## zmudz one year ago 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 $$f(n)$$.

1. freckles

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

2. Loser66

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

3. freckles

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

4. freckles

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

5. Loser66

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

6. freckles

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

7. Loser66

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

8. Loser66

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?

9. Loser66

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