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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  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: When 1^2 + ... + n^2 be perfect square Quote:
119^2 + 120^2 = 169^2  
September 8th, 2011, 02:35 PM  #4  
Senior Member Joined: Apr 2011 From: Recife, BR Posts: 352 Thanks: 0  Re: When 1^2 + ... + n^2 be perfect square Quote:
 
September 8th, 2011, 04:28 PM  #5  
Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3  Re: When 1^2 + ... + n^2 be perfect square Quote:
 
September 10th, 2011, 10:55 PM  #6 
Member Joined: Aug 2010 From: Vitória 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: Vitória 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. 

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