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I proved theres infinitely many n such that S_n has an element of order n^2

- Thread starter JoanBraidy
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I proved theres infinitely many n such that S_n has an element of order n^2

- #2

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Trivialish.

[itex]3^2+4^2+5^2<3.4.5[/itex].

Suppose [itex]3^{2i}+4^{2i}+5^{2i}<3^i.4^i.5^i[/itex] for [itex]i<n[/itex], then

[itex]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}[/itex], hence [itex]3^{2n}+4^{2n}+5^{2n}<3^n4^n5^n[/itex] by induction.

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

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

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

[itex]3^2+4^2+5^2<3.4.5[/itex].

Suppose [itex]3^{2i}+4^{2i}+5^{2i}<3^i.4^i.5^i[/itex] for [itex]i<n[/itex], then

[itex]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}[/itex], hence [itex]3^{2n}+4^{2n}+5^{2n}<3^n4^n5^n[/itex] by induction.

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

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

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

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similar to what I did thanksTrivialish.

[itex]3^2+4^2+5^2<3.4.5[/itex].

Suppose [itex]3^{2i}+4^{2i}+5^{2i}<3^i.4^i.5^i[/itex] for [itex]i<n[/itex], then

[itex]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}[/itex], hence [itex]3^{2n}+4^{2n}+5^{2n}<3^n4^n5^n[/itex] by induction.

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

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

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

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

- #4

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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 [itex]k=0[/itex] it is trivial that [itex]S_n[/itex] has an element of order [itex]n^k[/itex].

For any [itex]k\in\mathbb{N},k\geq1[/itex]

[tex]p_n^k+p_{n+1}^k+\dots+p_{n+r}^k+\dots+p_{n+k}^k[/tex]

[tex]\leq p_n^k(1+2^k+2^{2k}+\dots+2^{rk}+\dots+2^{k^2})\text{ (By Bertrand's postulate)}[/tex]

[tex]=p_n^k(2^{k(k+1)}-1)/(2^k-1)[/tex]

If [itex]n_k[/itex] is chosen so that

[tex]p_{n_k+k}\geq (2^{k(k+1)}-1)/(2^k-1)[/tex]

then for any [itex]n\geq n_k[/itex]

[tex]p_n^k(2^{k(k+1)}-1)/(2^k-1)[/tex]

[tex]\leq p_n^k(p_{n+k})[/tex]

[tex]\leq p_np_{n+1}\dots p_{n+r}\dots p_{n+k-1}p_{n+k}[/tex]

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

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 [itex]k=0[/itex] it is trivial that [itex]S_n[/itex] has an element of order [itex]n^k[/itex].

For any [itex]k\in\mathbb{N},k\geq1[/itex]

[tex]p_n^k+p_{n+1}^k+\dots+p_{n+r}^k+\dots+p_{n+k}^k[/tex]

[tex]\leq p_n^k(1+2^k+2^{2k}+\dots+2^{rk}+\dots+2^{k^2})\text{ (By Bertrand's postulate)}[/tex]

[tex]=p_n^k(2^{k(k+1)}-1)/(2^k-1)[/tex]

If [itex]n_k[/itex] is chosen so that

[tex]p_{n_k+k}\geq (2^{k(k+1)}-1)/(2^k-1)[/tex]

then for any [itex]n\geq n_k[/itex]

[tex]p_n^k(2^{k(k+1)}-1)/(2^k-1)[/tex]

[tex]\leq p_n^k(p_{n+k})[/tex]

[tex]\leq p_np_{n+1}\dots p_{n+r}\dots p_{n+k-1}p_{n+k}[/tex]

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

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