Find $f(1986)$ if $f(1)=1$ and $f(x + y) = f(x) + f(y) - 2f(xy)$

Question

If $f(x + y) = f(x) + f(y) - 2f(xy)$ for all real numbers $x$ and $y$ and $f(1) = 1$, compute $f(1986)$.

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I found that $f(2)=0$ and I could just plug in numbers from there on, but I can't think of an efficient way to solve this problem. Could someone please solve this problem? Thanks!

Answer 1

With $x=y=0$, we find $f(0)=0$. With $y=1$, we find $f(x+1)=f(1)-f(x)$, so $f(x+2)=f(1)-(f(1)-f(x))=f(x)$ and by induction $f(x+2n)=f(x)$. In particular, $$f(1986)=f(0+2\cdot 993)=f(0)=0. $$

Remark: The property that $f(1)=1$ was not needed.

Answer 2

Taking $y=1$ we get $$f(x+1)=f(x)+f(1)-2f(x)=f(1)-f(x)$$ In particular, taking $x=0$ in the above gives $f(1)=f(1)-f(0)$, so $f(0)=0$. Taking $x+1$ for $x$, we get $$f(x+2)=f(1)-f(x+1)=f(1)-(f(1)-f(x))=f(x)$$ Hence the function is periodic with period $2$.

Now, taking $y=2$ we get $$f(x)=f(x+2)=f(x)+f(2)-2f(2x)$$ which gives $$2f(2x)=f(2)$$ However, $f(2)=f(0)=0$. So $f(2x)=0$ for all $x$, and the only solution is the constant $$f(x)=0$$