Find $f(n)$ such that $f(m + n) + f(m - n) = 2f(m) + 2f(n)$ for all integers $m$ and $n$ and $f(4) = 16$

Question

The question is in the title itself- The function $f(n)$ takes the integers to the real numbers such that $f(m + n) + f(m - n) = 2f(m) + 2f(n)$ for all integers $m$ and $n$ and $f(4) = 16$. Find $f(n)$.

I tried substituting $m=4$ and $n=0$, but it did not help me in solving the actual function. Can I have a hint as to how to start, or am I on the right path and am missing something blatantly obvious? It would be greatly appreciated.

Answer 1

Find $f(0)$ by setting $m=n=0$.

Find $f(2)$ by setting $m=n=2$, and using $f(4)$ and $f(0)$.

Find $f(1)$ by setting $m=n=1$ and using $f(2)$ and $f(0)$.

Find $f(n+1)$ in terms of $f(n)$, $f(n-1)$, and $f(1)$, for $n\geq 0$.

Find a general formula for $f(n)$ for $n>0$.

Find $f(-n)$ in terms of $f(n)$ and $f(0)$ by setting $m=0$.

Answer 2

Setting $m = n$, we get $f(2n) + f(0) = 4f(n).$ In particular, if $n=0$, then we have $2f(0) = 4f(0) \Rightarrow f(0) = 0.$ So $f(2n) = 4f(n),$ and $f(2) = 4, f(1) = 1.$

Let us try to find a recursive expression by setting $n=1$: \begin{align*} f(m+1) + f(m-1) &= 2f(m) + 2f(1) \\ \Rightarrow \quad f(m+1) &= 2f(m) - f(m-1) + 2. \end{align*}

In particular, note that $f(m+1)-f(m) = f(m)-f(m-1) + 2.$ This finite difference indicates that $f$ is quadratic, and from the above 3 values we see that $f(n)$ must be $\boxed{ n^2 }.$ We can verify that this solution is also sufficient.