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Algebra Difficulty 4.8 AIME Prove it Belarus

It is given that integers aa, bb and cc satisfy the equality a+b+c=0a + b + c = 0. Denote S=ab+bc+caS = ab + bc + ca, A=a2+a+1A = a^2 + a + 1, B=b2+b+1B = b^2 + b + 1 and C=c2+c+1C = c^2 + c + 1.
Prove that the number (S+A)(S+B)(S+C)(S + A)(S + B)(S + C) is the square of an integer.

Solution

S+A=bc+a(b+c)+A=bc−a2+a2+a+1=bc−(b+c)+1=(b−1)(c−1). S + A = bc + a(b + c) + A = bc - a^2 + a^2 + a + 1 = bc - (b + c) + 1 = (b - 1)(c - 1).
Similarly S+B=(c−1)(a−1)S + B = (c - 1)(a - 1) and S+C=(a−1)(b−1)S + C = (a - 1)(b - 1). Hence,
(S+A)(S+B)(S+C)=((a−1)(b−1)(c−1))2 (S + A)(S + B)(S + C) = ((a - 1)(b - 1)(c - 1))^2
is the square of the integer (a−1)(b−1)(c−1)(a - 1)(b - 1)(c - 1).

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