Arithmetic Series Perfect Square

Frillth
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Homework Statement



I need to find all arithmetic sequences of integers with the property that the sum of the first n terms is a perfect square for all integers n.

Homework Equations



a_n = nth term of the sequence = a_1 + (n-1)d
d = common difference
Sum of the first n terms of the sequence = n[2a_1+(n-1)d]/2

The Attempt at a Solution



I know that the sequence 1, 3, 5, 7, 9... sums to a perfect square every time, as will 1x^2, 3x^2, 5x^2..., with x being any integer.

This is the only way I can find to make the sequence sum to a perfect square every time. If this isn't the only way, what are the others? If this is, how can I prove it is the only way?
 
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well you noticed that \sum^n_{m=0} 2m+1 = (n+1)^2. Note that (n+1)^2-n^2=2n+1. You have made a sequence where adding a successive term makes the sum equal the successive square. What is the (n+2)^2-n^2? Certainly you can set a_1 to be a perfect square, then you are guaranteed to have such a sequence that you want. I hope I made enough sense without giving away too much away.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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