Ponder This Challenge - February 2013 - Concatenating squares

Ponder This Challenge:

James Tanton tweeted ( https://twitter.com/jamestanton/status/293127359291330561) that "16 and 9 are each square numbers, and putting them together, 169, gives another square" and he then asked for other examples.

Our challenge this month is to find integers x, y, and z such that concatenating x^2 and y^2 gives z^2, and that z has at least four consecutive nines.

Update 2/6: the challenge stating that x,y and z should be non-zero.

We will post the names of those who submit a correct, original solution! If you don't want your name posted then please include such a statement in your submission!

We invite visitors to our website to submit an elegant solution. Send your submission to the ponder@il.ibm.com.

If you have any problems you think we might enjoy, please send them in. All replies should be sent to: ponder@il.ibm.com

Solution

• Click here to view the solution

  When preparing the challenge this month, I made a small mistake that made the question much easier than intended.

  Here is the solution, as I derived it:

  Let y^2 be a k-digits-long number, giving us 10^k x^2 + y^2 = z^2.

  Write z=10^k a + b and we get

  10^k x^2 + y^2 = (10^k a + b)^2 = 10^k a(10^k a+ 2b) + b^2.

  So, b^2=y^2 (mod 10^k). Here I made the mistake of concluding that b=y.

  Those who want to try to solve a more difficult version of the question, can stop here and ponder a little on solving this version.

  My mistake was, of course, that b can be -y mod(10^k), which lends itself to several trivial solutions like

  x = 5 * 10^(k-1) - 1; y = 10^k -1; and z = 5 * 10^(2 * k-1) - 10^k + 1.

  If we continue with the assumption that b=y, we get

  x^2 = a^2 10^k + 2ya

  so y = (x^2-a^2 10^k) / 2a.

  For every k and a, we can compute x as ceil(a*sqrt(10^k)) and check whether the resulting y is divisible by 2a.

  The fact that sqrt(10^37) is surprisingly close to an integer (~3162277660168379331.99889) gives us, for k=37, many possible solutions. Choosing a=463 gives us the intended solution of

  x=1464134556657959630716

  y=1619999831463380856

  z=4630000000000000000001619999831463380856

  J�nos Kram�r and Andreas Stiller sent us a solution with y=1, where one can solve Pell's equation: 10*x^2 + 1 = z^2 and get:

  x=44101535838578367767701517414261786147419230399948898912033566977281348406767427039616458154500555609301871792239537197062035119616191426015391050131988492071587920645046734772112341070906349353550324660996677696009221202338336

  y=1

  z=139461301561451525695691667071742666259644328693387043194785898131383963733064589327602699959968509145992213260946921467365082720725567556648743873803852310099994956836260701173999664267548159260952703090632330767251579838305281

  Jan Fricke also used Pell's equation to get to a solution with 60,503 digits.

Solvers

• Joseph DeVincentis (02/01/2013 01:44 PM EDT)
• Adel El-Atawy (02/01/2013 02:04 PM EDT)
• Dennis Langdeau (02/01/2013 03:17 PM EDT)
• Kevin Bauer (02/01/2013 03:33 PM EDT)
• Donald Dodson (02/01/2013 03:35 PM EDT)
• Martin Husar (02/01/2013 03:36 PM EDT)
• Sergey Grishaev (02/01/2013 03:59 PM EDT)
• Peter Gerritson (02/01/2013 04:07 PM EDT)
• Markus von Rimscha (02/01/2013 04:07 PM EDT)
• Vladimir Sedach (02/01/2013 04:27 PM EDT)
• Stéphane Higueret (02/01/2013 05:20 PM EDT)
• Todd Will (02/01/2013 05:33 PM EDT)
• Bartosz Budzanowski (02/01/2013 05:51 PM EDT)
• Rudy Cortembert (02/01/2013 05:54 PM EDT)
• Deane Stewart (02/01/2013 05:59 PM EDT)
• Anders Kaseorg (02/01/2013 06:36 PM EDT)
• Sunil Srivastava (02/01/2013 07:21 PM EDT)
• Deron Stewart (02/01/2013 11:32 PM EDT)
• Radu-Alexandru Todor (02/01/2013 10:51 PM EDT)
• Andrey Stepin (02/02/2013 12:47 AM EDT)
• Alex Wagner (02/02/2013 02:43 AM EDT)
• Oleg Zlydenko (02/02/2013 04:35 AM EDT)
• Leandro Araújo (02/02/2013 07:15 AM EDT)
• Tony Harrison (02/02/2013 09:07 AM EDT)
• Mark Stuckel (02/02/2013 10:35 AM EDT)
• Hakan Summakoglu (02/02/2013 11:30 AM EDT)
• Jesus Sanz (02/02/2013 12:48 PM EDT)
• Klaus Müller (02/02/2013 12:57 PM EDT)
• Øyvind Grotmol (02/02/2013 02:21 PM EDT)
• Andreas Stiller (02/02/2013 06:13 PM EDT)
• Florian Fischer (02/02/2013 06:21 PM EDT)
• Tao Cheng (02/02/2013 06:29 PM EDT)
• Dan Dima (02/03/2013 02:56 AM EDT)
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• Jerry Kearns (02/07/2013 02:35 PM EDT)
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• Priyanka Gagneja Dhingra (02/10/2013 11:02 PM EDT)
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• Rodrigo Viana Rocha (02/18/2013 08:42 PM EDT)
• Kerry M. Soileau (02/19/2013 08:07 PM EDT)
• Shmuel Menachem Spiegel (02/19/2013 11:38 PM EDT)
• Minghao (Tommy) Liang (02/20/2013 01:42 PM EDT)
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• Nyles Heise (02/20/2013 10:35 PM EDT)
• Michael Holschbach (02/21/2013 01:13 PM EDT)
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• (02/24/2013 04:06 AM EDT)
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• Dugan Schmid & Craig Oppenheim (02/28/2013 02:19 PM EDT)
• Benjamin Burch (02/28/2013 05:46 PM EDT)