Determine all possible triplets (p, q, n) of nonnegative integers that satisfy this equation:

                 2p + 2q = n! 

Provide valid argument for your answer.

*** Adapted from a problem which appeared at the Harvard/MIT math Olympiad.

A program finds only three such triplets for (p, q, n), if p <= q:
(0, 0, 2)
(1, 2, 3)

Completeness argument:
There can be no value of n greater than 6.  
Every factorial greater than or equal to 7! is divisible by 7.  But powers of 2 can only be {1,2,4} mod 7.  So there is no way to add two powers of 2 to get 0 mod 7. So the largest n! can be is 6! = 720.  Since 2^10 > 720, there is no need to test any p or q > 10
"""
def fac(n):
    """ Factorial """
    if n == 1 or n == 0:
        return 1
    else:
        return n*fac(n-1)

big = 100
factorials = [fac(i) for i in range(0,big)]

for p in range(11):
    for q in range(p,11):
        if (2**p + 2**q)%7 == 0:
            print(p,q)
        if 2**p + 2**q  in factorials:
            print('({}, {}, {})'.format    (p,q, factorials.index(2**p + 2**q))   )

Posted by Larry on 2023-02-27 16:02:12