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+14a+1, which is not very hard to prove.

Consider y2−1y^2-1, for some odd integer yy. Now, this can be factorized into (y+1)(y−1)(y+1)(y-1). Now, 2∣y+1,2∣y−12|y+1,2|y-1, since yy is odd. Thus, 4∣y2−14|y^2-1, or any odd y2y^2 can be expressed as 4a+14a+1.

Now, I was making a table of odd integers and their squares to find any patterns. Only after 2525, did I realise a pattern.

For 4949, a=12a=12.

For 8181, a=20a=20.

For 121121, a=30a=30.

For 169169, a=42a=42.

I realised a pattern, in which the difference between consecutive values of aa are in an arithmetic progression with difference of 22.

I tested my theory, and found it to comply, as for 225225, a=56a=56, which I predicted.

Is their a proof for this theory? Did I stumble upon some intricate pattern?

#Algebra #QuadraticEquations #ArithmeticProgression(AP) #PatternRecognition #RootsOfQuadraticEquation

Note by Nanayaranaraknas Vahdam
6 years, 12 months ago

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Comments

From the above question, a=x2−x=x(x−1)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,...2,6,12,20,30,42,...

Samuraiwarm Tsunayoshi - 6 years, 12 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.

Bogdan Simeonov - 6 years, 12 months ago
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