Choose any odd number. Square it and divide by two. Now get two numbers, one by rounding up and the other by rounding down. What you get is a Pythagorean Triple. I want to see if anyone can figure out
why this happens. I figured this out independently when I was thinking about difference of squares. I thought it was really cool, especially since it was something I found out myself without even trying to discover anything (lots of times I will try to think of why something someone is telling me is true, so I'll want to figure that out). The way I did it involves difference of squares. I'll start you out with a question: In your head, what's 96*104?
[A spoiler showing you how I figured out my pythagorean triple thing would be useful here. Any way you think you can allow spoilers in these fora Andy?]
Hint #1:
Well what is 96*104? Answer: 96=100-4, 104=100+4, so 96*104=(100-4)(100+4). Then, DOS (Difference of Squares), 100^2-4^2, which is 10000-16, which is 99984. Cool, right?
Hint #2:
You can do that (hint #1) with any 2 numbers equidistant from one hundred. Also, when they're equidistant from a number, n it would just be (n+d)(n-d), where d is the distance each number is from n. In fact, for any two numbers, a and b, there is one number that lies between them, which you can find through (a+b)/2, so you can use this for any product.
Hint #3:
If a is any natural number, what's a*1?
Hint #4:
When is the number between a and 1 whole? That would be when it's odd. Try change a*1 to a difference of squares multiplication for a few odd numbers: 1*1=?, 3*1=?, 5*1=?, 7*1=? For example, the number between 1 and 1 is 1, and the difference between those two numbers is zero, so you'd get (1-0)(1+0).
Hint #5:
For 1*a for the first few odd natural numbers a you'd get 1*1=(1+0), 1*3=(2-1)(2+1), 1*5=(3-2)(3+2), 1*7=(4-3)(4+3). The pattern here, obviously, is that the middle number between 1 and a goes by 1 when a goes up by two. This is because a's weight on the middle number is 1/2, so an increase of two would have 50% weight, making it an increase of 1 on the middle number. Since the middle number goes up one and 1 stays constant, the difference between 1 and that number goes up one each time a increases by two.
Hint #6:
So, how could we write 1*a in terms of a using difference of squares where a is an odd natural number?
Hint #7:
Well, since the middle number between the odd natural number a and 1 is (1+a)/2, which is 1/2+a/2, and the difference is always one less than the middle number, which is a/2+1/2-1, which is itself a/2-1/2, 1*a=[(a/2+1/2)-(a/2-1/2)]*[(a/2+1/2)+(a/2-1/2)]
Hint #8:
Don't simplify according to your instincts. Remember DOS!
Hint #9:
=(a/2+1/2)^2-(a/2-1/2)^2, and since an odd number divided by two plus a half is a whole number, and the same goes for subtracting a half, both of these are whole number when a is an odd whole number.
Hint #10:
What happens if a is a perfect square?
Final hint:
if a is a perfect square, you get sqrt(a)^2=(a/2+1/2)^2-(a/2-1/2)^2. And since, through the definition of perfect square, sqrt(a) is a perfect square, and because the other terms on the other side are whole numbers (we proved this earlier), we can, with any odd number x, square x to get an a that satisfies the pythagorean theorem using only whole numbers. If you want this all to look more pythagor-y to you, you can do this: sqrt(a)^2+(a/2-1/2)^2=(a/2+1/2)^2. QuantumElectricDynamics.
Since I'm out of time, I couldn't create a final answer spoiler all-in-one, but you can just go through the hints.
Nobody ever notices my signature. ):