6$.
From 1. and 2., we get:


*$xy + yz + zx = \frac{1}{2}((x + y + z)^2 - (x^2 + y^2 + z^2)) = -7$.


Therefore, the equations 1. 3. 4. tell us that $x, y, z$ are roots of the polynomial $T^3 - 7 T + 6$, which factors as $(T - 2)(T - 1)(T + 3)$.
Hence $x, y, z$ are all integers and we are done.
