wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> medium >> sum of the squares of three consecutive odd prime
        
(Message started by: Christine on Jan 21^st, 2017, 10:54am)

Title: sum of the squares of three consecutive odd prime
Post by Christine on Jan 21^st, 2017, 10:54am

83 is a prime number and 83 = 3^2 + 5^2 + 7^2

Can you prove that there are no other prime numbers which are the sum of the squares of the three consecutive odd primes?

Title: Re: sum of the squares of three consecutive odd pr
Post by rloginunix on Jan 21^st, 2017, 5:40pm

The triplet (3, 5, 7) is the only set of "three consecutive odd numbers that are all prime" or, equivalently, the only triplet of prime numbers comprising an arithmetic progression with a common difference equal to 2.

Proof.

Let x be an odd prime > 3.

Consider the numbers x, x + 2 and x + 4.

Case 1). Let x = 3q + 1 where q = 2, 4, 6, ... Then the term x + 2 = 3q + 3 = 3(q + 1) is composite.

Case 2). Let x = 3q + 2 where q = 1, 3, 5, ... Then the term x + 4 = 3q + 6 = 3(q + 2) is composite. What was required to prove.

From where it follows that no other set of "three consecutive odd numbers that are all primes" exist.

Title: Re: sum of the squares of three consecutive odd pr
Post by rmsgrey on Jan 22^nd, 2017, 8:09am

In the spirit of answering questions that weren't asked:

The stipulation that the primes be odd is not necessary - 4+9+25 is even so not a prime.

Title: Re: sum of the squares of three consecutive odd pr
Post by towr on Jan 22^nd, 2017, 12:01pm

I think I'd interpret the problem as finding p_n = p_k² + p_k+1² + p_k+2² where p_i is the ith prime and i>1.
I mean, otherwise, why bring squaring and summing to a prime into it?

I don't see any immediate reason why there wouldn't be other solutions, but for primes below a billion it's the only solution.

Title: Re: sum of the squares of three consecutive odd pr
Post by pex on Jan 22^nd, 2017, 1:23pm

Proof: [hide]Any such triplet would only include primes greater than 3, which are well known to be equal to +/- 1 mod 6. That means their squares are all 1 mod 6, making the sum equal to 3 mod 6 and hence divisible by 3, so not prime.[/hide]