2
$\begingroup$

I have to find the solutions of $2^a+2^b=c!,~a,b,c\in\mathbb{Z},~a,b,c\ge0$, but I don't know how to do that. I need some help.

  • 0
    You can assume WLOG that $a\ge b$ and then it becomes $2^a(1+2^{b'})=c!$ where $b'=b-a$. – 2017-01-11
  • 0
    You could try to look at it modulo several number. For example, modulo $3$, $2^k\equiv (-1)^k$ so $2^a(1+2^{b'})\equiv (-1)^a(1+(-1)^{b'})$ which can only be $\equiv 0$ if $b'$ is odd. So you can't have $b'$ even and $c \ge 3$. – 2017-01-11
  • 0
    One idea is that powers of $2$ and factorials are rare, so the only solutions will be small. Then use the other hint and reflect on the fact that $c!$ has about $c$ factors of $2$, so we need $a \approx c$ and $2^{b-a}+1$ has to have all the odd factors of $c!$ – 2017-01-11
  • 1
    @xavierm02: in your first comment it should be $b \ge a$ to work with the rest. – 2017-01-11
  • 1
    $a=2$, $b=1$ $c=3$ is a solution: $2^2+2=6=3!$ – 2017-01-11
  • 0
    @MonsieurGalois It is also $a=b=0,~c=2$. – 2017-01-11

1 Answers 1

3

Hint

Prove that $2^a+2^b$ is never divisible by $7$.

In order to see that write $a$ as $3k$ or $3k+1$ or $3k+2$ and see that $2^a$ just has $1$ or $2$ or $4$ as a remainder when divided by $7$. Do the same for $2^b$. So, $c<7$.

It is also easy to see that $c=0$ and $c=1$ don't have a solution.

Then just find solutions for $c \in \{2,3,4,5,6\}$ choosing $a$ and $b$ properly.

Can you finish?