Inspired by Kalpok Guha

I happened to see this problem. So I came up with a similar problem below.

How many integers $n>1$ are there such that $n,n+2,n+6$ are all prime numbers ?

Do post Awesome solutions with proofs!

#NumberTheory

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

@Kalpok Guha @Otto Bretscher @Calvin Lin @Pi Han Goh

Nihar Mahajan - 6 years ago

My intuition says there are infinitely many such $n$ .

Nihar Mahajan - 6 years ago

If there are infinitely many such n, then the twin prime conjecture is true. I suggest working on the twin prime conjecture first, because it is (likely) easier.

Calvin Lin Staff - 6 years ago

Yeah , I also think that.Thanks!

Nihar Mahajan - 6 years ago

Has anyone come close to proving the twin prime conjecture or do we need to solve the Riemann hypothesis first?

Sharky Kesa - 6 years ago

@Sharky Kesa – Zhang Yitang has proved a weakened form of the Twin Prime conjecture, namely that there exists an N such that there are infinitely many pairs of consecutive primes with difference < N.

Here is a good introductory read to the relationships between TP and RH. It is written by Dan Goldston (same as the above), though prior to Zhang's discovery. If you can understand through the first 5 chapters, that would be great

The conjecture that the distribution of twin primes satisfies a Riemann Hypothesis type error term is well supported empirically, but I think this might be a problem that survives the current millennium.

Calvin Lin Staff - 6 years ago

Yes, even as I was working out I found many numbers,

Here's my explanation,

First take prime numbers ending with $1$ ,

you can find that there can be many triplets satisfying the condition. For example: $\boxed{ 11, 13, 17}$ , $\boxed{41,43,47}$ , $\boxed{101,103,107}$ etc. . .

Now, let's take primes ending with $2$ ,

As there is only one possibility, but that proves to be wrong. $\boxed{2,4,8}$

Now, let's take primes ending with $3$ ,

We can say there's only no such possibility, because the second number i.e. $n+2$ yields us a number divisible by $5$ . Even this triplet proves wrong, $\boxed{3,5,9}$ as $9$ is not prime.

Now let's take primes ending with $5$ ,

Only one possibility, i.e. $\boxed{5,7,11}$ . In all other triplets, the first number i.e. $n$ is not prime

Now, let's take primes ending with $7$ ,

You can find that the can be many possibilities. For example: $\boxed{ 17, 19, 23}$ , $\boxed{107,109,113}$ , $\boxed{227,229,233}$ etc. . .

Now, let's take primes ending with $9$ ,

No such possibility, because the third number i.e. $n+6$ yields us a number divisivisible by $5$

Sravanth C. - 6 years ago

I didn't know 71+6=79. :P

Sharky Kesa - 6 years ago

@Sravanth Chebrolu I think you must develop a habit of always cross checking your work. :)

Nihar Mahajan - 6 years ago

Lol Btw Nice observation.

Aditya Chauhan - 6 years ago

197+6=203

Aditya Chauhan - 6 years ago

Where did you comment the 3 comments? I can only see one here.

pic

Nihar Mahajan - 6 years ago

Ah! Those were deleted by me :P

Sravanth C. - 6 years ago

@Sravanth C. – I guess U r the fastest one to get 200k points. Keep solving and making more problems.

Aditya Chauhan - 6 years ago

@Aditya Chauhan – Thanks! I'll be posting a new set keep your eyes on that!

Sravanth C. - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$