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

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

Answer 1

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.

Answer 2

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, $$f(x)=x$$

Answer 3

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 $x \in \mathbb{N}$ $\Box$.

Answer 4

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 $n \in \mathbb{Z^+}$ $\Box$.

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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 $n \geq 3$.