SOLUTION: Let a, b,c be positive integers with a>=b>=c such that a^2 - b^2 - c^2 +ab=2011 and a^2 + 3b^2 + 3c^2 -3ab -2ac-2bc=-1997 what is a?

Question 569241: Let a, b,c be positive integers with a>=b>=c such that
a^2 - b^2 - c^2 +ab=2011
and
a^2 + 3b^2 + 3c^2 -3ab -2ac-2bc=-1997
what is a?
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
It is generally best to avoid posting this question until the day after the AMC10/12A exams (yes, I took this exam as well). But since it's pretty late in the day, I'll show the solution. Make sure not to publicize it to anyone else.

Adding both equations (I should have done this myself during the exam...I tried adding three times the first equation to the second equation and got nowhere).

This factors to

Essentially we have three square numbers adding to 14, so the only possibility is {9,4,1}. Since a and c are the furthest apart, we have

So our three numbers are either {c,c+1,c+3} or {c,c+2,c+3}. Suppose the numbers are {c,c+1,c+3}. Plugging into the first equation we have

(this is actually a linear equation)

If c = 250, then a = 253 and we are done. For sake of completeness we can try the other case:

, however if you solved this, c would not be an integer. Therefore a = 253.

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