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Algebra Difficulty 4.8 AIME Prove it Ukraine

Numbers aa, bb, cc satisfy the conditions:
a2+2=b4,b2+2=c4,c2+2=a4. a^2 + 2 = b^4, \quad b^2 + 2 = c^4, \quad c^2 + 2 = a^4.
What values can take expression
(a2−1)(b2−1)(c2−1)(a^2 - 1)(b^2 - 1)(c^2 - 1)?

Solution

Let us subtract 11 from the left and right parts of equality and obtain:
a2+1=b4−1=(b2−1)(b2+1), a^2 + 1 = b^4 - 1 = (b^2 - 1)(b^2 + 1),

and analogously from other 22 equalities. Next multiply the resulting equalities:

(a2+1)(b2+1)(c2+1)=(a2−1)(a2+1)(b2−1)(b2+1)(c2−1)(c2+1) (a^2 + 1)(b^2 + 1)(c^2 + 1) = (a^2 - 1)(a^2 + 1)(b^2 - 1)(b^2 + 1)(c^2 - 1)(c^2 + 1)
(a2−1)(b2−1)(c2−1)=1. (a^2 - 1)(b^2 - 1)(c^2 - 1) = 1.
Note that this value is achieved when a=b=c=2a = b = c = \sqrt{2}.

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