If
$f(x-y) = f(x)f(y) $ (Call this equations (1) )
for all $x$ and $y$, and $f(x)\not = 0$ for any $x$, find all possible values for $f(1997)$.
My attempt:
Substitute $x=y=0$ in $(1)$ to obtain $f(0) = 1$. If you now substitute $x=y=1$ in (1), you get $f(1)^2 = f(0) \implies f(1) = \pm 1$. Now if $f(1)=1$, then you can see that $f(2)=1$ as well (by substituting $x=2,y=1$ in $(1)$ ). We can extend this to all natural numbers (assume that $f(k)=1 \forall k\in \mathbb{N} \leq n$:
\begin{align*} f((k+1)-1) &= f(k+1)f(1) \\ f(k) &= f(k+1)f(1) \\ f(k+1) &= 1 \end{align*}
On the other hand, if $f(1) = -1$, then (as before) $f(2) = 1$. This implies
\begin{align*} f(3-1) &= f(3)f(1) \\ f(2) &= f(3)f(1) \\ f(3) &= -1 \end{align*}
This makes us think that $f(n) = (-1)^n \forall n \in \mathbb{N}$. Indeed, this is the case (assume that it holds for all $k\leq n$):
\begin{align*} f((k+1)-1) &= f(k+1) f(1) \\ f(k) &= f(k+1) f(1) \\ (-1)^k &= f(k+1) f(1) \\ f(k+1) &= (-1)^{k-1} = (-1)^{k+1} \end{align*}
Therefore, all the possible values for $f(1977)$ are $\{1,-1\}$.
But apparently, $f(1977) = -1$ is not given in the solution to this problem (this problem is from Richard Rusczyk's and Mathew Crawford's Intermediate Algebra):
Setting $y=x$, we get $f(0) = f(x)^2)$. Setting $x=0$, we get $f(0) = f(0)^2$, so $f(0)[f(0) - 1] = 0$. Hence, $f(0) = 0$ or $f(0) = 1$. But $f(x)$ is never equal to $0$, so $f(0) = 1$. Then $f(x)^2 = 1$ for all x, so $f(x) = -1$ or $f(x) = -1$ for all $x$.
Setting $x=2y$ in the given equation, we get $f(y) = f(2y)f(y)$. Since $f(y)$ is nonzero, we may divide both sides by $f(y)$ to get $f(2y)=1$. Since this holds for all $y$, in particular, we have $f(1977) = \boxed{1}$.
What did I miss in my solution? Also, the fact that they did not mention the domain and co-domain of $f$ annoys me.
You start with $f(1) = -1$, then you get $f(n) = (-1)^n$.
But, the assumption is for all $x$ and $y$. Then you have $f(\frac{1}{2}) = f(1)f(\frac{1}{2}) = -f(\frac{1}{2})$ or $f(\frac{1}{2}) = 0$. Note that $f(x) \neq 0$ for all $x$.