Obviously $f$ is unbounded. 1) $f(m)\geq m,\forall m$. Suppose $f(x)<x$. From monotonicity, we have $f(x)f(n)+x>xf(n)$ or $f(n)<\frac{x}{x-f(x)},\forall n$. Absurd! 2) $f(1)=1$ Suppose $a=f(1)\geq2$. $P(m,1):f(af(m)+m)=f(am)+f(m)$. Using 1), we have, for all $m$, $f(am)\geq (a-1)f(m)+m$ $P(1,n):f(af(n)+1)=f(f(n))+a$ implies $f(f(n))+a\geq f(af(n))\geq (a-1)f(f(n))+f(n)$, where the last step is from the inequality above by replacing $m$ with $f(n)$. Since $a\geq2$, we have $a\geq (a-2)f(f(n))+f(n)\geq f(n),\forall n$. But $f$ is unbounded. Absurd! 3) $f(n)=n,\forall n$. $P(1,n):f(f(n)+1)=f(f(n))+1$. We can show $f(n)=n$ by induction.

xxp2000 wrote: Obviously $f$ is unbounded. 1) $f(m)\geq m,\forall m$. Suppose $f(x)<x$. From monotonicity, we have $f(x)f(n)+x>xf(n)$ or $f(n)<\frac{x}{x-f(x)},\forall n$. Absurd! 2) $f(1)=1$ Suppose $a=f(1)\geq2$. $P(m,1):f(af(m)+m)=f(am)+f(m)$. Using 1), we have, for all $m$, $f(am)\geq (a-1)f(m)+m$ $P(1,n):f(af(n)+1)=f(f(n))+a$ implies $f(f(n))+a\geq f(af(n))\geq (a-1)f(f(n))+f(n)$, where the last step is from the inequality above by replacing $m$ with $f(n)$. Since $a\geq2$, we have $a\geq (a-2)f(f(n))+f(n)\geq f(n),\forall n$. But $f$ is unbounded. Absurd! 3) $f(n)=n,\forall n$. $P(1,n):f(f(n)+1)=f(f(n))+1$. We can show $f(n)=n$ by induction. From where did you get that f u=is unbounded ... I beleive that that the solution above is not correct because the result $f(m)\geq m is not atteined unless f is strictly increasing which is not the case . And by the way f(m)=0 is also a solution .

blackman wrote: From where did you get that f u=is unbounded ... I beleive that that the solution above is not correct because the result $f(m)\geq m$ is not atteined unless f is strictly increasing which is not the case . And by the way f(m)=0 is also a solution . I think you missed the point $f : \mathbb N \to \mathbb N$.

Mahi wrote: blackman wrote: From where did you get that f u=is unbounded ... I beleive that that the solution above is not correct because the result $f(m)\geq m$ is not atteined unless f is strictly increasing which is not the case . And by the way f(m)=0 is also a solution . I think you missed the point $f : \mathbb N \to \mathbb N$. In my country 0 is included in $ \mathbb N $ but that doesn't change anything the problem is still unsolved

blackman wrote: Mahi wrote: I think you missed the point $f : \mathbb N \to \mathbb N$. In my country 0 is included in $ \mathbb N $ but that doesn't change anything the problem is still unsolved How does it not? $ f(f(m)f(n)+m)=f(mf(n))+f(m) $ and the set $\mathbb N$ does not contain $0$ directly implies $ f(f(m)f(n)+m) > f(m)$ for all $m,n$, i.e $f$ is unbounded.