I was going about finding the pattern of integers \(a\) such that \(x+a=x^2\) has at least one rational solution. Using the quadratic formula, the value boiled down to \(\frac{1\pm\sqrt{1+4a}}{2}\). Thus, this occurs when \(4a+1\) is the square of an integer.
I knew that odd squares can be expressed as , which is not very hard to prove.
Consider , for some odd integer . Now, this can be factorized into . Now, , since is odd. Thus, , or any odd can be expressed as .
Now, I was making a table of odd integers and their squares to find any patterns. Only after , did I realise a pattern.
For , .
For , .
For , .
For , .
I realised a pattern, in which the difference between consecutive values of are in an arithmetic progression with difference of .
I tested my theory, and found it to comply, as for , , which I predicted.
Is their a proof for this theory? Did I stumble upon some intricate pattern?
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Top NewestFrom the above question, a=x2−x=x(x−1). The examples you're interested in positive integers, that will be the common pattern of 2,6,12,20,30,42,...
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It's pretty easy to see that the difference of two consecutive a's is 2 by using that y must be odd and writing it as 2k-1.
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