# 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
4 years, 8 months ago

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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.

- 4 years, 8 months ago

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 $$2,6,12,20,30,42,...$$

- 4 years, 8 months ago