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 November 26th, 2008, 10:16 PM #1 Newbie   Joined: Nov 2008 Posts: 3 Thanks: 0 sum of squares of primes Does anyone know if the sum of squares of a set of distinct primes can be a perfect square or not? If yes, when is that sum square of a prime? In particular, if we add the squares of the first n prime numbers starting with 2 what is the smallest n for which the sum is a perfect square?
 November 26th, 2008, 11:33 PM #2 Newbie   Joined: Nov 2008 Posts: 3 Thanks: 0 Re: sum of squares of primes apparently there is plenty of answers to the first question when the set consists of 4 primes: e.g. [ 3 5 11 13 ] , [5 7 11 17] , [5 11 13 19] , [5 7 23 29], [ 5 11 13 19], [ 5 11 13 29] , etc. or five primes one of which is 2: e.g. [ 2 3 5 7 13] , [ 2 3 5 23 37] , [ 2 3 7 11 29] , [ 2 3 7 13 37] , etc. or six primes one of which is 2: e.g. [ 2 5 7 11 13 19 ], [ 2 5 7 11 17 59], [ 2 5 7 11 19 23] , etc. therefore the sum will be equal to an even number in all of these cases. How about adding squares of odd number of odd primes? Can the sum be a square? prime square?
 November 28th, 2008, 05:05 PM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: sum of squares of primes Prime squares come in two categories mod 4: 4, which is 0 mod 4, and all other prime squares, which are 1 mod 4. So: 0: there is one set that sums to a square: {} sums to 0^2. 1: {p^2} sums to p^2 for any prime p. 2: none 3: none 4: many possibilities, starting with 3^2 + 5^2 + 11^2 + 13^2 = 18^2. 5: many possibilities, starting with 2^2 + 3^2 + 5^2 + 7^2 + 13^2 = 16^2. 6: none 7: none 8: many possibilities, starting with 3^2 + 5^2 + 7^2 + 11^2 + 13^2 + 23^2 + 29^2 + 31^2 = 52^2 9: many possibilities, starting with 2^2 + 5^2 + 7^2 + 11^2 + 13^2 + 17^2 + 19^2 + 23^2 + 37^2 = 54^2 . . . Obviously the only possibilities are when the size of the set is 0 or 1 mod 4.
 November 28th, 2008, 09:31 PM #4 Senior Member   Joined: Oct 2008 Posts: 215 Thanks: 0 Re: sum of squares of primes In 2#, a set of six numbers have been found: [ 2 5 7 11 13 19 ] the sum of square is $27^2$
 November 29th, 2008, 01:00 AM #5 Senior Member   Joined: Oct 2008 Posts: 215 Thanks: 0 Re: sum of squares of primes If a set with k primes satisfy your condition,it must be k=0,1,4,5,6 ( mod 8 )
 November 29th, 2008, 01:14 AM #6 Senior Member   Joined: Oct 2008 Posts: 215 Thanks: 0 Re: sum of squares of primes If you want to get a set which has odd numbers of odd primes, you need at least 9 primes
 November 29th, 2008, 07:40 PM #7 Newbie   Joined: Nov 2008 Posts: 3 Thanks: 0 Re: sum of squares of primes I've got one set of 9 odd prime squares: 31^2+37^2+41^2+43^2+47^2+53^2+59^2+61^2+73^2=153^2 Is there for proof for the (mod 8 ) statement?
 December 1st, 2008, 03:31 PM #8 Senior Member   Joined: Oct 2008 Posts: 215 Thanks: 0 Re: sum of squares of primes It's easy. We know that for any odd integer x(so for odd prime too),$x^2=1(mod " />. So it will be very easy to get the result now. But it does not means that when the (mod condition is satisfied, there must be a solution. For example, we could not find a set with 12 numbers. (If 3 is in the set, the sum of square = 11=2 (mod 3); otherwise, the sum of sqaure=12=3(mod 9). Both of them are not square)

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