My Math Forum Problem with primes

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January 17th, 2018, 05:50 AM   #1
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Joined: Jan 2018

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Problem with primes

I have trouble solving this problem:

Quote:
 Let p, q and r be three consecutive prime numbers such that $\displaystyle p^{2}+q^{2}+r^{2}$ is also prime. How many such triples (p,q,r) are there?
I found that (3,5,7) is one of them. But I don't know whether it is the only one or there are more. I had two ideas:

1. Represent q as $\displaystyle p+2n_{1}$, and r as $\displaystyle p+2n_{2}$, but I didn't manage to solve;
2. I realise that 5 is the only prime ending with 5. It seemed to me this is a good idea, but it did not take me anywhere.

Does anybody have any idea?

Last edited by skipjack; January 17th, 2018 at 06:50 AM.

 January 17th, 2018, 07:23 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,888 Thanks: 1836 For any prime p greater than 3, pÂ² - 1 is a multiple of 3, so at least one of the primes is 3. The question is ambiguous as to whether p < q < r. If p < q < r, (p, q, r) = (3, 5, 7) is the only solution. Thanks from Ould Youbba and lua
 January 17th, 2018, 07:31 AM #3 Member     Joined: Aug 2011 From: Nouakchott, Mauritania Posts: 85 Thanks: 14 Math Focus: Algebra, Cryptography Good afternoon. If $\displaystyle x$ is an integer that is not divisible by $\displaystyle 3$, then we have : $\displaystyle x^2\equiv 1\pmod 3$. So, if all the numbers $\displaystyle p,q$ and $\displaystyle r$ are not equal to $\displaystyle 3$, then we get : $\displaystyle p^2+q^2+r^2\equiv 1+1+1\equiv 3\equiv 0\pmod 3.$And this means that sum is divisible by 3 and it is a prime number if only it is equal to 3. But, that is impossible. So one of the prime should be equal to 3. Hence, the possibilities are : $\displaystyle (3,5,7)$ and $\displaystyle (2,3,5)$. But since $\displaystyle 2^2+3^2+5^2$ is even, then the unique solution is $\displaystyle (3,5,7)$. Thanks from lua
January 17th, 2018, 07:35 AM   #4
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Joined: Aug 2011
From: Nouakchott, Mauritania

Posts: 85
Thanks: 14

Math Focus: Algebra, Cryptography
Quote:
 Originally Posted by skipjack For any prime p greater than 3, pÂ² - 1 is a multiple of 3, so at least one of the primes is 3. The question is ambiguous as to whether p < q < r. If p < q < r, (p, q, r) = (3, 5, 7) is the only solution.
I am really sorry . I didn't see your post before I posted my reply. It is the same idea.

 January 17th, 2018, 10:17 AM #5 Member   Joined: Jan 2018 From: Belgrade Posts: 54 Thanks: 2 I see. Number not divisible by 3 can be represented as 3n-1 or 3n-2, and we have: $\displaystyle (3n-1)^{2}=9n^{2}-6n+1=3(3n^{2}-2n)+1\equiv 1 \pmod 3$ $\displaystyle (3n-2)^{2}=9n^{2}-12n+4=3(3n^{2}-4n)+4\equiv 1 \pmod 3$ The rest is clear to me. Thank you both. Last edited by skipjack; January 17th, 2018 at 01:48 PM.

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