Find all $f: \mathbb{N} \rightarrow \mathbb{N}$ which satisfy $ f(m-n+f(n))=f(m)+f(n) $ Question -

Find all $f: \mathbb{N} \rightarrow \mathbb{N}$ which satisfy the relation -

$$
f(m-n+f(n))=f(m)+f(n)
$$
for all $m, n \in \mathbb{N}$ where  $N={1,2,3....}$

solution -

Observe $f(n) \geq n .$ Consider $F(n)=f(n)-n .$ Show that $F$ satisfies
$$
F(F(n)+m)=F(m)+n
$$
Using this, conclude that $F(1)=1$

and $F(n+1)=F(n)+$
$F(1)$ for all $n \geq 1 .$ Thus $F(n)=n F(1) .$ It follows that $F(n)=n$ and $f(n)=2 n$

Now i did not understand how they proved $F(1)=1$ ???

Any help will be appreciated 

thankyou
I don't understand how he proved it, but here's my solution:
$$f(m-n+f(n))=f(m)+f(n) \implies P(m,n)$$
$$P(0,0) \implies f(f(0))=2f(0)$$
Let $f(0)=c$  giving $f(c)=2c$, so
$$P(m,c)\implies f(m+c)=f(m)+2c $$
$$P(m,0) \implies f(m+c)=f(m)+c $$
So, $c=2c \implies c=0 \implies f(0)=0$
$$P(n,n) \implies f(f(n))=2f(n)$$
Let $f(1)=k$
$$P(1,1) \implies f(k)=2k$$
$$P(m,1) \implies f(m-1+k)=f(m)+k$$
$$P(m-1,k) \implies f(m-1-k+2k)=f(m-1+k)=f(m-1)+2k$$
So, combining the previous 2 equations, we get
$$f(m)+k=f(m-1)+2k \Leftrightarrow f(m)-f(m-1)=k \implies H(m) $$
$$H(2)\implies f(2)-f(1)=k \implies f(2)=2k$$
$$H(3)\implies f(3)-f(2)=k \implies f(3)=k+f(2)=k+2k=3k$$
and so on, by simple induction, and the fact that $f(0)=0$
$$f(x)=kx \text{ } \forall \text{ } x \in \mathbb{N}$$
Substituting in the original equation,
$$km-kn+{k^2}n=km+kn \Leftrightarrow {k^2}n=2kn \Leftrightarrow k^2=2k \Leftrightarrow k^2-2k=0 $$
$$ \Leftrightarrow k(k-2)=0$$
This gives $$(1) \text{  } k=0 \implies f(x)=0 \text{ } \forall \text{ } x \in \mathbb{N}$$ 
$$(2) \text{   } k=2 \implies f(x)=2x \text{ } \forall \text{ } x \in \mathbb{N}$$
the only 2 solutions $\Box$.

Note that when $0$ doesn't belong to $\mathbb{N}$, you can discard all the steps of relating to $0$ and the proof would still hold true.
let $F(n) = f(n) - n$. $F$ verify the functional equation:
$$F(m + F(n)) = F(m) + n$$
By choosing $m=0$, we have $F(F(n)) = F(0) + n$, so $F$ is injective.
By letting $n=0$, we have $F(m+F(0)) = F(m)$ and by injectivity we deduce
$F(0) = 0$.
Now we have $F(F(n)) = n$ so now $F$ is surjective. Let $n_0$ s.t. $F(n_0) = 1$.
We have now:
$$F(m + 1) = F(m) + n_0$$
so:
$$F(m) = n_0 m$$
by inserting this into last equation we deduce $n_0^2 = 1$ so $n_0 = 1$.
I have prouved $F(1) = 1$. Now by induction we deduce $F(n) = n$, i.e. $f(n) = 2n$