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January 17th, 2018, 04:50 AM   #1
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Joined: Jan 2018
From: Belgrade

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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 05:50 AM. January 17th, 2018, 06:23 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 For any prime p greater than 3, p² - 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, 06:31 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, 06:35 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, p² - 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, 09:17 AM #5 Member   Joined: Jan 2018 From: Belgrade Posts: 55 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 12:48 PM. Tags primes, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mobel Number Theory 2 September 21st, 2015 01:44 PM caters Number Theory 67 March 19th, 2014 04:32 PM PerAA Number Theory 4 October 18th, 2013 08:25 AM Al7-8Ex5-3:Fe#!D%03 Number Theory 26 September 30th, 2013 04:34 PM sophieT66 Elementary Math 5 December 2nd, 2010 04:55 PM

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