This can be written as $\zeta(1)$ (don't be afraid if you never saw this, search for Zeta Function).

\[1+\dfrac12+\dfrac13+\ldots=\prod_{p\in\mathbb P}\dfrac{p}{p-1}\]

where $\mathbb P$ is the set of primes.

Why does this hold? Think using unique prime factorization.

## Search found 592 matches

- Sat Oct 10, 2015 12:26 am
- Forum: Number Theory
- Topic: Sequence
- Replies:
**3** - Views:
**2486**

- Sat Oct 10, 2015 12:23 am
- Forum: Number Theory
- Topic: #Number Theory
- Replies:
**2** - Views:
**2192**

### Re: #Number Theory

See the article here.

viewtopic.php?f=26&t=3422

viewtopic.php?f=26&t=3422

- Sat Oct 10, 2015 12:22 am
- Forum: Number Theory
- Topic: Thue's Lemma
- Replies:
**0** - Views:
**1533**

### Thue's Lemma

Here is an article on Thue's Lemma.

- Sat Oct 10, 2015 12:20 am
- Forum: Number Theory
- Topic: Good positive integers
- Replies:
**1** - Views:
**1487**

### Re: Good positive integers

Aha. That problem of mine! It's quite easy actually. But it can be made a bit hard with more restrictions. Have you solved it?

- Mon Sep 21, 2015 10:42 pm
- Forum: Introductions
- Topic: Hi Everyone!
- Replies:
**1** - Views:
**3701**

### Re: Hi Everyone!

You are most welcome!

- Mon Sep 14, 2015 11:52 pm
- Forum: National Math Camp
- Topic: ONTC Final Exam
- Replies:
**34** - Views:
**18295**

### Re: ONTC Final Exam

now it should be correct

- Mon Sep 14, 2015 3:26 pm
- Forum: National Math Camp
- Topic: ONTC Final Exam
- Replies:
**34** - Views:
**18295**

### Re: ONTC Final Exam

You have to prove it.tanmoy wrote:$2^{n}+1$ is odd.So,$n^{2}-1$ is odd.So,$n$ is even.Let $n=2k$.Then $2^{n}+1=(2^{k})^{2}+1$.So,every divisor of $2^{n}+1$ is of the form $4m+1$.

So,$n^{2}-1 \equiv 1 (mod 4)$.

Or,$n^{2} \equiv 2 (mod 4)$,which is impossible.

So there is no solution.

- Sun Sep 13, 2015 11:16 pm
- Forum: National Math Camp
- Topic: ONTC Final Exam
- Replies:
**34** - Views:
**18295**

### Re: ONTC Final Exam

Try this: $n^2-1|2^n+1$

- Sat Sep 12, 2015 7:22 pm
- Forum: National Math Camp
- Topic: ONTC Final Exam
- Replies:
**34** - Views:
**18295**

### Re: ONTC Final Exam

Ya, this is the way to do it.

- Mon Sep 07, 2015 1:18 am
- Forum: National Math Camp
- Topic: ONTC Final Exam
- Replies:
**34** - Views:
**18295**

### Re: ONTC Final Exam

I posted the solutions of the first one in my new blog. But my laziness is getting the better of me again. When that passes, I will start writing again there. Stay tuned lol