# A Pattern in the Quadratic Formula!

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 $4a+1$, which is not very hard to prove.

Consider $y^2-1$, for some odd integer $y$. Now, this can be factorized into $(y+1)(y-1)$. Now, $2|y+1,2|y-1$, since $y$ is odd. Thus, $4|y^2-1$, or any odd $y^2$ can be expressed as $4a+1$.

Now, I was making a table of odd integers and their squares to find any patterns. Only after $25$, did I realise a pattern.

For $49$, $a=12$.

For $81$, $a=20$.

For $121$, $a=30$.

For $169$, $a=42$.

I realised a pattern, in which the difference between consecutive values of $a$ are in an arithmetic progression with difference of $2$.

I tested my theory, and found it to comply, as for $225$, $a=56$, which I predicted.

Is their a proof for this theory? Did I stumble upon some intricate pattern? Note by Nanayaranaraknas Vahdam
6 years, 5 months ago

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From the above question, $a = x^{2} - x = x(x-1)$. The examples you're interested in positive integers, that will be the common pattern of $2,6,12,20,30,42,...$

- 6 years, 5 months ago

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.

- 6 years, 5 months ago