Prime Problem

Number Theory Level 3

Take some time to read the statements below.

$[1]$ . It is impossible for $p$ , $p+2$ and $p+4$ to be all prime numbers where $p$ is a prime number greater than $3$ .

$[2]$ . It is possible for both $8p-1$ and $8p+1$ to be prime when $p$ is a prime number.

$[3]$ . If $p$ is a prime greater than $3$ , then $p^2-1$ is always divisible by $12$

Which of these statements are true?

Note : This problem is a part of the set "I Don't Have a Good Name For This Yet". See the rest of the problems here . And when I say I don't have a good name for this yet, I mean it. If you like problems like these and have a cool name for this set, feel free to comment here .

[1]

and

[2]

[1]

[1]

[2]

and

[3]

[1]

and

[3]

3 solutions

Mursalin Habib
Apr 9, 2014

Let's consider the statements one by one.

Is $[1]$ true? Yes, it is.

If $p$ is a prime greater than $3$ , then either it is of the form $3k+1$ , or it is of the form $3k-1$ where $k$ is a positive integer.

If $p=3k+1$ , then $p+2$ is divisible by $3$ .

If $p=3k-1$ , then $p+4$ is divisible by $3$ . So, it is impossible for all of them to be prime if $p>3$ .

The statement $[2]$ isn't true for $p=2, 3$ .

If $p>3$ , then either $p=3k+1$ or $p=3k-1$ where $k$ is a positive integer.

If $p=3k+1$ , then $8p+1=24k+9$ which is divisible by $3$ .

If $p=3k-1$ , then $8p-1=24k-9$ which is divisible by $3$ . So, it is impossible for both $8p-1$ and $8p+1$ to be prime. $[2]$ isn't true.

$[3]$ is true. Any prime number greater than $3$ is of the form $6k\pm 1$ . So, $p^2-1=36k^2\pm12k$ which is divisible by $12$ .

So, our final answer is $[1]$ and $[3]$ .

In fact, $p^{2}-1$ is divisible by $24$ for all primes $p \geq 5$ . This implies divisibility by $12$ but it is a stronger statement.

Caleb Townsend - 6 years, 4 months ago

Cool solution.

Soham Dibyachintan - 7 years, 2 months ago

can you please give a proof for the third statement, i love to collect data on primes ........ thanks:)

Abhinav Raichur - 6 years, 11 months ago

just straightforward method :) ,thus try all number 6k+n n=1,2,3,4,5,0

Alexander Freeman - 6 years, 10 months ago

I think this will help you to understand , (p-1)(p+1) where p is odd number(prime)

U Z - 6 years, 5 months ago

(p-1) and (p+1) are both even as p is odd ( for p > 3), so 4| $p^{2}$ -1. Now 3 divides a product of 3 consecutive numbers, but 3 doesn't divide ${p}$ , so it divides one of (p-1) or (p+1). Hence 12 | $p^{2}$ -1

Curtis Clement - 6 years, 5 months ago

Jubayer Nirjhor
May 3, 2014

$p\equiv 1,2 \pmod{3}$ . If $p\equiv 1\pmod3$ then $p+2\equiv 0 \pmod3$ . Again if $p\equiv 2\pmod 3$ then $p+4\equiv 0\pmod 3$ . So all of $p,p+2,p+4$ can't be primes.
Same logic as above. Checking both residues modulo $3$ shows that both can't be primes simultaneously.
$p\equiv \pm1\pmod {3,4}\Longrightarrow ~p^2-1\equiv 1-1=0\pmod {3,4}$ . So $12\mid p^2-1$ .

Hence, 1 and 3 are true.

Tahsin Ahmed
Apr 11, 2014

এটা লেভেল ২ এর হওয়া উচিত.

This should be a comment, not a solution.

Mursalin Habib - 7 years, 1 month ago

Bengali! wow! :)

Sagnik Saha - 7 years, 1 month ago

Prime Problem

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