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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 $1^2+2^2+\cdots +n^2= \frac{n(n+1)(2n+1)}{6}\ \text{is a perfect square.}$
 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, $18^{2} + 19^{2} + ... + 28^{2}= 77^{2}$ 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, $18^{2} + 19^{2} + ... + 28^{2}= 77^{2}$ 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

Quote:
 Originally Posted by elim Show that there is only one integer n ( > 1) such that $1^2+2^2+\cdots +n^2= \frac{n(n+1)(2n+1)}{6}\ \text{is a perfect square.}$
This fact was not proven until 1918.

 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: $n(n+1)(2n+1)= 6x^2$ Since $\gcd(n,\, n+1)= \gcd(n,\, 2n+1) = \gcd(n+1,\, 2n+1) = 1$, we have two 'big' cases: 1 - $\{n,\, n+1,\, 2n+1\}= \{a^2,\, b^2,\, 6c^2\}$ or 2 - $\{n,\, n+1,\, 2n+1\}= \{a^2,\, 2b^2,\, 3c^2\}$ that derives 9 'small' cases. One of them: $n= a^2, \ n+1 = b^2,\ 2n+1 = 6c^2$ In this one we have $a^2 + 1= b^2 \ \Rightarrow \ 1 = (b-a)(b+a) \ \Rightarrow \ b+a = b-a = \pm 1 \ \Rightarrow \ b = \pm 1 \ \Rightarrow \ a = 0$ Which is impossible because that we want $n > 1.$ So we have 5 remaining cases, one of then is the case that have the unique solution: $n= 6a^2$ where $a= 2.$ 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: $n= a^2, \ n+1 = 6b^2,\ 2n+1 = c^2$ From the second equation we have $n \equiv 2\ \pmod{3}$ and them $n$ cannot be a square. Similarlly(?), $n= a^2,\ n+1 = 3b^2,\ 2n+1 = 2c^2$ is impossible, just like $n= 2a^2,\ n+1 = b^2,\ 2n+1 = 3c^2$ and $n= 3a^2,\ n+1 = b^2,\ 2n+1 = 2c^2$ mixing first and last equations mod 3, $n= 3a^2,\ n+1 = 2b^2,\ 2n+1 = c^2$ mixing first and second mod 3, $n= 2a^2,\ n+1 = 3b^2,\ 2n+1 = c^2$ we have $2a^2+3b^2= c^2 \ \Rightarrow \ a \equiv c \equiv 0\ \pmod{3}$ and then $n \equiv 2n+1 \equiv 0\ \pmod{3}$ which is absurd, $n= 3a^2,\ n+1 = 2b^2,\ 2n+1 = c^2$ we have $1 \equiv 2b^2\ \pmod{3}$ which is absurd. Remaining cases: $n= a^2,\ n+1 = 2b^2,\ 2n+1 = 3c^2$ and $n= 6a^2,\ n+1 = b^2,\ 2n+1 = c^2$ which gives the solution.

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