Each of n positive integers x+100, x+200, ..., x+100*n, which are n consecutive terms of an arithmetic sequence with common difference of 100, 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.
I think the answer may be 5.
Numbers only seem to work if they have p^(2n) as a factor where p is a prime of the form 4k+3.
In other words multiples of 3 don't work unless they are multiples of 9.
Since adding 100 adds 1 to the residual mod 9, you'd want your numbers to be 7,8,0,1,2 mod 9.
They also can't be multiples of 7, 11 etc. but these can be avoided easily enough.
So I'm pretty sure no higher number is possible. I haven't found a case that shows n=5 is possible.

Posted by Jer
on 20130211 14:49:11 