Question

Given that a,b,c are positive integers solve the following equation:
a!b!=a!+b!+2^c

Only work with this if you like maths challenges.

Answer 1

With:
\[a!(b!-1)\ne b!\]
\[c=\frac{\ln (a!(b!-1)-b!)+2i\pi n}{\ln 2}\space \space :n\in\mathbb{Z}\]

Answer 2

I dont see how this helps at all.
HINT: divide through by b! first and think of what is there. (also you can assume without loss of generality that a is bigger than equal than b)

Answer 3

Basically, what the hint says is what I said it had to be.
The other way to think this one is by noting this is the same as:
\[\Gamma(1 + a) \Gamma(1 + b) = 2^c + \Gamma(1 + a) + \Gamma(1 + b)\]
which will yield:
\[c=\frac{\ln (\Gamma(1+a)(-1+\Gamma(1+b))-\Gamma(1+b)+2i\pi n}{\ln 2}\]
with n being an integer again.

Answer 4

\[(a,b,c)=\{(2,3,2),(3,2,2)\}\]

Answer 5

Very nice! Well done

Answer 6

I like solving things like this