Easy Peasy Prime Proof Problem

Given that pp and qq are primes such that qp=2q - p = 2, prove that there doesn't exist a non-negative integer kk such that

pq+10kpq + 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
4 years, 11 months ago

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1 vote

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I hope I can make myself expressible.

pp and qq are twin primes, that is, they differ by 22. Let rr be the only positive integer between pp and qq. Then prqprq, being a product of 33 consecutive integers, is divisible by 33. Since pp and qq are primes, 33 divides rr, and hence, p=3m+2p=3m+2 and q=3n+1q=3n+1, where mm and nn are positive integers.

Their product, thus, is of the form 3l+23l+2, where ll is a positive integer. Lastly, 10k=(9+1)k10^k=(9+1)^k is of the form 3o+13o+1, where oo is a positive integer.

Adding them together, pq+10kpq+10^k is of the form 3l+2+3o+13l+2+3o+1, which is of the form 3z3z, where zz is a positive integer.

This is divisible by 33, thus this cannot be a prime number.

Satvik Golechha - 4 years, 11 months ago

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

Tan Li Xuan - 4 years, 11 months ago

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

Satvik Golechha - 4 years, 11 months ago

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nice one, but I think that p=3m1p=3m-1 and q=3m+1q=3m+1 since you say they are twins

Trevor Arashiro - 4 years, 10 months ago

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Note that this doesn't hold for p=3,q=5 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 3pqr3r3 \mid pqr \rightarrow 3 \mid r .

Calvin Lin Staff - 4 years, 10 months ago

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

shivamani patil - 4 years, 10 months ago

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Thanks! Gave RMO?

Satvik Golechha - 4 years, 10 months ago

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@Satvik Golechha No.Did you give?

shivamani patil - 4 years, 10 months ago

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@Shivamani Patil Yeah!

Satvik Golechha - 4 years, 10 months ago

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Big hint: Consider the equation modulo 3.

Jordi Bosch - 4 years, 11 months ago

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