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Prove that there are infinitely many systems of $23$ consecutive integers whose sum of squares is a perfect square.

My try: $$(n-11)^2+\cdots+(n+11)^2=23n^2+1012=23(n^2+44)=m^2$$ so $m=23k$ , $n^2=23k^2-44$. From $\mod 23$, I see $n=23l+5$ or $n=23l+18$ but i don't know what now.

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  • $\begingroup$ Are you familiar with Pell equations? $\endgroup$
    – Erick Wong
    Jan 5, 2015 at 0:37
  • $\begingroup$ It's of "Pell type" but the right side isn't $1$. That is, it isn't as routine to solve $m^2-23n^2=1012$ as it is to solve the same with the right side $1$ $\endgroup$
    – coffeemath
    Jan 5, 2015 at 0:40
  • $\begingroup$ I'm not familiar with Pell equations. I look for elementary proof. $\endgroup$
    – Sinister
    Jan 5, 2015 at 0:48

2 Answers 2

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Dario Alpern's solver reports $n=18, k=4$ and $n=28, k=6$ as solutions, then if $(x,y)$ is a solution, so is $(24x+115y,5x+24y)$. It will show you the steps if you ask.

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  • $\begingroup$ There is also seed solution $n=28, k=6;$ there are two orbits giving solutions, under the automorphism group of the form. $\endgroup$
    – Will Jagy
    Jan 5, 2015 at 0:55
  • $\begingroup$ @WillJagy: The solver reported $n=18,k=-4$. I didn't notice that it generated a new list. $\endgroup$ Jan 5, 2015 at 1:03
  • $\begingroup$ Anyway, by Cayley-Hamilton, we get two sequences of $n$ values under $n_{j+2} = 48 n_{j+1} - n_j,$ one is $18, 892, 42798...$ the other is $28,1362,65348...$ $\endgroup$
    – Will Jagy
    Jan 5, 2015 at 1:04
  • $\begingroup$ Ross, yes, unless the target number is $\pm 1$ we expect to get a separate list from that, and here the target is $-44.$ If the target has more prime factors that are represented by some (primitive) form of the same discriminant, we expect even more lists if there are any. $\endgroup$
    – Will Jagy
    Jan 5, 2015 at 1:07
  • $\begingroup$ Checked some things, several orbits for $x^2 - 23 y^2 = 154$ $\endgroup$
    – Will Jagy
    Jan 5, 2015 at 1:15
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EDIT, March 2016. Based on what people seemed to want in a recent question on Pell's equation, I wrote a program that solves $x^2 - d y^2 = k$ quite quickly, and identifies the "fundamental" solutions, from which all others can be found by applying the automorphism group.

jagy@phobeusjunior:~$ ./Pell_Target_Fundamental

  24^2 - 23 5^2 = 1

 x^2 - 23 y^2 = 154

Thu Mar 31 10:59:54 PDT 2016

x:  19  y:  3 ratio: 0.157895  fundamental 
x:  27  y:  5 ratio: 0.185185  fundamental 
x:  73  y:  15 ratio: 0.205479  fundamental 
x:  111  y:  23 ratio: 0.207207  fundamental 
x:  801  y:  167 ratio: 0.208489
x:  1223  y:  255 ratio: 0.208504
x:  3477  y:  725 ratio: 0.208513
x:  5309  y:  1107 ratio: 0.208514
x:  38429  y:  8013 ratio: 0.208514
x:  58677  y:  12235 ratio: 0.208514
x:  166823  y:  34785 ratio: 0.208514
x:  254721  y:  53113 ratio: 0.208514
x:  1843791  y:  384457 ratio: 0.208514
x:  2815273  y:  587025 ratio: 0.208514
x:  8004027  y:  1668955 ratio: 0.208514
x:  12221299  y:  2548317 ratio: 0.208514

Thu Mar 31 11:00:14 PDT 2016

 x^2 - 23 y^2 = 154

jagy@phobeusjunior:~$

I decided to draw the complete diagram of the Conway topograph, first the river, then the two extensions (trees) away from the river that, together, give all orbits for representing $x^2 - 23 y^2 = 154,$ those four seed pairs being $$ (19,3); \; (27,5); \; (73,15); \; (111,23). $$

As far as the original posted problem, the seeds for representing $x^2 - 23 y^2 = -11$ all occur along the river. Note that, as $x^2 - 23 y^2 \equiv x^2 + y^2 \pmod 4,$ whenever $x^2 - 23 y^2 \equiv 0 \pmod 4,$ it follows that both $x,y$ are even. That is, the seeds for $-11$ are $$ (9,2); \; (14,3), $$ therefore the only seeds for $-44$ are $$ (18,4); \; (28,6). $$

You can see Ross's formula $(24x+115y, 5x+24y)$ at the far right of the river diagram, on graph paper. We see a representation of $1$ with column vector $(24,5)^T,$ below it and all the way to the edge of the paper, a representation of $-23$ with column vector $(115,24)^T.$ Put them side by side and we get the two by two matrix $$ \left( \begin{array}{cc} 24 & 115 \\ 5 & 24 \end{array} \right) $$ of determinant $+1.$ That matrix, applied to a column vector $(x,y)^T,$ gives Ross's mapping.

I used a pink pen for the represented numbers in the two tree diagrams, it is a bit hard to read; next time, always red for the represented numbers.

I put four explanatory documents at OTHER with prefix indefinite_binary. For that matter, Conway's entire book is available at PDF

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