2 cand: heavy one... I have no idea how did you figure out that 2011 is prime! Do you remember the table of prime numbers from 2 to 10000?? I easily found trivial solution b=0 a=+-1, and then got stucked. however, later I found that this is unique solution, and (3)=c^2= 1006^2. I've used your approach. You've got a 4 possibilities, +-1 +-2011 and +-2011 +-1, but 2 of them lead to complex roots. You see, sum (1)+(2)+(3) gives you (a+b+c)^2=1+2012+x, thus x>=-2013, so (b-a) may be only +-1, not +-2011. The answer is (3)=1012036

In the equation a^2 + 2bc = 1, obviously, a is either 1 or -1 and either b or c is zero. Then if you derive b from the second equation, b = sqrt(2012 - 2ac), c can't be zero, because sqrt(2012) is not an integer. Therefore b = 0. So, in order for b to be 0, 2012 - 2ac should be zero or 2ac = 2012. => it's either a = -1 and c = -606 or a =1 and c = 606. In any case, because b is zero, 2bc term in the 3rd equation disappears and c^2 gives an answer 367236.

To the comment above, 2ab term in the 3rd equation, of course

simple eq: (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca. from given eq, we know: (a+b+c)^2 = (2012 + 1 + UNKNOWN) = K^2 with K being anything larger than 45. 45*45 = 2025 46*46 = 2116 .. so you can figure out the missing term by: (K^2 - 2013)