Question about perfect squares and a pattern.

Factual Questions
1

OK, so we all know that the first 10 perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100. This is important to my question because there is a certain pattern that I observed a long time ago and I was wondering if it had a name.

The pattern is thus: the next perfect square is the result plus 2 times the number squared plus 1. For instance, go from 4 to 9. 9-4 is 5, which is 2 times 2 plus 1. Now go from 9 to 16. 16-9 is 7, which is 2 times 3 plus 1. And so on and so on.

First of all, why is this the case? Second of all, is there a name for this relationship? Third, is there some elaborate and easily understandable to the layman proof for this? Last, does the pattern hold indefinitely, or is it just coincidence at the more common numbers?

2

Not very difficult to demonstrate if you just write down what you are saying:

n[sup]2[/sup]+2n+1 = (n+1)[sup]2[/sup]

3

Even easier is to look at a picture. Here’s 1[sup]2[/sup]:

X

To get to 2[sup]2[/sup], we want to add a new column on the side, and a new row on the bottom, and one in the corner:

XX
XX

That’s adding 2*1+1 squares. Now to get to 3[sup]2[/sup], we do the same thing:

XXX
XXX
XXX

so 3[sup]2[/sup] = 2[sup]2[/sup] + 2*2+1

Jumping ahead, in the same way, if we have 9[sup]2[/sup] and we want to get to 10[sup]2[/sup], we need to add one column of 9 along the side, and a new row of 9 on the bottom, and one in the corner:

XXXXXXXXXX
XXXXXXXXXX
XXXXXXXXXX
XXXXXXXXXX
XXXXXXXXXX
XXXXXXXXXX
XXXXXXXXXX
XXXXXXXXXX
XXXXXXXXXX
XXXXXXXXXX

That’s an addition of 2*9+1, and so on.

4

:smack:

I haven’t done math in a long time. Sorry for bothering you guys with such an obvious question.

5

Define [symbol]D[/symbol][sub]h[/sub]f(x) = f(x + h) - f(x). Then [symbol]D[/symbol][sub]h[/sub]x[sup]2[/sup] = 2hx + h[sup]2[/sup]. In particular, [symbol]D[/symbol][sub]1[/sub]x[sup]2[/sup] = 2x + 1. This may seem a bit overly formal, but the [symbol]D[/symbol][sub]h[/sub] operator is pretty important in various applications.

6

Any function x[sup]n[/sup] can be expressed as a summation of a function whose… blanking on a term here, order? is (n-1).

I worked out a lot of them for fun while bored at work a few years back.

7

Degree.

8

So, x[sup]2[/sup] = Σ 2x - 1 , as x goes from 1 to n, where n is the value for which you wish to evaluate x[sup]2[/sup].

Caveat : To make it work for negative numbers requires some extra manipulation.

Caveat the second: You can also express it as Σ 2x + 1, but then you must evaluate it from 0 to n-1.

9

THANK you. For the life of me, I could not remember that today.

10

Sounds like you guys are alluding to the Calculus of Finite Differences. An easy introduction to the subject is printed as Chapter 2 of Martin Gardner’s Colossal Book of Mathematics.

11

You’d probably enjoy this page:

12

This patter also explains why there is an infinate number of Pythagorean triples that are integers; i.e., there is an infinite number of non-proportional integer triples that satisfy the Pythagorean Theorem: a[sup]2[/sup] + b[sup]2[/sup] = c[sup]2[/sup].

The difference between any two adjacent square of integers is an odd number: 2[sup]2[/sup]-1[sup]2[/sup]=3, 3[sup]2[/sup]-2[sup]2[/sup]=5. The difference increases by 2, so that every odd number is the difference between two adjacent squares.

Additionally, every square of an odd number is another odd number.

So, since every odd number is a difference of two squares, and there are an infinite number of perfect squares that are odd, there will be an infinite number of integer triples that satisfy the Pythagorean equation.

In general:
(2n+1)[sup]2[/sup]+([(2n+1)[sup]2[/sup]-1]/2)[sup]2[/sup]=([(2n+1)[sup]2[/sup]+1]/2)[sup]2[/sup]
where n is in the set of natural numbers

This produces the triples:
3 4 5
5 12 13
7 24 25
9 40 41
11 60 61
13 84 85

13

Could be. Like I said, this was just something I was wasting time with one day.

To achive the summation equations for higher exponents, relatively quickly, you can use Pascal’s Triangle.

14

I think a simpler way to say this is that the differences between the terms follows the pattern 3, 5, 7, 9, 11, 13, …, in other words, the odd numbers. I remember in high school algebra 30 years ago another kid in my class noticed that pattern.

Difference in the square series terms:
n[sup]2[/sup] - (n-1)[sup]2[/sup]
n[sup]2[/sup] - [sup]2[/sup] + 2n - 1
2n - 1, which is how to describe a series of odd numbers.