Let m^2 be the known square in the sequence and d be the difference between successive members. Show that there is a subsequent square, n^2, also in the sequence:
m^2 + kd = n^2
kd = n^2  m^2 = (n  m)(n + m)
let (nm) be any multiple of d, so nm = i*d
kd = i*d * (i*d + 2m)
k = i*d*(i*d+2m) / d = i(i*d + 2m)
k is the difference between m and n, but i*d is the difference of the squares that is the multiple of the difference between successive members of the series.
Now n^2 can be used as the new m^2 in another iteration of this process, and so on ad infinitum.
As an example, take m^2 = 25 and it's part of an arithmetic series with a difference, d of 17. Let i be 1. Then k = 17+10=27, so n = 5+27*17 = 464 and 464^2 = 215296, which is 25 + 12663*17, another member of the sequence. Keep repeating from there with m now equal to 464, etc.

Posted by Charlie
on 20180328 10:14:13 