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Let $\mathbb{Z}$ denote the set of integers. Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be such that $f(x) f(y)=f(x+y)+f(x-y)$ for all $x, y \in \mathbb{Z}$. If $f(1)=a >2$, then find all such possible functions .

What i considered was let the function relation be called as $P(x,y)$ , then $P(x,x)$ we get $f(2x) =f^2(x) - f(0)$ and $P(x,0)$ gives $f(x)(f(0)-2) = 0$ , which leads us to $f(0) = 2$ since if $f(x) = 0 \forall x$ then we have $f(0) = 0$ which is contradicting the given condition . Hence $f(0) = 2$ , now we check injectivity $P(x_1,x_2)$ ,such that $f(x_2)= f(x_1)$ but that doesnt lead me to $x_1 =x_2$ conclusion also how do we check surjectivity in this case ?

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Define the function $$ P(x,y) := -f(x) f(y)+f(x+y)+f(x-y) \tag{1} $$ and $\,P(x,y)=0\,$ is assumed true for all $\,x,y\in\mathbb{Z}$ while $\,a := f(1) >2\,$ is also assumed.

Now $\,P(n,1)=0\,$ is equivalent to $$ f(n+1) = a\,f(n) - f(n-1). \tag{2} $$ This is a linear recursion. Since $\,f(0)=2 ,f(1)=a\,$ are given there is a unique solution which is $$ f(n) = 2\,T_n(a/2) \tag{3} $$ where $\,T_n\,$ is the Chebyshev polynomial of the first kind.

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