Using the Pythagorean Theorem,
[tex]a^2+b^2=c^2[/tex]
(See attachment)
Let us substitute the values and solve for b in terms of [tex]x[/tex].
[tex](x^2-1)^2+b^2=(x^2+1)^2[/tex]
Grouping the [tex]x[/tex] terms on one side gives,
[tex]b^2=(x^2+1)^2-(x^2-1)^2[/tex]
We now apply difference of two squares to obtain,
[tex]b^2=((x^2+1-(x^2-1))(x^2+1+(x^2-1))[/tex]
[tex]b^2=((x^2+1-x^2+1))(x^2+1+x^2-1))[/tex]
[tex]b^2=(2)(2x^2)[/tex]
[tex]b^2=4x^2[/tex]
[tex]b=\sqrt{4x^2}[/tex]
[tex]b=2x[/tex],
Testing for some few values greater than 1, we can generate the Pythagorean triples as follows;
When [tex]x=2[/tex]
[tex]a=2^2-1=3[/tex]
[tex]b=2(1)=2[/tex]
[tex]c=2^2 +1=5[/tex]
When [tex]x=3[/tex]
[tex]a=3^2 -1=8[/tex]
[tex]b=2(3)=6[/tex]
[tex]c=3^2 +1=10[/tex]
