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Consecutive Contemplation (Posted on 2012-11-08) Difficulty: 3 of 5
Each of the n consecutive positive integers x+1, x+2, ...., x+n is expressible as the sum of squares of two distinct positive integers.

Determine the maximum value of n and prove that no higher value of n is possible.

No Solution Yet Submitted by K Sengupta    
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possible solution | Comment 5 of 6 |

... or possibly I've misunderstood the problem, which seems quite simple.

The sum of squares of two distinct positive integers can equal 4k+0, or 1 or 2, but cannot equal 4k+3. So every number of the form 4k+3 is NOT the sum of two distinct positive integral squares, and accordingly the answer is that n<=3.

This theoretical n is reached, for example, at x=71, since 72, 73, 74 are all sums of two squares. However, 72=36+36, and those squares are not distinct. But 232=14^2+6^2, 233=13^2+8^2, 234=15^2+3^2, hence if x=231, n=3, confirming that the maximum value of n is 3.

Compare Squares by the dozen, which I posted a while ago.

Addendum: Having wasted a bit more time on this out of curiosity, I found A064715 in Sloane which contains a partial list of the terms: 232, 520, 584, 800, 808, 1096, 1224, etc.

Interestingly, if the requirement for distinct positive integers is relaxed (i.e. one of the sums of squares = 2k^2) then taking the triangular numbers (3,6,10,15..) doubling them, squaring them and doubling them again will produce the smallest of 3 numbers that satisfies the condition, as can easily be shown.

It's not so easy to produce a single formula that solves the original problem, though.


Edited on November 9, 2012, 8:00 am
  Posted by broll on 2012-11-09 02:04:01

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