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(a + b + 1) (a + b - 1) (a - b + 1) (- a + b + 1) constant if a =\sqrt{1 + x^2}

Source: 2000 Moldova JBMO TST p1

February 20, 2021
algebraSum

Problem Statement

Show that the expression (a+b+1)(a+b−1)(a−b+1)(−a+b+1)(a + b + 1) (a + b - 1) (a - b + 1) (- a + b + 1), where a=1+x2a =\sqrt{1 + x^2}, b=1+y2b =\sqrt{1 + y^2} and x+y=1x + y = 1 is constant ¸and be calculated that constant value.
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