# 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

• Is this function well defined? How do you know that $$m-n+f(n)>0$$ is true for all $m$ and $n$? Apr 22, 2020 at 10:12

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.

• i have clearly mention in question that N=1,2,3.....!!! Apr 22, 2020 at 3:19
• @User88463 I edited that: note that you can discard all the steps relating to zero and the proof would still hold true for $f(x)=2x$ only Apr 22, 2020 at 23:26

$$f(0)+f(0)=f(f(0))$$

$$f(0)+f(f(0))=f\bigg(0-f(0)+f(f(0))\bigg)=f\bigg(0-f(0)+f(0)+f(0)\bigg)=f(f(0))$$

Subtracting $$f(f(0))$$ each side we get $$f(0)=0$$

$$f(0)+f(n)=f(n)=f(0-n+f(n))\iff f(n)=f(f(n)-n)$$

$$f(m)+f(n)=f(m)+f(f(n)-n)=f\bigg(m-f(n)+n+f(f(n)-n)\bigg)=f\bigg(m-f(n)+n+f(n)\bigg)=f(m+n)$$

Thus $$f$$ is linear with $$f(0)=0$$ so $$f(n)=an$$.

Reporting in equation $$f(n)+f(n)=f(f(n))\iff 2an=a^2n\iff a=2, a=0$$

Thus $$f(n)=2n$$ or $$f(n)=0$$

• but N does not contain 0 !!! Apr 20, 2020 at 12:17

I cannot comment, so I will do it here for @zwim: you are using the functional equation for m negative (in $$f(0)+f(n)=f(n)=f(0-n+f(n))\iff f(n)=f(f(n)-n)$$)

My solution: Let us consider $$h(n) = f(n) - 2n$$. $$h$$ verify the functional equation: $$h(m+n+h(n)) = h(m) - h(n), \text{ where } h: \mathbb{N}_{>0} \to \mathbb{Z}$$ we extend $$h$$ to $$\mathbb{N}$$ by $$h(0) = 0$$. In particular we have: $$h(2n + h(n)) = 0$$. So if there exists $$n_0 > 0$$ s.t. $$h(n_0) = 0$$, we have $$h(m + n_0) = h(m)$$ so $$h$$ is $$n_0$$-periodic. Let $$0 < i < n_0$$, we put $$h_i = h(i)$$. We have $$h(n+i+h_i) = h(n) - h_i$$ and by induction: $$h(n+k(i+h_i)) = h(n) - kh_i$$. By choosing $$k = n_0$$, we conclude $$h_i = 0$$, so by periodicity we have $$h(n) = 0$$, so $$f(n) = 2n$$. Otherwise, if there isn't $$n_0>0$$ s.t. $$h(n_0) = 0$$, then from $$h(2n+h(n)) = 0$$ we conclude $$f(n) = 0$$.

• $h$ is from $\mathbb{N} \to \mathbb{Z}$. If you exclude $0$ from $\mathbb{N}$ (which is $\mathbb{N}_{>0}$) then the only solution is $f(n) = 2n$ Apr 20, 2020 at 14:47
• ok, but can you pls tell me how F(1)=1 in my original doubt that actually i have....i just wanted to know that how they write $F(1)=1$ after substituting $F(n)=f(n)-n$ ? if this doubt will be cleared then i have very easy solution for this problem... Apr 20, 2020 at 15:39

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. $$f(n) = 2n$$

• are you serious ??? i have told many time in this post that N=1,2,3.... Apr 23, 2020 at 15:51