 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  2011 Square Sum Start (Posted on 2011-12-31) Determine the smallest positive integer, expressible as the sum of two nonzero perfect squares, whose base ten representation begins with 2011 (reading left to right). What is the next smallest integer with this property?

How about the smallest positive integer, expressible as the sum of two positive perfect cubes, whose base ten representation begins with 2011 (reading left to right)?

*** For an extra challenge, solve this puzzle without using a computer program.

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Possible solution Comment 1 of 1

1. Generally, check the prime factors for congruence to 1,3 mod 4. 2011 is prime. 2011 mod 4 is 3, so 2011 is not a sum of two squares, and hence neither is 20110. 20111= 7�13^2�17 which is not a sum of two squares. 20112= 2^4�3�419 which is not a sum of two squares. 20113 is a prime of the form 4k+1 and hence the sum of two squares (87^2+112^2). 20114=2�89�113 which is a sum of two squares (67^2+125^2), (83^2+115^2).

2. I am not aware of a similar rule for cubes. However, 201159= 46^3+47^3 (a 'primitive' solution) and 2011024= 22^3+126^3= 8*(11^3+63^3); if a^3+b^3=x then (a^3+b^3)y^3 is clearly also a sum of two cubes.

Edited on January 4, 2012, 10:55 pm
 Posted by broll on 2012-01-04 22:45:10 Please log in:
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