Given two numbers, a perfect square 's' and any positive integer 'd' is there a situation where s, s+d, s+2d, and s+3d are all perfect squares?

Question

I have been messing around with a question over the last few months and haven't been able to make any headway.

Say you have a square number 's' and another positive integer (square or otherwise) 'd'. Is there a situation where s, s+d, s+2d, and s+3d are all square numbers?

Using a python script, I have found mounds of solutions up to the s+2d case, but haven't found any solutions that also satisfy s+3d. Am I in over my head here?

Example:

• s = 2500
• d = 60000
• s = $50^2$
• s + d = 2500 + 60000 = 62500 = $250^2$
• s + 2d = 2500 + 120000 = 122500 = $350^2$
• s + 3d = 2500 + 180000 = 182500 = 427.2001... (fail)

Answer 1

There are no solutions with 4 squares in an arithmetic progression. The proof seems involved. See https://www.mathpages.com/home/kmath044/kmath044.htm

Answer 2

According to Wikipedia, you're in good company. The non-existence of $4$ squares in arithmetic progression was conjectured by Fermat and proved by Euler.