abhinavzandubalm wrote: Does There Exist A Function $ f : N \to N $ $ \forall n \ge 2$ $ f(f(n-1)) = f (n+1) - f(n) $ $ f(x) = 0$ for all $ x \in N $ is a solution.

truongtansang89 wrote: $ f(x) = 0$ for all $ x \in N $ is a solution. I think it said $ f : N \to N $ ...

abhinavzandubalm wrote: Does There Exist A Function $ f : N \to N $ $ \forall n \ge 2$ $ f(f(n-1)) = f (n+1) - f(n) $ $f(n+1)-f(n)\ge 1$ $\forall n\ge 2$ and so $f(n)\ge f(2)+n-2\ge n-1$ $\forall n\ge 3$ So : $\forall n\ge 5$ : $f(n-1)\ge n-2\ge 3$ and so $f(f(n-1))\ge f(n-1)-1\ge n-3$ and so $f(n+1)-f(n)\ge n-3$ Adding these lines for $n=5,6,7$, we get $f(8)-f(5)\ge 9$ and so $f(8)\ge 10$. Let then $a=f(8)\ge 10$ Adding then the lines $f(f(n-1))=f(n+1)-f(n)$ for $n=2\to a-1$, we get $f(a)-f(2)=\sum_{k=1}^{a-2}f(f(k))$ And, since $a\ge 10$, we can write $f(a)-f(2)=f(f(8))+\sum_{k=1,k\ne 8}^{a-2}f(f(k))$ and so, since $f(8)=a$, this becomes : $-f(2)=\sum_{k=1,k\ne 8}^{a-2}f(f(k))$, clearly impossible since $LHS<0$ while $RHS>0$ And so no solution.

Let $f(n_{0})$ has the minimum value of the function (because $f : {\mathbb{N}}\to{\mathbb{N}}$ we can suppose that). It easily can be considered that ${1}\leqslant{f(n_{0})}\leqslant{f(n)}$. Now we have two situation: 1)${n_{0}}\geqslant{3}$: We have $f(f(n-1)) +f (n) = f (n+1)$ and by putting ${n}\to{n_{0}-1}$ we have: $f(f(n_{0}-2)) + f(n_{0}-1) = f(n_{0})$ but we know that ${f(f(n_{0}-2))}\geqslant{f(n_{0})}$ and ${f(n_{0}-1)}\geqslant{f(n_{0})}$ $\Longrightarrow$ ${f(n_{0}) = f(f(n_{0}-2)) + f(n_{0}-1)}\geqslant{2f(n_{0})}$ $\Longrightarrow$ ${f(n_{0})}\leqslant{0}$ $\Longrightarrow$ there not exist such a function. 2)$f(2)$ has the minimum value: Let $f(n-1) = a$ we have $f(a) = f(n+1) - f(n)$ and we have ${f(a)}\in{\mathbb{N}}$ so ${f(n+1) - f(n)}\geqslant{1}$ $\Longrightarrow$ ${f(n+1)}$>${f(n)}$ $\Longrightarrow$ ${f(2)}$>${f(1)}$ but we supposed $f(2)$ has the minimum value. $\Longrightarrow$ there not exist such a function.

Sansa wrote: Let $f(n_{0})$ has the minimum value of the function (because $f : {\mathbb{N}}\to{\mathbb{N}}$ we can suppose that). It easily can be considered that ${1}\leqslant{f(n_{0})}\leqslant{f(n)}$. Now we have two situation: 1)${n_{0}}\geqslant{3}$: We have $f(f(n-1)) +f (n) = f (n+1)$ and by putting ${n}\to{n_{0}-1}$ we have: $f(f(n_{0}-2)) + f(n_{0}-1) = f(n_{0})$ but we know that ${f(f(n_{0}-2))}\geqslant{f(n_{0})}$ and ${f(n_{0}-1)}\geqslant{f(n_{0})}$ $\Longrightarrow$ ${f(n_{0}) = f(f(n_{0}-2)) + f(n_{0}-1)}\geqslant{2f(n_{0})}$ $\Longrightarrow$ ${f(n_{0})}\leqslant{0}$ $\Longrightarrow$ there not exist such a function. 2)$f(2)$ has the minimum value: Let $f(n-1) = a$ we have $f(a) = f(n+1) - f(n)$ and we have ${f(a)}\in{\mathbb{N}}$ so ${f(n+1) - f(n)}\geqslant{1}$ $\Longrightarrow$ ${f(n+1)}$>${f(n)}$ $\Longrightarrow$ ${f(2)}$>${f(1)}$ but we supposed $f(2)$ has the minimum value. $\Longrightarrow$ there not exist such a function. Two remarks : 1) If $n_0<3$, we have two cases : $n_0=2$ and $n_0=1$. You looked only at $n_0=2$ b) in 2), you say $f(n+1)>f(n)$ but you forget that this is only true for $n\ge 2$, and so you cant apply $n=1$ to this inequality.

pco wrote: Two remarks : 1) If $n_0<3$, we have two cases : $n_0=2$ and $n_0=1$. You looked only at $n_0=2$ b) in 2), you say $f(n+1)>f(n)$ but you forget that this is only true for $n\ge 2$, and so you cant apply $n=1$ to this inequality. For the first one in the problem we have $\forall{n}\geqslant{2}$ so it doesn't need to check $n=1$. But for the second one I have no idea. do anybody have any idea about that???