Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ which satisfy $ f\left(m^{2}+m n\right)=f(m)^{2}+f(m) f(n) $ Question -
Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ which satisfy the equation
$$
f\left(m^{2}+m n\right)=f(m)^{2}+f(m) f(n)
$$
for all natural numbers $m, n$
by putting $m=1$ and $f(1)=k$ we get 
$f(n+1)=k^2 + kf(n)$
then hint says use $3^2 + 3.1 = 2^2 +2.4$ to get polynomial relation for k..
i am not getting how to use this hint ...i think i am missing some very easy tricks to get at this which i have not lernt yet ...
any help will be appreciated
thankyou
 A: Putting $n=1$ in the condition for $f$ gives
$$f(m^2+m)=f(m)^2+kf(m)$$
Now set $m=3$. By the hint, we have
$$f(3^2+3)=f(2^2+2\cdot 4)=f(2)^2+f(2)f(4)$$
which gives us the condition
$$f(3)^2+kf(3)=f(2)^2+f(2)f(4)$$
You should be able to find $f(2)$,$f(3)$ and $f(4)$ in terms of $k$ by using your condition for $f(n+1)$.
Hope this helps.
A: I just wanted to give a full solution to the problem in case anyone needed it. (This solution only works if $0 \in \mathbb{N}$, and I've posted another solution if $0 \notin \mathbb{N}$)
$$f(m^2+mn)=f(m)^2+f(m)f(n) \implies P(m,n)$$
$$P(0,0) \implies f(0)=0$$
$$P(m,0) \implies f(m^2)=f(m)^2 \tag{1}$$
At this point we can switch squares inside and outside as we like, 
$$P(m,m) \implies f(2m^2)=2f(m)^2=2f(m^2) \tag{2}$$
Let $f(1)=k$,
$$P(1,1) \implies f(2)=2k^2$$
and by $(2)$
$$f(2)=f(2\cdot 1^2)=2f(1^2)=2k$$
$$\implies 2k=2k^2$$
Case 1: $k=0$
$$P(1,m) \implies f(m+1)=0 \implies f(n)=0 \text{   }\forall \text{  } n \in \mathbb{N}$$
Case 2: $k=1$
$$P(1,m) \implies f(m+1)-f(m)=1 \tag{3}$$
$$P(1,1) \implies f(2)=2$$
$$P(1,2) \implies f(3)-f(2)=1 \implies f(3)=3$$
and by simple induction, and the fact that $f(0)=0$
$$f(x)=x$$
for all $x \in \mathbb{N}$ $\Box$.
A: Now this is my full solution if $0 \notin \mathbb{N}$
$$f(m^2+mn)=f(m)^2+f(m)f(n) \implies P(m,n)$$
Let $f(1)=k$
$$P(1,m) \implies f(m+1)=kf(m)+k^2$$
$$P(1,1) \implies f(2)=kf(1)+k^2=2k^2$$
$$P(1,2) \implies f(3)=kf(2)+k^2=k(2k^2)+k^2=2k^3+k^2$$
$$P(1,3) \implies f(4)=kf(3)+k^2=k(2k^3+k^2)+k^2=2k^4+k^3+k^2$$
and so on, by induction,
$$f(n)=2k^n+k^{n-1}+k^{n-2}+\dots+k^2 \tag{1}$$
for $n \geq3$.
Now $n=6$ in $(1)$ gives
$$f(6)=2k^6+k^5+k^4+k^3+k^2$$
while
$$P(2,1) \implies f(6)=f(2)^2+kf(2)=(2k^2)^2+k(2k^2)=4k^4+2k^3$$
So,
$$2k^6+k^5+k^4+k^3+k^2=4k^4+2k^3 $$
$$\Leftrightarrow 2k^6+k^5-3k^4-k^3+k^2=0$$
and since $k \neq 0$ we can divide by $k^2$:
$$2k^4+k^3-3k^2-k+1=0$$
We can easily find by the rational roots theorem that $k=1$ is the only possible root, and checking back, it works.
Thus, by $(1)$,
$$f(n)=2(1^{n})+\underbrace{1^{n-1}+\dots+1^{2}}_\text{$(n-2)$ terms}=2+n-2=n $$
for all $n \geq 3$. Since $f(1)=k=1$ and $f(2)=2k^2=2$, we can extend the definition:
$$f(n)=n$$
for all $n \in \mathbb{Z^+}$ $\Box$.


I'll make the induction proof of $(1)$ here. Our base case $n=3$ works, now 
$$f(n)=2k^n+k^{n-1}+k^{n-2}+\dots+k^2$$
and 
$$P(1,m) \implies f(m+1)=kf(m)+k^2$$
So
$$f(n+1)=kf(n)+k^2=k(2k^n+k^{n-1}+k^{n-2}+\dots+k^2)+k^2$$
$$=2k^{n+1}+k^{n}+k^{n-1}+\dots+k^3+k^2$$
So, indeed statement $(1)$ is true for all $n \geq 3$.
A: The function which you are searching for has the following properties.
$$f(x+y) = f(x)+f(y)$$
and
$$f(x*y) = f(x)*f(y)$$
The only function which fits into these constraints is identity function, i.e,
$$f(x)=x$$
