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.

## 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. – Satvik Golechha · 2 years, 4 months ago

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– Trevor Arashiro · 2 years, 3 months ago

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

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 · 2 years, 3 months ago

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– Tan Li Xuan · 2 years, 4 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 :)Log in to reply

– Satvik Golechha · 2 years, 4 months ago

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

@Satvik Golechha and @Sharky Kesa – Shivamani Patil · 2 years, 3 months ago

Good one!!!Log in to reply

– Satvik Golechha · 2 years, 3 months ago

Thanks! Gave RMO?Log in to reply

– Shivamani Patil · 2 years, 3 months ago

No.Did you give?Log in to reply

– Satvik Golechha · 2 years, 3 months ago

Yeah!Log in to reply

Big hint: Consider the equation modulo 3. – Jordi Bosch · 2 years, 4 months ago

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