New Algorithm - Pythagoras Generator

[Naman]

I've written an algorithm which when given input, will return you a pythagorean triplet including that number.
It works for EVERY number (provided n > 2). Do let me know if you guys are interested, I'll make a PR with my code attached!

[Subham Kumar Singh]

I am new to open-source.Please assign me.

[Rutuja Kawade]

I can complete it. Please assign it to me

[Abhijit Kailash Gawai]

Bro, so you are saying you have written pythagoras genrator. And you will make PR. Then why you are listing this as issue. Please tell me . I am new to open source, i dont dont how this works

[Driptanil Datta]

I can complete this Pythagoras Generator, Please assign this to me
I am new to opensource and I am willing to learn and contribute.

[Kasjan Chilarski]

Bro, so you are saying you have written pythagoras genrator. And you will make PR. Then why you are listing this as issue. Please tell me . I am new to open source, i dont dont how this works

I think @Naman-Singla777 is asking the maintainers to accept his proposal so he can make a pullrequest. He is not asking anyone to collaborate with him or solve this issue.

[Sadegh Pouriyan Zadeh]

Bro, so you are saying you have written pythagoras genrator. And you will make PR. Then why you are listing this as issue. Please tell me . I am new to open source, i dont dont how this works

I think @Naman-Singla777 is asking the maintainers to accept his proposal so he can make a pullrequest. He is not asking anyone to collaborate with him or solve this issue.

So where is his proposal? I can't find anything here!

[rajadhruwe-ui]

I am new on python

[Ashwanth K]

I can do this work!
I love python.
I am new to open source
I am a college student pursuing CSE

[Ashwanth K]

"""
Description: triplet(n) function displays all the possible pythagorean triplets where one of the number is n

>>> triplet(5)
5 12 13
3 4 5
Count:  2
>>> triplet(17)
17 144 145
8 15 17
Count:  2
>>> triplet(35)
35 84 91
35 120 125
12 35 37
35 612 613
21 28 35
Count:  5
>>> triplet(31)
31 480 481
Count:  1
>>> triplet(30)
30 40 50
30 224 226
16 30 34
30 72 78
18 24 30
Count:  5
>>>

"""

def triplet(n):
    ans = set()
    for i in range(1,int(n)+1):                              # triplets where a^2 + n^2 = b^2
        if (n*n)%i == 0:
            c = i
            d = (n*n)//i
            a = (c+d)/2
            b = (d-c)/2
            if a%1 == 0 and b%1 == 0 and a>0 and b>0:
                ans.add(tuple(sorted([a,b,n])))

                
    for a in range(1 , n+1):                                # triplets where a^2 + b^2 = n^2
        b = (n**2 - a**2)**0.5
        if b%1 == 0 and b>0:
            ans.add(tuple(sorted([a,b,n])))
    c = 0
    for (x,y,z) in ans:
        print(int(x) , int(y)  , int(z))
        c += 1
    print("Count: ",c)
    

I've written an algorithm which when given input, will return you a Pythagorean triplet including that number. It works for EVERY number (provided n > 2). Do let me know if you guys are interested, I'll make a PR with my code attached!

[stale]

This issue has been automatically marked as stale because it has not had recent activity. It will be closed if no further activity occurs. Thank you for your contributions.

[Farhaan Nazirkhan]

Hello,
Can you assign this issue to me?