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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 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)," />. 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) Tags primes, squares, sum 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 caters Number Theory 67 March 19th, 2014 04:32 PM proglote Number Theory 13 March 20th, 2012 05:39 PM smslca Number Theory 3 April 7th, 2011 04:43 PM WillingSponge Number Theory 5 July 9th, 2010 06:52 AM wiley Calculus 1 June 22nd, 2010 03:35 PM

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