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?

## Comments

Sort by:

TopNewestFrom 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 \(2,6,12,20,30,42,...\) – Samuraiwarm Tsunayoshi · 2 years, 11 months ago

Log in to reply

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 · 2 years, 11 months ago

Log in to reply