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# 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
2 years, 2 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. · 2 years, 2 months ago

nice one, but I think that $$p=3m-1$$ and $$q=3m+1$$ since you say they are twins · 2 years, 1 month 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 · 2 years, 1 month 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 :) · 2 years, 2 months ago

Yeah! Thanks, I just wanted to use simple divisibility rules, the hint Sharky gave. · 2 years, 2 months ago

Good one!!! @Satvik Golechha and @Sharky Kesa · 2 years, 1 month ago

Thanks! Gave RMO? · 2 years, 1 month ago

No.Did you give? · 2 years, 1 month ago

Yeah! · 2 years, 1 month ago