*continuing my little self-indulgent trip through my math journal...*

Once I knew that 8, 15, 17 was a

**Pythagorean Triple**, I knew that there had to be others waiting to be found. And, naturally, 4, 3, 5 was just a little too obvious for me to notice. The flip side of this was if, for example, I wanted a triangle where the shortest side was 12, could I find the other sides easily.

a^{2} + b^{2} = (b + 2)^{2}a ^{2} + b^{2} = b^{2} + 4b + 4a ^{2} = 4b + 4a ^{2} = 4(b + 1) | c = b + 2 c = (1/4)a ^{2} - 1 + 2c = (1/4)a ^{2} + 1 |

Because a is even, a

^{2}is even and a multiple of 4. (2n)

^{2}= 4n

^{2}.

Therefore (1/4)a

^{2}is a Whole Number.

This works for every value of a that is an even number. However, when (1/4)a

^{2}is an odd number, b and c will also be even numbers. That means that they are not relatively prime and can be reduced by at least a factor of 2.

Only multiple of 4 produce reduced triples.

Since a is a multiple of 4, then a = 4n and

I prefer the latter form it is easier to halve and square then to square a bigger number and then divide by four.

If I see __, 63, 65, I know that I can start with 64, take the square root (which is 8) and double it to get 16.

Granted, I'm not really solving any of these. I'm just looking for problems for my students to solve the old-fashioned way.

What about triples in the form of a, b, b+3 or a, b, b+4 or higher?

That will be the next installment.

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