Easy Peasy Prime Proof Problem

Given that $p$ and $q$ are primes such that $q - p = 2$ , prove that there doesn't exist a non-negative integer $k$ such that

$pq + 10^k$

is also prime.

Hint:

If you want this hint, it is that you may prove the problem using divisibility rules.

#NumberTheory #Sharky

Easy Math Editor

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Math	Appears as
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`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

I hope I can make myself expressible.

$p$ and $q$ are twin primes, that is, they differ by $2$ . Let $r$ be the only positive integer between $p$ and $q$ . Then $prq$ , being a product of $3$ consecutive integers, is divisible by $3$ . Since $p$ and $q$ are primes, $3$ divides $r$ , and hence, $p=3m+2$ and $q=3n+1$ , where $m$ and $n$ are positive integers.

Their product, thus, is of the form $3l+2$ , where $l$ is a positive integer. Lastly, $10^k=(9+1)^k$ is of the form $3o+1$ , where $o$ is a positive integer.

Adding them together, $pq+10^k$ is of the form $3l+2+3o+1$ , which is of the form $3z$ , where $z$ is a positive integer.

This is divisible by $3$ , thus this cannot be a prime number.

Satvik Golechha - 6 years, 6 months ago

The first part can also be proved using the fact that all primes $\geq 5$ are of the form $6k + 1$ or $6k + 5$ , where $k$ is a positive integer. Anyway, nice solution :)

Tan Li Xuan - 6 years, 6 months ago

Yeah! Thanks, I just wanted to use simple divisibility rules, the hint Sharky gave.

Satvik Golechha - 6 years, 6 months ago

nice one, but I think that $p=3m-1$ and $q=3m+1$ since you say they are twins

Trevor Arashiro - 6 years, 6 months ago

Note that this doesn't hold for $p = 3, q = 5$ .

This case has to be dealt with separately. (obvious fix).

The error in the initial logic is making the wrong assumption that $3 \mid pqr \rightarrow 3 \mid r$ .

Calvin Lin Staff - 6 years, 6 months ago

Good one!!! @Satvik Golechha and @Sharky Kesa

shivamani patil - 6 years, 6 months ago

Thanks! Gave RMO?

Satvik Golechha - 6 years, 6 months ago

@Satvik Golechha – No.Did you give?

shivamani patil - 6 years, 6 months ago

@Shivamani Patil – Yeah!

Satvik Golechha - 6 years, 6 months ago

Big hint: Consider the equation modulo 3.

Jordi Bosch - 6 years, 6 months 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}`	$\boxed{123}$