- #1
Char. Limit
Gold Member
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- 22
I was pondering square numbers today, and I noticed something interesting: every natural number contains information to construct a Pythagorean triple. Let me show what I mean for an odd natural number, also using the binomial theorem for square quadratic equations (equations of the form [tex](x+y)^2[/tex])
Firstly, for a square number to be odd, it must satisfy the equation [tex]m^2=2n+1=n+(n+1)[/tex]. In other words, an odd square number is always equal to two integers one number apart. Seems obvious, huh? Well, it's important too. Bear with me. Now, here's another equation:
[tex]n^2+2n+1=(n+1)^2[/tex]
Now, let's rearrange that last equation:
[tex]2n+1+n^2=(n+1)^2[/tex]
Now, is it not true that any positive number is equal to the square of its square root ([tex]x=(\sqrt{x})^2[/tex])? So, this is true:
[tex]2n+1+n^2=(n+1)^2=(\sqrt{2n+1})^2+n^2[/tex]
Also, if [tex]x^2=2n+1[/tex], then [tex]x=\sqrt{2n+1}[/tex] and [tex]n=\frac{x^2-1}{2}[/tex]. Substituting some values gives us:
[tex]x^2+(\frac{x^2}{2}-\frac{1}{2})^2=(\frac{x^2}{2}-\frac{1}{2}+1)^2=(\frac{x^2}{2}+\frac{1}{2})^2[/tex]
In summary, a pythagorean triple can be formed with any odd number, half of its square minus one-half, and half of its square plus one-half. I have a marvelous equation for the even natural numbers, but it is too narrow to fit in the margin, and besides, I'm sure everyone here has done this before. Lastly, here are 7 examples:
[tex]3^2+4^2=5^2[/tex]
[tex]5^2+12^2=13^2[/tex]
[tex]7^2+24^2=25^2[/tex]
[tex]9^2+40^2=41^2[/tex]
[tex]11^2+60^2=61^2[/tex]
[tex]13^2+84^2=85^2[/tex]
[tex]15^2+112^2=113^2[/tex]
Firstly, for a square number to be odd, it must satisfy the equation [tex]m^2=2n+1=n+(n+1)[/tex]. In other words, an odd square number is always equal to two integers one number apart. Seems obvious, huh? Well, it's important too. Bear with me. Now, here's another equation:
[tex]n^2+2n+1=(n+1)^2[/tex]
Now, let's rearrange that last equation:
[tex]2n+1+n^2=(n+1)^2[/tex]
Now, is it not true that any positive number is equal to the square of its square root ([tex]x=(\sqrt{x})^2[/tex])? So, this is true:
[tex]2n+1+n^2=(n+1)^2=(\sqrt{2n+1})^2+n^2[/tex]
Also, if [tex]x^2=2n+1[/tex], then [tex]x=\sqrt{2n+1}[/tex] and [tex]n=\frac{x^2-1}{2}[/tex]. Substituting some values gives us:
[tex]x^2+(\frac{x^2}{2}-\frac{1}{2})^2=(\frac{x^2}{2}-\frac{1}{2}+1)^2=(\frac{x^2}{2}+\frac{1}{2})^2[/tex]
In summary, a pythagorean triple can be formed with any odd number, half of its square minus one-half, and half of its square plus one-half. I have a marvelous equation for the even natural numbers, but it is too narrow to fit in the margin, and besides, I'm sure everyone here has done this before. Lastly, here are 7 examples:
[tex]3^2+4^2=5^2[/tex]
[tex]5^2+12^2=13^2[/tex]
[tex]7^2+24^2=25^2[/tex]
[tex]9^2+40^2=41^2[/tex]
[tex]11^2+60^2=61^2[/tex]
[tex]13^2+84^2=85^2[/tex]
[tex]15^2+112^2=113^2[/tex]