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One square implies many (Posted on 2018-03-28) Difficulty: 2 of 5
Prove the following:
If there is one perfect square in an arithmetic progression, then there are infinitely many.

No Solution Yet Submitted by Ady TZIDON    
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Extension to rationals. Comment 6 of 6 |
(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^2-1) = 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 2018-03-29 04:37:32

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