January 17th, 2018, 05:50 AM  #1  
Member Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2  Problem with primes
I have trouble solving this problem: Quote:
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: 20,089 Thanks: 1902 
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. 
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 $\displaystyle (3,5,7)$. 
January 17th, 2018, 07:35 AM  #4 
Member Joined: Aug 2011 From: Nouakchott, Mauritania Posts: 85 Thanks: 14 Math Focus: Algebra, Cryptography  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: 55 Thanks: 2 
I see. Number not divisible by 3 can be represented as 3n1 or 3n2, and we have: $\displaystyle (3n1)^{2}=9n^{2}6n+1=3(3n^{2}2n)+1\equiv 1 \pmod 3$ $\displaystyle (3n2)^{2}=9n^{2}12n+4=3(3n^{2}4n)+4\equiv 1 \pmod 3$ The rest is clear to me. Thank you both. Last edited by skipjack; January 17th, 2018 at 01:48 PM. 

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