How to find $f(2013)$ if $f(5)=45$ and $f(m)+f(n)= f(m+n)$ for all $m,n\in\mathbb N$?

Question

$f: \mathbb N\to\mathbb N$, $f(m)+f(n)=f(m+n)$ for all $m,n\in\mathbb N$, and $f(5)=45$. Find $f(2013)$.

I messed up my original posting, its fixed now. I changed $m+ n$ to $f(m+n)$.

Answer 1

The question now says that $f$ is linear (which is very different!).

Hint: $kf(1)=f(1)+\ldots +f(1)=f(1+\cdots +1)=f(k)$ and so if $f(5)=45$ then $5f(1)=f(5)=45$. Can you find $f(1)$ from this?

---

(This was the contradiction that appears using the previous, unedited question.)

No such $f$ exists. If $\mathbb{N}$ includes $0$ then $f(m)=m$ for all $m\in\mathbb{N}$ as shown in the comments above.

If $\mathbb{N}$ does not contain $0$ then $2f(1)=f(1)+f(1)=1+1=2$ and so $f(1)=1$. Now, $f(m)+1=f(m)+f(1)=m+1$ and so $f(m)=m$ for all $m\in\mathbb{N}$. This contradicts $f(5)=45$.

Answer 2

If $f(n+m)=f(n)+f(m)$, you notice that $$ f(2)=f(1+1)=2f(1), $$ $$ f(3)=f(2+1)=f(2)+f(1)=\cdots $$ With an induction argument you should be able to show that $f(2013)=18117$.

Answer 3

Since we know $f(5)$, we can find $f(10)$, $f(15)$, and so on - that is, it's easy to evaluate $f(5k)$ for any $k$ by a brief induction argument.

So the question is reduced to computing $f(3)$, since $2013 = 5(...) + 3$. But note that

$$3 f(5) = f(15) = f(3 + 3 + 3 + 3 + 3) = f(3) + f(3) + f(3) + f(3) + f(3) = 5 f(3)$$

Answer 4

We can prove by induction that if $x_i \in \mathbb{N}$, $$f(\sum_{i=1}^{n}x_i)=\sum_{i=1}^{n}f(x_i)$$.

Hence $f(5) = 5f(1)=45$. Thus $f(1)=9$.

Fix $x_i=1$ for all $1\leq i\leq n$ and $n=2013$ thus,

$$f(2013) = 2013\times 9$$