If $a=b+c$ prove that $S=a^4+b^4+c^4$ is twice the square of a positive integer

Question

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.

Answer 1

$$K=(b^2+bc+c^2)^2$$ You can find it after noticing it should have both square terms.

Answer 2

Given the hint from @didgogns, I found a way using Newton identities. Using the notation $s_k=b^k+c^k$, $\sigma_1=b+c$ and $\sigma_2=bc$ it holds that

$$S=\sigma_1^4 + s_4$$

but as $s_4=\sigma_1^4-4\sigma_2\sigma_1^2+2\sigma_2^2$ (Newton identity),

$$S=2\sigma_1^4-4\sigma_2\sigma_1^2+2\sigma_2^2=2(\sigma_1^4-2\sigma_2\sigma_1^2+\sigma_2^2)$$ or $$S=2(\sigma_1^2-\sigma_2)^2,$$ as required.

Answer 3

$(b+c)^4=b^4+4b^3c+6b^2c^2+4bc^3+c^4=(b^4+c^4)+4bc(b^2+c^2)+6b^2c^2=>$ $$ S=a^4+b^4+c^4=a^4+a^4-4bc(b^2+c^2)-6b^2c^2=2a^4-4bc(b+c)^2+2b^2c^2=2a^4-4bca^2+2b^2c^2=2(a^4-2bca^2+b^2c^2)=2(a^2-bc)^2 $$ $=>S=2(a^2-bc)^2$ $$ Q.E.D $$

Answer 4

We have:

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