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 September 7th, 2011, 01:51 PM #1 Senior Member   Joined: Dec 2006 Posts: 167 Thanks: 3 When 1^2 + ... + n^2 be perfect square Show that there is only one integer n ( > 1) such that September 7th, 2011, 02:05 PM #2 Senior Member   Joined: Apr 2011 From: Recife, BR Posts: 352 Thanks: 0 Re: When 1^2 + ... + n^2 be perfect square Though this isn't the answer to the problem, it is a related fact that n > 10, being n = 11 the least possible number of consecutive integers whose sum of squares is also a square. For example, Your answer is 24, but I still can't prove that there is no other such n. September 8th, 2011, 01:54 PM   #3
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Re: When 1^2 + ... + n^2 be perfect square

Quote:
 Originally Posted by proglote Though this isn't the answer to the problem, it is a related fact that n > 10, being n = 11 the least possible number of consecutive integers whose sum of squares is also a square. For example, Your answer is 24, but I still can't prove that there is no other such n.
I'm not sure I understand what you mean.

119^2 + 120^2 = 169^2 September 8th, 2011, 02:35 PM   #4
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Re: When 1^2 + ... + n^2 be perfect square

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 Originally Posted by agentredlum I'm not sure I understand what you mean. 119^2 + 120^2 = 169^2
Obviously n > 2 , there are infinitely many pythagorean triples. September 8th, 2011, 04:28 PM   #5
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Re: When 1^2 + ... + n^2 be perfect square

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 Originally Posted by elim Show that there is only one integer n ( > 1) such that
This fact was not proven until 1918. September 10th, 2011, 10:55 PM #6 Member   Joined: Aug 2010 From: Vit�ria ES, Brazil Posts: 71 Thanks: 0 Re: When 1^2 + ... + n^2 be perfect square Some ideas: We fall in this diofantine equation: Since , we have two 'big' cases: 1 - or 2 - that derives 9 'small' cases. One of them: In this one we have Which is impossible because that we want So we have 5 remaining cases, one of then is the case that have the unique solution: where I think that its possible to use Pell equation in some cases to prove that theres no solution, but I don't know very much about them. September 10th, 2011, 11:00 PM #7 Member   Joined: Aug 2010 From: Vit�ria ES, Brazil Posts: 71 Thanks: 0 Re: When 1^2 + ... + n^2 be perfect square The remaining case with one of them being 6 times a square is impossible too: From the second equation we have and them cannot be a square. Similarlly(?), is impossible, just like and mixing first and last equations mod 3, mixing first and second mod 3, we have and then which is absurd, we have which is absurd. Remaining cases: and which gives the solution. Tags perfect, square 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 redount2k9 Algebra 3 January 1st, 2013 07:57 PM harrypham Number Theory 8 July 19th, 2012 03:12 PM greg1313 Number Theory 5 November 28th, 2011 06:16 AM PRO Number Theory 6 August 3rd, 2011 05:38 PM calligraphy Number Theory 4 February 10th, 2011 05:34 AM

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