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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

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up vote 11 down vote accepted

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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@ThomasAndrews : thanks for the correction. – mercio Mar 23 '13 at 13:41

After simplification of the equation you will get $$(a+\frac{3}{2})^2+(b+\frac{3}{2})^2+(c+\frac{3}{2})^2=\frac{7}{4}.$$Note that LHS$\geq 3.(\frac{3}{2})^2>RHS$, hence not possible.

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What if they are negative? – Thomas Andrews Mar 23 '13 at 13:31
$a, b, c$ could be negative. – Ivan Loh Mar 23 '13 at 13:32
It's definitely nice to reduce it to solving $p^2+q^2+r^2=7$ for $p,q,r\in\mathbb Q$. – Thomas Andrews Mar 23 '13 at 13:35
Yes, I should follow mercio's way. – Easy Mar 23 '13 at 13:40

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