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, 7 months ago
Easy Math Editor
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
Sort by:
Top NewestFrom 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 \(2,6,12,20,30,42,...\)
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.
Log in to reply