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September 7th, 2011, 01:51 PM   #1
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When 1^2 + ... + n^2 be perfect square

Show that there is only one integer n ( > 1) such that
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September 7th, 2011, 02:05 PM   #2
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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.
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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
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September 8th, 2011, 02:35 PM   #4
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Re: When 1^2 + ... + n^2 be perfect square

Quote:
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.
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September 8th, 2011, 04:28 PM   #5
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Re: When 1^2 + ... + n^2 be perfect square

Quote:
Originally Posted by elim
Show that there is only one integer n ( > 1) such that
This fact was not proven until 1918.
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September 10th, 2011, 10:55 PM   #6
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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.
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September 10th, 2011, 11:00 PM   #7
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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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