Maths Problem

[Kuciwalker]

Do tell.

[Snowflake]

This is pretty cool though for one to memorize the triplets. For example if we want to see what is the triplets that has 9 as the shortest side. We then get 9²=81=40+41. We then know the triplet is 9, 40, 41, without having to check if it fits a²+b²=c².

[Snowflake]

Originally posted by Kuciwalker
Do tell.

a²=c²-b²=(b+c)(c-b)
Therefore a²/(c-b)=b+c

It's very simple, actually, I just have never thought of it this way.

[Urban Ranger]

Originally posted by Snowflake
This is pretty cool though for one to memorize the triplets. For example if we want to see what is the triplets that has 9 as the shortest side. We then get 9²=81=40+41. We then know the triplet is 9, 40, 41, without having to check if it fits a²+b²=c².

Yeah, but many pairs of integers add up to 81.

[Snowflake]

You need to make it b+c=81 and b-c=1 though.

[Urban Ranger]

Originally posted by Snowflake
You need to make it b+c=81 and b-c=1 though.

Yes.

Is this a special case or the general case? IOW, is b-c always 1?

[Snowflake]

No, but this is the easy one to get. For example, if a=11, you get 121=60+61. If a=13, you get 169=84+85. If a = 15, you get 225=112+113 ...
So your triplets would be 11, 60, 61; 13, 84,85; 15, 112, 113 ...

For even numbers it seems you need to make b-c=2 or something ...

[Snowflake]

Ok for even numbers you cut the square into half before you figure the b and c out. For example, if a=10, you get 100/2=50=24+26, so the triplet is 10, 24,26. If a=12, you get 144/2=72=35+37, so the triplets is 12, 35, 37. If a =14, you get 196/2=98=48+50, and the triplets is 14, 48, 50 ...

[KrazyHorse]

Originally posted by Snowflake
Ok this is another strange thing that I never thought about before. Remember some of the Pythagorean triplets? 3,4,5; 5,12,13; 7,24,25 ... Well obviously we know that for them a²+b²=c². However, it seems that there's something else that are true: 3²=4+5, 5²=12+13, 7²=24+25 ...

I wonder if this is merely coincidental, or is there something there that could be proved. I also tried the set 8,15,17, while 8²is not equal to 15+17, if you cut them by half, 4²=15/2+17/2 ... Can we prove that for each of the triplets there always exist a factor where we could make this relationship to be true?

Yes. Duh.

(n+1)^2 - n^2 = 2n + 1 = n + (n+1)

Simple algebra...

[KrazyHorse]

In fact, it's quite easy to show that up to a scale factor all pythagorean triplets are equivalent to a = (2n + 1), b = (2n^2 + 2n), c = (2n^2 + 2n + 1)

[Snowflake]

Originally posted by KrazyHorse


Yes. Duh.

(n+1)^2 - n^2 = 2n + 1 = n + (n+1)

Simple algebra...

You are assuming the two longer sides are continous integers, which may or may not be the case, as shown above. However you are right the algebra is very simple. I'm just happy that I found a way to get the Pythagorean triplets for any number with ease.

[KrazyHorse]

It is the case up to a scale factor, as I've said. For all intents and purposes 3 4 5, 5 12 13, 7 24 25 etc. are the only pythagorean triplets. All others are direct multiples of these.

[Snowflake]

Originally posted by KrazyHorse
In fact, it's quite easy to show that up to a scale factor all pythagorean triplets are equivalent to a = (2n + 1), b = (2n^2 + 2n), c = (2n^2 + 2n + 1)

Good thinking. If you got an even number you can simply scale it back to an odd number before you get the triplets. Although this would actually give you different solutions sometimes ... Hmmm

For example for 12, I would get 12,35,37, and you would get 12,16, 20 ...

[Snowflake]

In view of your last post, the one I'd get may be a pythagorean triplet and the one you'd get would be a direct multiple.

The thing is for your method it is easy to get a real pythagorean if a is an odd number, but not if it is an even number.

[Snowflake]

Looks to me if a is an odd number, or multiples of four, then you'd be able to get a real pythagorean triplet, otherwise they would only be multiples ...