<s>[INST] 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.
 [/INST] $(b+c)^4=b^4+4b^3c+6b^2c^2+4bc^3+c^4=(b^4+c^4)+4bc(b^2+c^2)+6b^2c^2=>$
$$
S=a^4+b^4+c^4=a^4+a^4-4bc(b^2+c^2)-6b^2c^2=2a^4-4bc(b+c)^2+2b^2c^2=2a^4-4bca^2+2b^2c^2=2(a^4-2bca^2+b^2c^2)=2(a^2-bc)^2
$$
$=>S=2(a^2-bc)^2$
$$
Q.E.D
$$
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