<s>[INST] Proving $ab(a+b)+ac(a+c)+bc(b+c)$ is even 
Prove that $\forall a,b,c\in \mathbb N: ab(a+b)+ac(a+c)+bc(b+c)$ is even

I tried to simplify the expression to something that would always yield an even number: $ (a+b+c)(ab+ac+bc)-3abc$ but that's just a sum of numbers that are divisible by $3$...
Is there a way to do this without checking every combination of the parity of $a,b,c$?
 [/INST] $a$ and $a^2$ are either both even or both odd. So you don't change parity when you replace $a^2$ with $a$:
$$\begin{align}
ab(a+b)+bc(b+c)+ca(c+a)
&=a^2b+ab^2+b^2c+bc^2+ac^2+a^2c\\
&\equiv ab+ab+bc+bc+ac+ac\\
&= 2(ab+bc+ac)\\
\end{align}$$
which is even.
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