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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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Solution solution assuming all members of the series are integers | Comment 3 of 6 |
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 (n-m) be any multiple of d, so n-m = 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 2018-03-28 10:14:13
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