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# Find all integer solutions for $a_1(a_1+k)(a_1+2k)...(a_1+(2k+1)d)+1 = n^2$ [closed]

+1
−3

Closed as not constructive by beloh‭ on Dec 22, 2022 at 10:47

This question cannot be answered in a way that is helpful to anyone. It's not possible to learn something from possible answers, except for the solution for the specific problem of the asker.

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Find all integer solution for the equation below: $$a_1(a_1+k)(a_1+2k)...(a_1+(2k+1)d)+1 = n^2$$

=$$1+\prod_{k=0}^{2k+1} (a_1 + kd) = n^2$$

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Is there supposed to be +1 in the title? (1 comment)
Video (1 comment)

+3
−0

The initial observation is explained by the identity $$(x^2 + 3x + 1)^2 = x(x+1)(x+2)(x+3) + 1$$

The generalisation is messed up in the question as it currently stands, but the only smallish similar identity which I can find is the uninteresting $$(x+1)^2 = x(x+2) + 1$$

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