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

f (1986) = 0

Step-by-step explanation:

f (a + b) = f (a) + f (b) - 2f (ab)

We need to find f (1986)

f (1986) = f (1 + 1985). Using the above formula, we can write:

f (1 + 1985) = f (1) + f (1985) - 2f (1 x 1985)

f (1986) = 1 - f (1985) Equation 1

Applying the same formula again on f (1985), we get:

f (1985) = f (1 + 1984) = f (1) + f (1984) - 2f (1984)

f (1985) = 1 - f (1984)

Using this value in Equation 1, we get:

f (1986) = 1 - (1 - f (1984))

f (1986) = f (1984)

Continuing this, we can observe,

f (1986) = f (1984) = f (1982) = f (1980) ... = f (4) = f (2)

So,

f (1986) = f (2)

f (2) = f (1 + 1) = f (1) + f (1) - 2 (1 x 1) = f (1) + f (1) - 2f (1)

f (2) = 1 + 1 - 2 = 0

Therefore,

f (1986) = f (2) = 0