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

Question

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

Answer 1

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.

Answer 2

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

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

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 $f(n)=0$

Answer 3

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 $f(n) = 0$.

Answer 4

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. $f(n) = 2n$