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

1 vote185 views1 answer
Flatten image python numpy
Python
19 votes210 views6 answers
Pip install django
Python Django
0 vote162 views2 answers
Converting list of arrays with same size to single array python
Python
0 vote170 views1 answer
Sorted key len python
Python
44 votes187 views2 answers
Repl.it python
Python
1 vote177 views1 answer
How to pronounce aesthetic
Python
4 votes169 views2 answers
Is python easy or hard to learn
Python
0 vote171 views1 answer
Python not readable file
Python
3 votes191 views1 answer
Streamlit ssl error
Python
2 votes178 views1 answer
How to move a button lower on a gui tkinter
Python