Proving $ab(a+b)+ac(a+c)+bc(b+c)$ is even

Question

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 $a,b,c$?

Answer 1

Below we derive that it is equivalent to $\, a(a\!+\!1)\,$ is even.

Lemma $\,\ ab(a\!+\!b)+ac(a\!+\!c)+bc(b\!+\!c)$ even $\,\iff ab(a\!+\!b)\,$ even

Proof $\ \ (\Rightarrow)\ \ $ Let $\,c = 0.\quad (\Leftarrow)\ \ $ Permute $\,a,b,c\,$ and add.

Lemma $\,\ ab(a\!+\!b)\,$ even $\iff \color{#c00}{a(a\!+\!1})\,$ even

Proof $\ \ (\Rightarrow)\ \ $ Let $\,b = 1.\quad (\Leftarrow)\ \ \ \color{#c00}a\color{#0a0}b(a\!+\!b)\, =\, \color{#c00}a\,\color{#0a0}b\,(\color{#c00}{a\!+\!1}\, + \,\color{#0a0}{b\!+\!1}\, -\, 2)$

Answer 2

We will use that for any $A$ the numbers $A$, $3A$ and $A^3$ all have the same parity. Then writing $S$ for our expression, we see $$\begin{align*} S &= ab(a+b)+bc(b+c)+ca(c+a) \\ &\equiv 3ab(a+b) + 3bc(b+c)+3ca(c+a) \\ &= (a+b)^3 - a^3 - b^3 + (b+c)^3 - b^3 - c^3 + (c+a)^3 - c^3 - a^3 \\ &\equiv (a+b)^3 + (b+c)^3 + (c+a)^3 \\ &\equiv (a+b) + (b+c) + (c+a) \equiv 0. \end{align*} $$ Here $\equiv$ denotes equality modulo $2$, i.e. $M \equiv N$ means that $M$ and $N$ have the same parity. So $S$ has the same parity as $0$, hence it is always even.

Answer 3

$a$ 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.