I bring this sample in order to ilustrate

$$x! = 2^x + 8$$

I know the answer is $x=4$ but I dunno how to prove it. I mean, if i put the number 4 by observation, tryal and error, I can get the results, but I dunno how to solve it isolating x like this:

(1) $x (x-1)! = 2^x +8$

(2) $x = \dfrac{2^x+8}{(x-1)!}$

(3) $x = \dfrac{2^x + 2^3}{(x-1)!}$

(4) $x = \dfrac{2\cdot2^{x-1}+2^3}{(x-1)!}$

From that point on I dunno how to procee using algebra

I would not know how to proceed if I come across another equation that its resolutions is not so easy to solve, like that one by trial and error.

  • $\begingroup$ You can use the fact that $x!$ has a greater growth rate than $2^x$ i.e. $x! > 2^x +8$ for sufficiently large x. Then you can show that there are no more solutions other than 4. $\endgroup$
    – Jacob
    Apr 7, 2016 at 4:02

2 Answers 2


Nothing much changes, even if you ask $$ x! = 2^y + 8. $$ As soon as $x \geq 6,$ we have $x!$ divisible by $16.$ As soon as $y \geq 4,$ we know $2^y + 8$ is not divisible by $16.$ Since $6! = 720,$ we would need $y \geq 9,$ guaranteed failure.

So $x \leq 5.$

As pointed out by @marty the same reasoning applies to $$ x! = 2^y - 8, $$ with solution $x=5, y=7,$ also $x=4, y=5.$

  • 1
    $\begingroup$ Nice reasoning. Works well for $x! = a^y\pm b$ by considering the prime factors of $a$ and $b$. $\endgroup$ Apr 7, 2016 at 4:20
  • 1
    $\begingroup$ @martycohen added in your $\pm$ idea $\endgroup$
    – Will Jagy
    Apr 7, 2016 at 4:31
  • $\begingroup$ So, you mean that example there is no solution using algebra? $\endgroup$
    – Luis P.
    Apr 11, 2016 at 19:45

I think trial and error is the way to go. You can guess that the solution should be small since factorial dominates exponentials.


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