If $a=b+c$ prove that $S=a^4+b^4+c^4$ is twice the square of a positive integer 
If $a=b+c$, and $a$,$b$,$c\in \Bbb N$, prove that $S=a^4+b^4+c^4$ is twice the square of a positive integer.

Source: a list of problems used in the preparation to math contests.
My attempt:
By making the substitution $a=b+c$ in $S$ and developing $(b+c)^4$, it is easy to show that 
$$S=2(b^4+c^4+bc(2b^2+3bc+2c^2))$$
an expression with the form $S=2K$. The problem now is how to prove that K is a square of a positive integer. I also tried to use Newton Identities but with no luck (Note: later, after a hint, I found a way to solve using this approach, see below). 
Hints and answers are welcomed. Sorry if this is a dup.

We have:
$$S=(b+c)^4+b^4+c^4=2b^4+4b^3c+6b^2c^2+4bc^3+2c^4$$
$$=2(b^4+2b^3c+3b^2c^2+2bc^3+c^4)=2 (b^2+bc+c^2)^2$$
