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Pythagorean Triples

When math textbooks need an example of a right triangle, they frequently use a triangle with sides of length 3, 4, and 5, since the numbers work out so nicely: {3^{2}+4^{2}=5^{2}} by the Pythagorean theorem. If that gets tiresome, 12, 5, 13 might be used: {5^{2}+12^{2}=13^{2}}. Clearly, multiplies of these numbers work also, e.g. {6^{2}+8^{2}=10^{2}.}

Such groups of integers are called pythagorean triples. Recently, I remembered that my high school algebra teacher once challenged the class to find another triple besides the two basic ones above. This was in the days before everyone had a calculator, so just trying out numbers would have been very tedious.

However, I found a way to create triples without trial and error, which I will describe below. It’s nothing profound, and is certainly not original to me, but the reader may find it interesting.

We want:

{x^{2}+y^{2}=z^{2},}where x, y, and z are integers

Rewriting

{x^{2}=z^{2}-y^{2}=\left(z+y\right)\left(z-y\right)}

or

{x=\sqrt{\left(z+y\right)\left(z-y\right)}=\sqrt{z+y}\bullet\sqrt{z-y}}

Now, if x is to be an integer, {\sqrt{z+y}} and {\sqrt{z-y}} must be integers also, so that {z+y} and {z-y} must be perfect squares:

{\begin{array}{l} z+y=a^{2}\\ z-y=b^{2}\end{array}}

Where a and b are integers, and x will be {ab}. Solving these equations for z and y:

{\begin{array}{l} z=\frac{1}{2}\left(a^{2}+b^{2}\right)\\ y=\frac{1}{2}\left(a^{2}-b^{2}\right)\\ x=ab\end{array}}

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This entry was posted on Wednesday, July 21st, 2010 at 12:30 pm and is filed under Mathematics History, Number Theory, Recommended Books. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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