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Find all function $f:\mathbb{N}\rightarrow\mathbb{N}\cup\{0\}$ satisfying

$$ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = (f(1) + f(2) + \ldots + f(n))^2$$

$\forall n \in \mathbb{N}$.

Thank you, Batominovski and Guy Fabrice..

Is my understanding correct ? Please let me know if there is any mistake.

Substitute $n=1$, $f(1)^3 =f(1)^2$, so $f(1)=0$ or $1$

$f(1)=0$ : substitute $n=2$, $f(2)^3 =f(2)^2$, so $f(2)=0$ or $1$

If $f(2) = 0$, let $l$ be the maximal value such that $f(l) =0$.

so $ f(1)^3 + f(2)^3 + \ldots + f(l+1)^3 = (f(1) + f(2) + \ldots + f(l+1))^2$

then $f(l+1)^3=(f(l+1))^2$ so $f(l+1) = 1$

If $f(2) = 1$, then $0+1+f(3)^3=(f(1)+f(2)+f(3))^2$ , so $f(3) = 0$ or $2$

Since $ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = (f(1) + f(2) + \ldots + f(n))^2$

and $ f(1)^3 + f(2)^3 + \ldots + f(n+1)^3 = (f(1) + f(2) + \ldots + f(n+1))^2$

so $f(n+1)^3 = 2(f(1)+f(2)+\ldots+f(n))f(n+1)+f(n+1))^2$

so $f(n+1)^2-f(n+1)=2(f(1)+f(2)+\ldots+f(n))$

i.e., if $f(n) \not= 0$, then $f(n)=n, \forall n \in \mathbb{N}$

so if $f(x)=0, \forall x \in \mathbb{N}$ then $f(n) = 0, \;\forall n \in \mathbb{N}$

If $f(x)$ be the value such that Max$\{f(1), f(2), \ldots ,f(x)\}=k$ then $f(x+1) = k+1$ or $0$ ---[1]

By induction, let $P(n)$ denotes [1]

Basic step , it's obvious that if $f(1) = 0$, then $f(l_i)=1$, $1\leq l_i\leq x$

so $1+f(l_i+1)^3 = (f(l_i+1) +1)^2$ so $f(l_i+1)=2$ or $0$

Inductive step, suppose that $P(k)$ is true.

$1^3+2^3+\ldots+k^3+f(x+1)^3=(1+2+\ldots+k+f(x+1))^2$,

so $f(x+1)=k+1$ or $0$, so $P(n)$ is true.

The sequence of $f$ is $\underbrace{0\ldots0}_{\text{$k_1$}}1\underbrace{0\ldots0}_{\text{$k_2$}}2\ldots$, where $k_1, k_2, \ldots \in \mathbb{N}$

Answer check :

$ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = 1^3+2^3+\ldots+m^3=\left(\frac{m(m+1)}{2}\right)^2=(f(1) + f(2) + \ldots + f(n))^2$

$\blacksquare$

user403160
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    Such a function is a sequence of integers. Now, every sequence of the form $$0, \dots , 0, 1 , 0, \dots , 0, 2, 0, \dots , 0, 3 , 0, \dots , 0, 4, 0, \dots , 0, 5, \mathrm{ etc.}$$ (where $0, \dots , 0$ denotes finitely many zeroes) satisfies the functional equation. Alternatively, you can truncate such a sequence, and then add infinitely many zeroes. – Crostul Aug 19 '17 at 09:31
  • It seems that you got the idea correctly but you have to write it in a better way. For example you can start claiming that every solution is of the form @Crostul has mentioned (and prove that every function of that form is indeed a solution; that is what you've called "answer check"). Then for proving that claim, you can use induction. But if you want it to be clearer, I think it's better to do it in the way Batominovski has suggested. – Mohsen Shahriari Aug 21 '17 at 17:35
  • @Mohsen Shahriari. Thank you for your suggestion. I'll try. – user403160 Aug 22 '17 at 14:17

2 Answers2

1

Find all function $f:\mathbb{N}\rightarrow\mathbb{N}\cup\{0\}$ satisfying

$$ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = (f(1) + f(2) + \ldots + f(n))^2$$ This at $n+1$ gives

$$ f(1)^3 + f(2)^3 + \ldots + f(n+1)^3 = (f(1) + f(2) + \ldots + f(n+1))^2$$ i.e

$$ (f(1) + f(2) + \ldots + f(n))^2+ f(n+1)^3 = (f(1) + f(2) + \ldots + f(n+1))^2$$ hence we have $$ f(n+1)^3 = (f(1) + f(2) + \ldots + f(n+1))^2-(f(1) + f(2) + \ldots + f(n))^2= 2f(n+1)(f(1) + f(2) + \ldots + f(n)) +f(n+1)^2$$

we then get the relationships $$ f(n+1)^2 -f(n+1)=2(f(1) + f(2) + \ldots + f(n)) $$ Hence $f(n+1)(f(n+1)-1) $ must be an even number . this way you check the values of $f(n)$ by induction.

Guy Fsone
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Hint: Prove by induction on $n$ that, if $f(n)\neq 0$, then

(1) we have $f(n)=M_n+1$, where $M_n:=\max\big\{f(1),f(2),\ldots,f(n-1)\big\}$ for $n>1$, and $M_1:=0$;

(2) for $n>1$, each the numbers $1,2,\ldots,M_n$ appears only once among $f(1),f(2),\ldots,f(n-1)$ in an increasing order (i.e., $i$ appears before $i+1$).

Both statements should be proven simultaneously (i.e., not with separate inductive proofs).

Batominovski
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