Is the following result trivial?

[JoanBraidy]

I proved there's infinitely many n such that S_n has an element of order n^2

 

[Martin Rattigan]

Trivialish.

3^2+4^2+5^2<3.4.5.

Suppose 3^{2i}+4^{2i}+5^{2i}<3^i.4^i.5^i for i<n, then
3^{2(i+1)}+4^{2(i+1)}+5^{2(i+1)}<25(3^i.4^i.5^i)<3^{i+1}4^{i+1}5^{i+1}, hence 3^{2n}+4^{2n}+5^{2n}<3^n4^n5^n by induction.

It follows that S_{3^n4^n5^n} has an element of order (3^n4^n5^n)^2 for all n\in \mathbb{N}.

Similarly 5^{3n}+7^{3n}+9^{3n}+11^{3n}<5^n7^n9^n11^n, so there are an infinite number of k such that S_k contains an element of order k^3.

I think its probably true that there are an infinite number of n such that S_n contains an element of order n^k for any k\in \mathbb{N}.

 

[JoanBraidy]

Trivialish.

3^2+4^2+5^2<3.4.5.

Suppose 3^{2i}+4^{2i}+5^{2i}<3^i.4^i.5^i for i<n, then
3^{2(i+1)}+4^{2(i+1)}+5^{2(i+1)}<25(3^i.4^i.5^i)<3^{i+1}4^{i+1}5^{i+1}, hence 3^{2n}+4^{2n}+5^{2n}<3^n4^n5^n by induction.

It follows that S_{3^n4^n5^n} has an element of order (3^n4^n5^n)^2 for all n\in \mathbb{N}.

Similarly 5^{3n}+7^{3n}+9^{3n}+11^{3n}<5^n7^n9^n11^n, so there are an infinite number of k such that S_k contains an element of order k^3.

I think its probably true that there are an infinite number of n such that S_n contains an element of order n^k for any k\in \mathbb{N}.

similar to what I did thanks

if I proved the general case, do you think that would be trivial?

 

[Martin Rattigan]

There are straightforward proofs quoting Bertrand's postulate or the prime number theorem or other theorems on prime distribution. Most people wouldn't call the quoted theorems trivial, so you'd probably want to distinguish between a trivial result and a trivial proof.

I've appended a proof (I tried to put this under a spoiler but it doesn't seem to work with latexed code - you'll just have to avoid looking if you're still doing your own proof). Whether you'd call the proof trivial is really subjective.

For k=0 it is trivial that S_n has an element of order n^k.

For any k\in\mathbb{N},k\geq1

p_n^k+p_{n+1}^k+\dots+p_{n+r}^k+\dots+p_{n+k}^k
\leq p_n^k(1+2^k+2^{2k}+\dots+2^{rk}+\dots+2^{k^2})\text{ (By Bertrand's postulate)}
=p_n^k(2^{k(k+1)}-1)/(2^k-1)

If n_k is chosen so that
p_{n_k+k}\geq (2^{k(k+1)}-1)/(2^k-1)
then for any n\geq n_k

p_n^k(2^{k(k+1)}-1)/(2^k-1)
\leq p_n^k(p_{n+k})
\leq p_np_{n+1}\dots p_{n+r}\dots p_{n+k-1}p_{n+k}

Hence for any such n, S_{p_np_{n+1}\dots p_{n+r}\dots p_{n+k}} contains an element of order (p_np_{n+1}\dots p_{n+r}\dots p_{n+k})^k.