There's an easy way to come up with your own Pythagorean triples. Set a value for a, then c-b and c+b are factors of a^2. For example, if you set a to 119, you could set c+b to 289 and c-b to 49 and solve for b and c. 119-120-169 right triangle.
To limit the results to the relatively prime, just consider the cases where m and n are relatively prime themselves. (Notice that while a is always even, if b and c are also even the result is no longer relatively prime, so if b and c are odd m and n must have one odd and the other even.)
Your answer to this classical problem, namely: (a, b, c) = (2mn, m^2 - n^2, +-(m^2 + n^2)) or (a, b, c) = (m^2 - n^2, 2mn, +-(m^2 + n^2)) misses the solution (a, b, c) = (9, 12, 15). It does find, among others, those solutions where GCD(a, b) = 1. A correct answer is: (a, b, c) = k(2mn, m^2 - n^2, +-(m^2 + n^2)) or (a, b, c) = k(m^2 - n^2, 2mn, +-(m^2 + n^2)) , where k, m, and n are integers. Note that (k, m, n) = (3, 2, 1) yields (a, b, c) = (9, 12, +-15).
@@someperson188 Someone else already commented on this. "Sybermath" implicitly assumed that m and n have no common factors. (6, 8, 10) is just a multiple of one of those solutions.
You made a mistake: your listed solutions do not include all Pythagorean triples. For example, a=12, b=9, c=15 is a Pythagorean triple, being 3 times the famous triple (4, 3, 5). But it is easy to see that c=15 is not the sum of two squares, so that triple is left out of your list of solutions. The problem is that you seem to assume that m and n are relatively prime; i.e. that their greatest common divisor is 1. But that leaves out my triple. That can be fixed by adding one more positive integer parameter: let's call it k. Then all Pythagorean triples are given by (2kmn, k(m²-n²), k(m²+n²)) or by switching the first two numbers (a and b). Then my triple is given by k=3, m=2, n=1. The redundancy in that list of solutions can be reduced if you impose these restrictions on the parameters: k is any positive integer. m and n are positive integers such that: m > n m and n have opposite parity: one of them is even and the other is odd m and n are relatively prime: their greatest common divisor is 1. I remember solving this problem way back in math camp (aka SSTP or the Ross Mathematics Program) when I was in high school. My method of solution took more steps but was more straightforward than your method. You have fewer steps but some of those steps are "tricks." Thanks for bringing back the memories.
![เปิดบ้าน นาราเครปกระเทย 1 ไร่ ขายขาดทุนหลัก10ล้าน เอามาใช้หนี้!!! l [Nickynachat]](http://i.ytimg.com/vi/0jz84VFWVP4/mqdefault.jpg)
Nice video Sybermath as always! Always glad to collaborate with you friend!
Is your video out yet? I didn't find it.
Thank you, man! Likewise 🥳😍😄
This was cool. Thank you
Euclid already found the formula :
Let 2 integers u et v with 1 ≤ v < u .
Then a = 2uv ; b = u`^2 - v^2 ; c = u^2 + v^2.
I’ll be sharing that with teenagers, as it’s digestible and fascinating. Thanks
There's an easy way to come up with your own Pythagorean triples. Set a value for a, then c-b and c+b are factors of a^2. For example, if you set a to 119, you could set c+b to 289 and c-b to 49 and solve for b and c. 119-120-169 right triangle.
Reminds me of a joke where someone says they have CDO. That's OCD, but the letters are in the proper order.
I can clearly see an influence of BlackPenRedPen videos on this idea. :-)
To limit the results to the relatively prime, just consider the cases where m and n are relatively prime themselves.
(Notice that while a is always even, if b and c are also even the result is no longer relatively prime, so if b and c are odd m and n must have one odd and the other even.)
Nice!
Your answer to this classical problem, namely:
(a, b, c) = (2mn, m^2 - n^2, +-(m^2 + n^2)) or (a, b, c) = (m^2 - n^2, 2mn, +-(m^2 + n^2))
misses the solution (a, b, c) = (9, 12, 15). It does find, among others, those solutions where GCD(a, b) = 1. A correct answer is:
(a, b, c) = k(2mn, m^2 - n^2, +-(m^2 + n^2)) or (a, b, c) = k(m^2 - n^2, 2mn, +-(m^2 + n^2)) ,
where k, m, and n are integers.
Note that (k, m, n) = (3, 2, 1) yields (a, b, c) = (9, 12, +-15).
Another method of solution uses Gaussian integers. That might fit with you a + bi channel.
In other words, either A or B have to be odd, while the other is even, resulting in C, which always has to be odd.
How about (a, b, c) = (6, 8, 10)?
@@someperson188 This can be reduced to (a, b, c) = (3, 4, 5). C is odd.
@@someperson188 Someone else already commented on this. "Sybermath" implicitly assumed that m and n have no common factors. (6, 8, 10) is just a multiple of one of those solutions.
If it's assumed that GCD(a,b) =1, then Sybermath's solution must say that
GCD(m, n) = 1 and m and n are of opposite parity.
a = 3, b = 4, c = 5
✌️😎👍👏
c=(a×b)/2-1
If a=5 and b=12, c=13
You made a mistake: your listed solutions do not include all Pythagorean triples. For example, a=12, b=9, c=15 is a Pythagorean triple, being 3 times the famous triple (4, 3, 5). But it is easy to see that c=15 is not the sum of two squares, so that triple is left out of your list of solutions.
The problem is that you seem to assume that m and n are relatively prime; i.e. that their greatest common divisor is 1. But that leaves out my triple. That can be fixed by adding one more positive integer parameter: let's call it k. Then all Pythagorean triples are given by
(2kmn, k(m²-n²), k(m²+n²)) or by switching the first two numbers (a and b).
Then my triple is given by k=3, m=2, n=1.
The redundancy in that list of solutions can be reduced if you impose these restrictions on the parameters:
k is any positive integer.
m and n are positive integers such that:
m > n
m and n have opposite parity: one of them is even and the other is odd
m and n are relatively prime: their greatest common divisor is 1.
I remember solving this problem way back in math camp (aka SSTP or the Ross Mathematics Program) when I was in high school. My method of solution took more steps but was more straightforward than your method. You have fewer steps but some of those steps are "tricks." Thanks for bringing back the memories.
Wow! Thanks for sharing
"If and only if" assumes proof in 2 ways.