# 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. Note by Sharky Kesa
6 years, 10 months ago

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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.

- 6 years, 10 months ago

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

- 6 years, 9 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 $3 \mid pqr \rightarrow 3 \mid r$.

Staff - 6 years, 9 months ago

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

- 6 years, 9 months ago

Thanks! Gave RMO?

- 6 years, 9 months ago

No.Did you give?

- 6 years, 9 months ago

Yeah!

- 6 years, 9 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 :)

- 6 years, 10 months ago

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

- 6 years, 10 months ago

Big hint: Consider the equation modulo 3.

- 6 years, 10 months ago