$f:\{1,2,3,\ldots\}\to\{1,2,3,\ldots\},$
such
that for all $n=1,2,3,\ldots$
the sum $f(1)+f(2)+\cdots+f(n)$
is equal to
a perfect cube that is less than or equal to $n^3.$

What values can
$f(1)$
take? 
What value must
$f(1) + f(2)$
take? 
Find an expression for
$\sum_{i=1}^{n}{f(i)}$
in terms of$n.$

Prove your hypothesis using induction.

What is the value of
$f(n)$
if$\sum_{i=1}^{n}{f(i)} = n^3$
and$\sum_{i=1}^{n1}{f(i)} = (n1)^3?$

By trying small values of
$n,$
we find that$f(1) \leq 1^3,$
which means that necessarily$f(1)=1.$
The next case gives us$f(1)+f(2) = 8.$
This can give us an intuition that the function is unique:$f(n) = n^3  (n1)^3.$
To prove this formally we use induction.Let
$\Phi(n)$
be the induction hypothesis “If$S(i)=\sum_{j=1}^i{f(j)}$
is a perfect cube less than or equal to$i^3$
for all$i \leq n$
then$S(i) = i^3.$
”Base case:
$f(1) = 1$
as it is the only perfect cube less than or equal to$1^3,$
so$\Phi(1)$
is true.Inductive case: Assuming
$\Phi(k)$
is true, we can prove$\Phi(k+1)$
as follows: Suppose$S(i)$
is a perfect cube less than or equal to$i^3$
for all$i \leq k+1.$
Using the induction hypothesis$\Phi(k)$
,$S(i) = i^3$
for all$i \leq k.$
Since$S(k) = k^3$
and$S(k+1) \leq (k+1)^3,$
$S(k+1)$
must equal$(k+1)^3$
as there are no cubes between$k^3$
and$(k+1)^3.$
Therefore$\Phi(k+1)$
is true.By induction,
$S(n) = n^3$
for all$n,$
and so$f(n) = \sum_{j=1}^{n}{f(j)}  \sum_{j=1}^{n1}{f(j)}$
$= n^3  (n1)^3.$