Reply To: Maths

[ByteInBits]

I agree with @Astralica
\\ PARI-gp Code
\\ a! + b! = 2^n, Find all pairs for a,b and n.
{
\\ In true all the for()’s only need go from 1 to 10
for(n=1,1000,for(a=1,1000,for(b=1,1000,
if(a!+b!==2^n,
print(“a! = ” a! ” b! = ” b! ” 2^n = ” 2^n ” (where a:b:n was “a”:”b”:”n”)”)))));
}
a! = 1 b! = 1 2^n = 2 (where a:b:n was 1:1:1)
a! = 2 b! = 2 2^n = 4 (where a:b:n was 2:2:2)
a! = 2 b! = 6 2^n = 8 (where a:b:n was 2:3:3)
a! = 6 b! = 2 2^n = 8 (where a:b:n was 3:2:3)

I do not think there are any others mostly due to the imbalance of the sides
2^n doubles for each n, the factorials mostly end in many zeros unlike 2^n