<s>[INST] Sum of $k^\text{th}$ Powers of the First $n$ Natural Numbers is a Perfect Square. 
Question:
If for some $k$ and $\forall$ $n>2017, n\in \mathbb{N}$, there exist an $x\in \mathbb{N}$ such that
$$1^k+2^k+3^k+\cdots+n^k=x^2$$
then $k=3$.


Minor clarification:
The question says : $\forall n$. Now, we know that $1^3+2^3+..+n^3=\left(\frac{n(n+1)}{2}\right)^2$.
But, for all $n\in \mathbb{N}$ and $n>2017$, you won't get some positive integer $x$ such that $1^k+2^k+\cdots+n^k=x^2$. You will get such $x$ only when $k=3$. I don't think there's any particular significance of $2017$ here.

How do we even start this? It looks so weird.
How to proceed after that?
My idea is to use:
$$\sum_{x=0}^n (x+1)^{k+1}-x^{k+1}=n^k \left({k \choose 0} + {k-1 \choose 0} + \cdots + {0 \choose 0}\right) + n^{k-1} \left({k \choose 1} + {k-1 \choose 1} + \cdots + {1 \choose 1}\right) + \cdots + n^{k-k}\left(k\choose k \right)$$.
The $\text {LHS}$ equals $\left(n+1\right)^{k+1}-1$. And we have the $1^k+2^k+3^k+\cdots+n^k$ term in the $\text {RHS}$
But I don't know if  his helps to show that the sum of the $k^\text {th}$ powers will be a non perfect sqaure $\forall n>2017$
We can also do something like this-
If $1^k+2^k+\cdots+n^k=x^2$, then $x^2+(n+1)^k=y^2$. And then, $y^2+(n+2)^k=z^2$ and so on.
First of all, is $a^2+n^k=b^2$ possible for all $n>2017$ (though $2017$ hasn't got any special significance) and for a particular $k$ (and needless to mention,  $a,b$ are not constants since $n$ is variable)?
But, how do we show that it won't hold unless $k=3$?
 [/INST] Answer posted by OP (solved by some AoPSer)
$f(n)=1^m+...n^m$, obvious, that $m$ is odd.
$f(n)<n^{m+1}$
Let $p$ -big prime.
$f(p-1)=x^2,f(p)=y^2 \to p^m=y^2-x^2 \to m=a+b;b>a,x=\frac{p^b-p^a}{2}$
$4x^2=(p^b-p^a)^2=4f(p-1)<4(p-1)^{m+1}$
$p^{2b}+p^{2a}-2p^{a+b}<4p^{a+b}(p-1)$
$p^{2b-2a}+1-2p^{b-a}<4p^{b-a}(p-1)$
Let $p^{b-a}=q$
$q^2-q(4p-2)+1<0 \to q<4p-2<p^2 \to b-a=1$
So if $m=2k+1$ then $x=\frac{p^k(p-1)}{2}$ and so $f(p-1)=(\frac{p^k(p-1)}{2})^2$
We can prove, that for $k\geq 2$ is true, that $f(n)<\frac{(n+1)^{2k+2}}{6}$
$(\frac{p^k(p-1)}{2})^2<\frac{p^{2k+2}}{6}$
$3p^{2k}(p-1)^2<2p^{2k+2}$
$p^2-6p+3<0$ -wrong for big $p$
So $k=0,1 \to m=1,3$ and easy to check, that $m=3$ is answer
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