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 $f(n)$.

Question

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 $f(n)$.

freckles · Answer

so far I have come up with a value for f(0)
and I came up with the function f is even:
how I did this:
...
m+n=4
m-n=4
------
2m=8
m=4 when n=0
so 
$$f(4+0)+f(4-0)=2f(4)+2f(0) \ f(4)+f(4)=2f(4)+2 f(0) \ 0=2 f(0) \ f(0)=0 \ \text{ now I set } m=0 \ \text{ and so we have } \ f(0+n)+f(0-n)=2f(0)+2f(n) \ f(n)+f(-n)=0+2f(n) \ f(-n)=f(n) \text{ which says } f \text{ is even }$$
so maybe we can use these two facts somehow
or maybe not

loser66 · Answer

me too, i got what you got but didn't see the link of f(4) =16 to the problem :(

freckles · Answer

well we can come with f(8) using that fact with the already found stuff

freckles · Answer

$$f(4+4)+f(4-4)=2f(4)+2f(4) \ f(8)+f(0)=4 f(4) \ f(8)+0=4(16) \ f(8)=64$$

loser66 · Answer

It seems the function is f(x) = x^2

freckles · Answer

lol let's test
$$(m+n)^2+(m-n)^2=2m^2+2n^2  \ m^2+2nm+n^2+m^2-2nm+n^2=2m^2+2n^2 \ $$
the equation holds  for f(x)=x^2

loser66 · Answer

You are well organized. I just look at f(0) =0 , f(4) =16, f(8) =64 to conclude that f(x) =x^2

loser66 · Answer

oh, we are done. hahaha. it asked us to find f(n) and we get f(x) = x^2 hence f(n) = n^2 
right?

loser66 · Answer

yeah!! I think so, with your argument!!! we prove that f(n) = n^2 . And by your testing, we have the expression hold for any m, n in Z