<s>[INST] Does there exist $a,b,c\in \mathbb Q$ such that $(a+b+c)^2 + 3(a+b+c)+5=2(ab+bc+ca)$ 
Does there exist $ a,b,c\in \mathbb Q$ such that $(a+b+c)^2 + 3(a+b+c)+5=2(ab+bc+ca)$

I think the answer is no
 [/INST] After doing a variable change and multiplying by a common denominator, you are left with solving $a^2+b^2+c^2 = 7d^2$ with integers $a,b,c,d$.
However, squares are congruent to either $0,1$ or $4$ modulo $8$.
So $a^2+b^2+c^2 \not\equiv 7 \pmod 8$ and $7d^2 \equiv 0,4,7 \pmod 8$
Hence $d$ must be even, which implies that $a,b,c$ are also even, and then $a/2,b/2,c/2,d/2$ are new integers satisfying the same equation.
Thus $a,b,c,d$ are infinitely divisible by $2$, so they are all $0$.
So there aren't any solution to the equation.
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