(In reply to
quite easy solution for integers by armando)
To take a simple example:
17/31+14n/31= m^2
(14n+17)/31 = m^2, which is square for m=1.
It is also a whole number whenever n = (31k+1); 1, 32, 63, etc
Now we need only x^2 = (31k+1)
(x^21) = 31k, with a solution at, e.g. k = 31n^2  60n + 29, x = 30  31n, etc.
Generalising:
if D/N+dn/N=m^2, where N is the common denominator, D is the numerator of the first term, d is the denominator of the common difference, n is the number of repeats of the common difference, and m^2 is a square appearing in the sequence, then
(1) since m^2 is a whole number, there must be some common denominator, N.
(2) For the same reason, D/N+dn/N is also a whole number whenever n = (Nk+m^2).
Further,
(3) since m^2 is a square, it must be of the form (4k+1) or 4k. This eliminates cases where D/N+dn/N is an integer only for values (4k+2) or (4k+3), which could never be square.
Now we only need x^2 = (Nk+m^2) for some new k, which accedes to the same reasoning given in earlier posts.
Edited on March 29, 2018, 4:47 am

Posted by broll
on 20180329 04:37:32 