If \(f(a + b) = f(a) + f(b) - 2f(ab)\) for all nonnegative integers \(a\) and \(b\), and \(f(1) = 1\), compute \(f(1986)\).

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\[f(n) = f(1) + f(n-1) - 2f(1) f(n-1)\]
\[ \implies f(n) + f(n-1) = 1\]
f(2) = 0, f(3) = 1, f(4) = 0

and 1986 is even

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