Just prove $f(n)=n$ by induction over $n$.

$n=1$ is true, but how to prove with $n=k+1$?

IMOStarter wrote: $n=1$ is true, but how to prove with $n=k+1$? Replace $f(k)=k, \forall k\leq n $ to $(f(1))^3+(f(2))^3+...+(f(n+1))^3=(f(1)+f(2)+...+f(n+1))^2$, , slove equation to find $f(n+1)$

Here is the solution from my problem book, but it is in Hungarian. But the proof can be understanded, because these there are formulas. Quote: Jelölje $f(1)+f(2)+f(3)+...+f(n)=g(n)$, tehát ${{f}^{3}}(1)+{{f}^{3}}(2)+{{f}^{3}}(3)+...+{{f}^{3}}(n)=g{{(n)}^{2}}$. Cseréljük ki $n$ -et $(n+1)$ -re, és azt kapjuk, hogy ${{f}^{3}}(1)+{{f}^{3}}(2)+{{f}^{3}}(3)+...+{{f}^{3}}(n)+{{f}^{3}}(n+1)={{\left( g(n)+f(n+1) \right)}^{2}}$, és elvégezve a négyzetre emelést, behelyettesítve az eredeti egyenletet azt kapjuk, hogy ${{f}^{2}}(n+1)=2g(n)+f(n+1)$ vagyis ${{f}^{2}}(n+1)=2(f(1)+f(2)+...+f(n))+f(n+1)$, és ismét cseréljük ki $n$ -et $(n+1)$ -re, és azt kapjuk, hogy ${{f}^{2}}(n+2)=2(f(1)+f(2)+...+f(n)+f(n+1))+f(n+2)$és ebből kivonva az előbbi egyenletet azt kapjuk, hogy ${{f}^{2}}(n+2)-{{f}^{2}}(n+1)=f(n+2)+f(n+1)$ vagyis $f(n+2)-f(n+1)=1$ , ezért $\sum\limits_{k=0}^{n-2}{\left( f(n+2)-f(n+1) \right)}=n-1$ ahonnan $f(n)=n$ és ez valóban teljesíti is az adott függvényegyenletet.

We use strong induction. Base case. $n=1\Rightarrow f(1)^3=f(1)\Rightarrow f(1)=1$ Induction hypothesis. $f(n+1)=n+1$ if $f(k)=k\forall k\le n$. Induction step. The assertion becomes $1^3+2^3+\ldots+n^3+f(n+1)^3=(1+2+\ldots+n+f(n+1))^2$, or $f(n+1)^3=2f(n+1)(1+\ldots+n)+f(n+1)^2$. This is simply $f(n+1)^2-f(n+1)-n(n+1)=0$. As a quadratic in $f(n+1)$, we get $f(n+1)=\frac{1\pm(2n+1)}2$, so either $f(n+1)=-n$ (absurd) or $f(n+1)=n+1$, which completes the proof. $\square$

What happens if we change the condition into: Find all function $f:\mathbb{N}^*\rightarrow \mathbb{N}$ that satisfy: $$(f(1))^3+(f(2))^3+...+(f(n))^3=(f(1)+f(2)+...+f(n))^2$$ Here is my solution Suppose there are some $k$ (maybe infinitely many) such that $f(k)=0$. Remove all of them and rearrange the sequence $1,2,3,...$ as ${a_1}<{a_2}<{a_3}<....$ which $f({a_i}) \ne 0$ for every $i$ Then by induction we can easily get $f(a_i)=i$ for every $i$ Then the solution is $f(a_i)=i$,$f(m)=0$ for all $ m \ne {a_j}\forall j $

SagiriIzumi wrote: IMOStarter wrote: $n=1$ is true, but how to prove with $n=k+1$? Replace $f(k)=k, \forall k\leq n $ to $(f(1))^3+(f(2))^3+...+(f(n+1))^3=(f(1)+f(2)+...+f(n+1))^2$, , slove equation to find $f(n+1)$ how to type these math characters?

jasperE3 wrote: We use strong induction. Base case. $n=1\Rightarrow f(1)^3=f(1)\Rightarrow f(1)=1$ Induction hypothesis. $f(n+1)=n+1$ if $f(k)=k\forall k\le n$. Induction step. The assertion becomes $1^3+2^3+\ldots+n^3+f(n+1)^3=(1+2+\ldots+n+f(n+1))^2$, or $f(n+1)^3=2f(n+1)(1+\ldots+n)+f(n+1)^2$. This is simply $f(n+1)^2-f(n+1)-n(n+1)=0$. As a quadratic in $f(n+1)$, we get $f(n+1)\frac{1\pm(2n+1)}2$, so either $f(n+1)=-n$ (absurd) or $f(n+1)=n+1$, which completes the proof. $\square$ There exist light induction?

StarSky07 wrote: SagiriIzumi wrote: IMOStarter wrote: $n=1$ is true, but how to prove with $n=k+1$? Replace $f(k)=k, \forall k\leq n $ to $(f(1))^3+(f(2))^3+...+(f(n+1))^3=(f(1)+f(2)+...+f(n+1))^2$, , slove equation to find $f(n+1)$ how to type these math characters? It is $\LaTeX$. You can learn them here.

Thank you!