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.

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## Comments

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TopNewestI 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.

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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 :)

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

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

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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 \(3 \mid pqr \rightarrow 3 \mid r \).

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

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Thanks! Gave RMO?

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Big hint: Consider the equation modulo 3.

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