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Denis
 Post subject: Square deal!
PostPosted: Sun, 9 Oct 2011 18:15:48 UTC 
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1^2 + 2^2 + 3^2 +...+ 20^2 + 21^2 + [22^2 + 23^2 +...+ 67^2 + 68^2]

Sum of bracketed portion of above consecutive squares = 103823
The number of terms is 68 - 21 = 47
103823 / 47 = 2209 = 47^2

In other words, the average of n consecutive squares = n^2

u^2 = the square preceeding the n consecutive squares (u = 21 in example)
v^2 = last square (v = 68 in example)
So n = v - u

What is v in terms of u ?

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Shadow
 Post subject: Re: Square deal!
PostPosted: Sun, 9 Oct 2011 22:10:51 UTC 
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Denis wrote:
1^2 + 2^2 + 3^2 +...+ 20^2 + 21^2 + [22^2 + 23^2 +...+ 67^2 + 68^2]

Sum of bracketed portion of above consecutive squares = 103823
The number of terms is 68 - 21 = 47
103823 / 47 = 2209 = 47^2

In other words, the average of n consecutive squares = n^2

u^2 = the square preceeding the n consecutive squares (u = 21 in example)
v^2 = last square (v = 68 in example)
So n = v - u

What is v in terms of u ?


Spoiler:
We have:

$\sum_{k=u+1}^vk^2={v(v+1)(2v+1)\over 6}-{u(u+1)(2u+1)\over 6}=(v-u)^3

Now this is the beautiful part:

We get: 2v^3+3v^2+v-2u^3-3u^2-u=6(v-u)(v^2-2uv+u^2) which is GREAT, because it means our curve secretly has genus 0 and we can actually NAME ALL THE SOLUTIONS.

2(v-u)(u^2+uv+v^2)+3(v-u)(u+v)+(v-u)=6(v-u)(u^2+uv+v^2) and clearly u\ne v is the non-trivial case, so let's get at the others:

4v^2+4u^2-14uv-3u-3v-1=0\iff 4v^2-v(14u+3)+(4u^2-3u-1)=0

Now we can be boring and answer the original question here:

u=v (trivial case) or -- by the quadratic formula -- $v={1\over 8}\left(14u+3\pm\sqrt{132u^2+132u+25}\right), and checking against the one solution we already know, we see we should take the negative square root, because v>u was stipulated. And that's OK, if you're into that thing, but I want to do stereographic projection off of (0,1) -- the point I get when I let u=0 and let u=s(v-1) so that I can find all solutions to your equation just by plugging in some rational numbers.

This yields:

I.e. $(v-1)\left(4v+1+6s^2(v-1)-12vs-2vs-2s^2(v-1)-3s\right)=0

and this is MUCH cooler, because you can see the quadratic formula in it again, only this time it's much better because we already know one solution, namely v=1, and the other is given by solving:


$v(4+6s^2-12s-2s-2s^2)+1-6s^2+2s^2-3s=0

which finally is an easy task:

$v={2(2s^2-7s+2)\over (s+1)(4s-1)}\quad u={s(5-17s)\over (s+1)(4s-1)}

I personally think this is even better than the original answer because if you put in any rational s you can get all the rational points on the curve, and if you think about it this gives you all integer solutions to your original question. It's also easy to see that you only need rationals so that u,v\ge 0, and it's easy to solve for the appropriate intervals. Fun problem all in all.


Edit: Calculation fixed.

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Denis
 Post subject: Re: Square deal!
PostPosted: Mon, 10 Oct 2011 02:35:03 UTC 
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Agree; good stuff Shadow...

Only 5 solutions keeping u < 100 million:

u = 21; 22^2 to 68^2 : n = 47
u = 988; 989^2 to 3149^2 : n = 2,161
u = 45,449; 45450^2 to 144808^2 : n = 99,359
u = 2,089,688; 2089689^2 to 6658041^2 : n = 4,568,353
u = 96,080,221; 96080221^2 to 306125100^2 ; n = 210,044,879

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outermeasure
 Post subject: Re: Square deal!
PostPosted: Mon, 10 Oct 2011 04:39:28 UTC 
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Shadow wrote:
Spoiler:
...$v={1\over 8}\left(14u+3\pm\sqrt{132u^2+132u+25}\right), and checking against the one solution we already know, we see we should take the negative square root, because v>u was stipulated. And that's OK, if you're into that thing, but I want to do stereographic projection off of (0,1) -- the point I get when I let u=0 and let u=s(v-1) so that I can find all solutions to your equation just by plugging in some rational numbers.

This yields:

I.e. $(v-1)\left(4v+1+6s^2(v-1)-12vs-2vs-2s^2(v-1)-3s\right)=0

and this is MUCH cooler, because you can see the quadratic formula in it again, only this time it's much better because we already know one solution, namely v=1, and the other is given by solving:


$v(4+6s^2-12s-2s-2s^2)+1-6s^2+2s^2-3s=0

which finally is an easy task:

$v={2(2s^2-7s+2)\over (s+1)(4s-1)}\quad u={s(5-17s)\over (s+1)(4s-1)}

I personally think this is even better than the original answer because if you put in any rational s you can get all the rational points on the curve, and if you think about it this gives you all integer solutions to your original question. It's also easy to see that you only need rationals so that u,v\ge 0, and it's easy to solve for the appropriate intervals. Fun problem all in all.



I'd diverge starting there
Spoiler:
While rational points is cool, we actually want integer points...

Concentrating on the condition \sqrt{132u^2+132+25}\in\mathbb{Z}, with the substitution y=2u+1, we get the Pell-like equation x^2-33y^2=-8. Now the fundamental unit of \mathcal{O}_{\mathbb{Q}(\sqrt{33})}=\mathbb{Z}[\frac{1+\sqrt{33}}{2}] is 23+4\sqrt{33}, and we know (x,y)=(5,\pm1) is a solution, so general theory gives x+\sqrt{33}y=\pm(5\pm\sqrt{33})(23+4\sqrt{33})^m, m\in\mathbb{Z}. The four combinations of the \pms give 4 family of solutions, and now we are tasked with rejecting those with v\leq u or u<0 or v<0.

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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Shadow
 Post subject: Re: Square deal!
PostPosted: Mon, 10 Oct 2011 06:07:55 UTC 
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outermeasure wrote:
Shadow wrote:
Spoiler:
...$v={1\over 8}\left(14u+3\pm\sqrt{132u^2+132u+25}\right), and checking against the one solution we already know, we see we should take the negative square root, because v>u was stipulated. And that's OK, if you're into that thing, but I want to do stereographic projection off of (0,1) -- the point I get when I let u=0 and let u=s(v-1) so that I can find all solutions to your equation just by plugging in some rational numbers.

This yields:

I.e. $(v-1)\left(4v+1+6s^2(v-1)-12vs-2vs-2s^2(v-1)-3s\right)=0

and this is MUCH cooler, because you can see the quadratic formula in it again, only this time it's much better because we already know one solution, namely v=1, and the other is given by solving:


$v(4+6s^2-12s-2s-2s^2)+1-6s^2+2s^2-3s=0

which finally is an easy task:

$v={2(2s^2-7s+2)\over (s+1)(4s-1)}\quad u={s(5-17s)\over (s+1)(4s-1)}

I personally think this is even better than the original answer because if you put in any rational s you can get all the rational points on the curve, and if you think about it this gives you all integer solutions to your original question. It's also easy to see that you only need rationals so that u,v\ge 0, and it's easy to solve for the appropriate intervals. Fun problem all in all.



I'd diverge starting there
Spoiler:
While rational points is cool, we actually want integer points...

Concentrating on the condition \sqrt{132u^2+132+25}\in\mathbb{Z}, with the substitution y=2u+1, we get the Pell-like equation x^2-33y^2=-8. Now the fundamental unit of \mathcal{O}_{\mathbb{Q}(\sqrt{33})}=\mathbb{Z}[\frac{1+\sqrt{33}}{2}] is 23+4\sqrt{33}, and we know (x,y)=(5,\pm1) is a solution, so general theory gives x+\sqrt{33}y=\pm(5\pm\sqrt{33})(23+4\sqrt{33})^m, m\in\mathbb{Z}. The four combinations of the \pms give 4 family of solutions, and now we are tasked with rejecting those with v\leq u or u<0 or v<0.


I forget, do we know that
Spoiler:
\mathbb{Z}\left[{1+\sqrt{33}\over 2}\right] is a UFD?
.

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outermeasure
 Post subject: Re: Square deal!
PostPosted: Mon, 10 Oct 2011 07:10:16 UTC 
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Shadow wrote:
I forget, do we know that
Spoiler:
\mathbb{Z}\left[{1+\sqrt{33}\over 2}\right] is a UFD?
.


Spoiler:
Yes. In fact it is a PID.

We can go through the Minkowski bound as always, only needing to check up to \dfrac{1}{2}\sqrt{33}=2.872\dots, and 2=\dfrac{\sqrt{33}+5}{2}\cdot \dfrac{\sqrt{33}-5}{2} does it.

Or you can cheat and look Sloanne A003656 or similar.

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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Shadow
 Post subject: Re: Square deal!
PostPosted: Mon, 10 Oct 2011 14:15:52 UTC 
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Wonderful, then that's great. And of course we get that since Dedekind domain's only lacked that one quality before being PIDs. Thanks for the clarification, Google wasn't wanting to tell me the answer to my question.

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outermeasure
 Post subject: Re: Square deal!
PostPosted: Mon, 10 Oct 2011 14:25:57 UTC 
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Hmm... just realised I missed a step in the solution.
Spoiler:
There is the possibility of using (\frac{5\pm\sqrt{33}}{2})^3 instead of (5\pm\sqrt{33}), these are the only ideals of \mathbb{Z}[\frac{1+\sqrt{33}}{2}] of norm 8, but of course (\frac{5\pm\sqrt{33}}{2})^3=\frac{1}{2}(155+27\sqrt{33})\notin\mathbb{Z}[\sqrt{33}] kills that, since the fundamental unit lies in \mathbb{Z}[\sqrt{33}].

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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Shadow
 Post subject: Re: Square deal!
PostPosted: Mon, 10 Oct 2011 14:29:47 UTC 
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outermeasure wrote:
Hmm... just realised I missed a step in the solution.
Spoiler:
There is the possibility of using (\frac{5\pm\sqrt{33}}{2})^3 instead of (5\pm\sqrt{33}), these are the only ideals of \mathbb{Z}[\frac{1+\sqrt{33}}{2}] of norm 8, but of course (\frac{5\pm\sqrt{33}}{2})^3=\frac{1}{2}(155+27\sqrt{33})\notin\mathbb{Z}[\sqrt{33}] kills that.


:)

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outermeasure
 Post subject: Re: Square deal!
PostPosted: Mon, 10 Oct 2011 15:23:32 UTC 
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Also, I should say a word or two about generating solutions so Denis can run on the computer without the tiresome exhaustive search.
Spoiler:
There is the recurrence generated by x+y\sqrt{33}=(blah)\cdot(23+4\sqrt{33})^m, namely
\left\{\begin{aligned} x_{m+1}&=23x_m+132y_m\\ y_{m+1}&=23y_m+4x_m \end{aligned}\right., or equivalently \left\{\begin{aligned} x_{m-1}&=23x_m-132y_m\\ y_{m-1}&=23y_m-4x_m \end{aligned}\right..
Remembering (u,v)=\left(\dfrac{y-1}{2}, \dfrac{7y-4+x}{8}\right), you can translate that into recurrence relation involving the u,v:
\left\{\begin{aligned} u_{m+1}&=-5u_m+16v_m+5\\ v_{m+1}&=-16u_m+51v_m+17 \end{aligned}\right. or \left\{\begin{aligned} u_{m-1}&=51u_m-16v_m+17\\ v_{m-1}&=16u_m-5v_m+5 \end{aligned}\right.

Now the starting conditions:- although there are four point (x_0,y_0)=(\pm 5,\pm 1), in the (u,v) space the \pm(5,-1) can be rejected since they will never give you u,v that are integers (they give you starting point (u_0,v_0)=(0,-3/4),(-1,-1/4), which the recurrence will always keep v strictly quarter-integer and u integer). So you only need to check with the starting conditions (u_0,v_0)=(0,1),(-1,-2).

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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