Find the smallest possible value of $a^4+b^4+c^4-136abc$

Asked Jan 21 '16 at 22:02

Active Jan 21 '16 at 22:32

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Let $a$, $b$, and $c$ be real numbers such that $a+b+c=-68$ and $ab+bc+ca=1156$. The smallest possible value of $a^4+b^4+c^4-136abc$ is $k$. Find the remainder when $k$ is divided by $1000$.

I would solve this question by first noting that $(a+b+c)^2 = a^2+b^2+c^2 +2*1156 = 68^2 \implies a^2+b^2+c^2 = 68^2-2*1156$. Then I am wondering what I should do next. I could square $a^2+b^2+c^2$ to get $a^4+b^4+c^4 +2(a^2b^2+b^2c^2+a^2c^2) = 2312^2$. I am not sure what to do next.

asked Jan 21 '16 at 22:02

Puzzled417

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Real numbers ? You probably mean integers. – Peter Jan 21 '16 at 22:05
@Peter No, I think real numbers is right (at least that's how the question was written). – Puzzled417 Jan 21 '16 at 22:07
But what should the remainder of $k$ divided by $1000$ mean if $k$ is not an integer ? – Peter Jan 21 '16 at 22:07
@Peter It should be an integer otherwise you are right. – Puzzled417 Jan 21 '16 at 22:08
Perhaps, the given conditions imply that $k$ must be an integer ... – Peter Jan 21 '16 at 22:09
Have you tried calculus? – Matt Samuel Jan 22 '16 at 02:20
Note that $a^2+b^2+c^2=2ab+2bc+2ca$. – Chad Shin Jan 22 '16 at 03:32

Find the smallest possible value of $a^4+b^4+c^4-136abc$

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