demonstration number theory

Question

Let be $a, b, c$ real numbers such that: $$a+b+c \neq 0 \mbox{ and } ab+bc+ca \mbox{ is rational. }$$

Prove that $a^4+b^4+c^4$ is rational if and only if $abc$ is rational

Answer 1

A simple counter-example:
Let $(a,b,c)=(0,\pi,1/\pi)$. Then $a+b+c=\pi+1/\pi\neq0$, $ab+bc+ac=bc=1\in Q$, and $abc=0\in Q$, but $a^4+b^4+c^4=\pi^4+1/\pi^4\notin Q$.(Here $\pi$ could be replaced by any transcendental real number.)
A converse counter-example:
Let $(a,b,c)=(\sqrt2,\sqrt2,\sqrt2)$. Then $a+b+c\neq0$, $ab+bc+ac=6\in Q$, and $a^4+b^4+c^4=12\in Q$, but $abc=2\sqrt2\notin Q$.
So, under your conditions, both implications are invalid.
However, if one puts on the condition that $a+b+c\in Q$, then the assertion is true, as shown in what follows.
Let $n_1=a+b+c$, $n_2=ab+bc+ac$, and $n_3=abc$. If $n_3\in Q$, then
$a^4+b^4+c^4=n_1^4+2n_2^2-4n_1^2n_2+4n_1n_3$,
and hence it is also rational. Conversely, if $a^4+b^4+c^4 \in Q$, then $n_3$ also $\in Q$. QED.