Pe039

def number_of_solution(p):
    num = 0
    for a in range(1,p//3):
        if (p**2-2*p*a)%(2*p-2*a) == 0:
            num += 1
    return num

def main(n=1000):
    d = {p:number_of_solution(p) for p in range(2,n+1,2)}
    ans = max(d,key=d.get)
    return ans

Source: pe.metaquant.org

Similar codes

0 vote355 views1 answer
Python lane angle detection
Python
0 vote368 views1 answer
Root = tk() python 3
Python
0 vote365 views1 answer
Padnas check if string is in list of strings
Python
0 vote370 views1 answer
Code folding vim python
Python
0 vote373 views1 answer
Django update replace regex
Python Django
0 vote328 views1 answer
Pandas read csv skip rows
Python
0 vote361 views1 answer
Arrange array in ascending order python
Python
4 votes385 views4 answers
Recursion python examples
Python
0 vote389 views1 answer
How to compare values in dictionary with same key python
Python
0 vote372 views1 answer
This branch is for urllib3 v1.22 and later
Python