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

### a^{2} + b^{2} = (b + 3)^{2}

My original notes (not in my journal) had a mistake or an omission in them, but my discovery in part:

^{2}+ b

^{2}= b

^{2}+ 6b + 9 and

a

^{2}= 6b + 9 = 3(2b + 3)

So a

^{2}, and therefore a, is a multiple of 3.

The omission is proof that b (and therefore c, which is 3 more than b) is also a multiple of 3.

This would mean that the triple can be reduced.

All examples that I can find in this form can be reduced. That isn't proof. I need a new approach to showing that this is always the case.

### a^{2} + b^{2} = (b + 4)^{2}

This one is actually easier to show than the previous one.

^{2}+ b

^{2}= b

^{2}+ 8b + 16 and

a

^{2}= 8b + 16 = 8(b + 2)

b = (1/8)a ^{2} - 2 | c = (1/8)a ^{2} + 2 |

The only way for (1/8)a

^{2}to be a whole number is if a is a multiple of 4.

If a = 4n then

(1/8)a

^{2}= (1/8)(4n)

^{2}= (1/8)(16n

^{2}= 2n

^{2},

which is even.

So a is even, b, which is 2 less than an even number, must be even and c, which is 2 more than an even number, must be even.

Therefore, any Pythagorean Triple in this form can be reduced by a factor of 2.

I am content to believe at this point that checking any Pythagorean Triple of the form

**a**,

^{2}+ b^{2}= (b + n)^{2}where n is a Natural Number and n > 2

would be a waste of time. Particularly because I've crunched hundreds of numbers on an Excel spreadsheet. Not proof, naturally. But if any triples exist, they will be so high as to be useless on an Algebra test.

However, I am not quite done. There is one more kind of triple that I hadn't been aware of until (relatively) recently. And I found it in a fifty-year-old book. I guess I never encountered triangles like these when I was in school because I think I would have remembered!

### For my students

Can you guess what the next set of right triangles will look like?

Maybe your previous teachers taught you more than mine did.

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