Find with proof all solutions to $2^n = a! + b!$, where $a$, $b$, $n$ are positive integers and $a \leq b$.

Question

So far I have looked at $n=1$ with $a=1$ and $b=1$ which is $$2^1 = 1! + 1! = 2 $$ and $n=2$ with $a=2$ and $b=2$, $$2^2 = 2!+ 2! = 4$$ and finally $n=3$ with $a=3$ and $b=2$, $$2^3 = 3! + 2! = 8$$

I have also tried the positive integers up to $n=14$ and found that none of them work. I also tried random values such as $n=30$ and it doesn't seem to work either.

For example, $n=11$, $2^{11} = 2048$. $7! = 5040$ so this cannot be used. However, since $6! = 720$, the highest number we can make is $6! + 6! = 1440$.

However, exhaustively listing out the answers doesn't seem like a concrete enough proof to say that it cannot work for any other numbers.

Can anybody give a more concrete reason why this wouldn't work for positive integers above $n=3$?

Answer 1

• If $a,b \geq 3$, we have $3$ divides $a!$ and $3$ divides $b!$. Hence, $3$ divides $a!+b!$. However, $3$ doesn't divide $2^n$.
• $a \geq 3$, $b=1 \text{ or }2$. If $b=1$, then $a!+1$ is odd and cannot be $2^n$. If $b=2$, then $a!+2 = 2(a!/2+1)$, where $a!/2+1$ is odd for $a\geq 4$, and hence cannot equal $2^n$. For $a=3$, we see that $3!+2 = 8 = 2^3$.
• $a = 1 \text{ or }2$, $a = 1 \text{ or }2$. This we can check by brute-force and we see that $1!+1! = 2$ and $2! + 2! = 2^2$.

Hence, the only solutions are $(a,b,n) = (1,1,1)$, $(a,b,n) = (2,2,2)$, $(a,b,n) = (3,2,3)$ and $(a,b,n) = (2,3,3)$.