$f(n)$ takes the integers to the reals with $f(m + n) + f(m - n) = 2f(m) + 2f(n)$. 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 $f(n)$.
 A: Follow the steps


*

*Compute $f(0)$ (use $m=n=0$), $f(2)$ (use $m=n=2$) and deduce $f(1)$ ($m=n=1$).

*Find a relation between $f(n+1)$ in terms of $f(n)$, $f(n−1)$, and $f(1)$, for $n\geq0$.

*Guess  a formula for $f(n)$ for positive $n$ and prove it by induction.

*Link f(−n) to $f(n)$ .

A: We have:
$$
n=m\implies f(2m)+f(0)=4f(m)\implies f(2m)=4f(m)-f(0),\\
n=0\implies f(m)+f(m)=2f(m)+2f(0)\implies f(0)=0.
$$
Thus, we have
$$
f(2m)=4f(m)\implies 16=f(4)=4f(2)=16f(1)\implies f(1)=1.
$$
We conclude that $f(2^m)=4^m$. At this point, it's suggestive to guess that $f(x)=x^2$ and we can indeed verify that this is sufficient.
To prove necessity, one approach is induction. We have already see that $f(1)=1^2$ and $f(0)=0^2$. Suppose that $f(m)=m^2$ for $1\leq m\leq k$. Then,
$$
f(k+1)=2f(k)+2f(1)-f(k-1)=2k^2+2-(k^2-2k+1)=k^2+2k+1=(k+1)^2.
$$
This proves $f(m)=m^2$ for $m\geq 0$. The induction goes in the negative direction too: suppose that $f(m)=m^2$ for $m\geq k$ for $k\leq 0$, then
$$
f(k-1)=2f(k)+2f(1)-f(k+1)=2k^2+2-(k^2+2k+1)=k^2-2k+1=(k-1)^2.
$$
This completes the proof.
