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An interesting little problem:

I have found solutions $[a=3,b=2,c=1,d=5,e=4]$ but not able to find proof that these are all that exist.

Find, with proof, all integers $a$, $b$, $c$, $d$ and $e$ such that:

$a^2 = a + b - 2c + 2d + e - 8$

$b^2 = -a - 2b - c + 2d + 2e - 6$

$c^2 = 3a + 2b + c + 2d + 2e - 31$

$d^2 = 2a + b + c + 2d + 2e - 2$

$e^2 = a + 2b + 3c + 2d + e - 8$

It seems simple, but I cannot find a proof.

nikolai

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this seems like an opportunity to use groebner basis! –  Peter Sheldrick Oct 9 '11 at 6:49
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1 Answer 1

Calculate the sum $(a-3)^2+(b-2)^2+(c-1)^2+(d-5)^2+(e-4)^2$.

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+1: Looks like Nikolai's teacher was having a little bit fun. –  Jyrki Lahtonen Oct 9 '11 at 8:59
    
I like this; concise and beautiful. –  J. M. Oct 9 '11 at 9:55
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