# Infinitely many systems of $23$ consecutive integers

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.

• Are you familiar with Pell equations? – Erick Wong Jan 5 '15 at 0:37
• 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 $1$ – coffeemath Jan 5 '15 at 0:40
• I'm not familiar with Pell equations. I look for elementary proof. – Sinister Jan 5 '15 at 0:48

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.

• There is also seed solution $n=28, k=6;$ there are two orbits giving solutions, under the automorphism group of the form. – Will Jagy Jan 5 '15 at 0:55
• @WillJagy: The solver reported $n=18,k=-4$. I didn't notice that it generated a new list. – Ross Millikan Jan 5 '15 at 1:03
• 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 $28,1362,65348...$ – Will Jagy Jan 5 '15 at 1:04
• 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. – Will Jagy Jan 5 '15 at 1:07
• Checked some things, several orbits for $x^2 - 23 y^2 = 154$ – Will Jagy Jan 5 '15 at 1:15

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   