# demonstration number theory

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

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Do you mean that $a+b+c$ and $ab+bc+ca$ are rational and $a+b+c\neq 0$? If you only require $ab+bc+ca$ rational, I don't think this is true. – Thomas Andrews Feb 13 '13 at 19:51
After you have fixed the list of assumptions (as suggested by Thomas Andrews) look up Newton(-Girard) identities. I'm fairly sure that you can use those to derive your conclusion :-) – Jyrki Lahtonen Feb 13 '13 at 19:53
I have developed remarkable products of $a+b+c$-square and the fourth power and $ab+bc+ca$ squared, but I have not gotten anywhere – user62189 Feb 13 '13 at 20:00
Consider the following counter example: $a=b=\sqrt{2}-1, c=3+2\sqrt{2}$, then $a+b+c=4\sqrt{2}+1 \not =0, ab+ac+bc=5 \in \mathbb{Q}, abc=1 \in \mathbb{Q}$, but $a^4+b^4+c^4=611+384\sqrt{2} \not\in \mathbb{Q}$. – Ivan Loh Feb 14 '13 at 8:50
So now the question arises, why do you think the problem is correct? – Gerry Myerson Feb 14 '13 at 11:49

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.)
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 $abc=2\sqrt2\notin Q$.
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.