Prove no values $x, y \in \mathbb{N} \cup \{0\}$ satisfy the following $$\frac{x!}{y}=(x+1)^y-1.$$

  • $\begingroup$ Does it help to substitute $a=x-1$? This gives $a^y=(a-1)!/48+1$. I think you will be able to quickly eliminate all possible odd $x$ because the left hand side is almost certainly even. And then study what you know about the powers of $2$ in each side. That's my suggestion assuming you're in $\mathbb{N}$. If not, then ignore! $\endgroup$ – samerivertwice Mar 1 '17 at 19:39
  • $\begingroup$ Write $x=n-1$, $y=m$ and do the same idea of proof as in this question. $\endgroup$ – Dietrich Burde Mar 1 '17 at 19:40
  • $\begingroup$ We can say, $$\frac{x!}{48}=(x+1)^y-1=(x+1)^y-1^y=x(x^{y-1}+x^{y-2}+\cdots+x+1)$$ $$=x^{y}+x^{y-1}+\cdots+x^2+x=\frac{1-x^{y+1}}{1-x}-1=\frac{x-x^{y+1}}{1-x}=\frac{x^{y+1}-x}{x-1}$$ Then, $$\frac{x!}{48}=\frac{x^{y+1}-x}{x-1}\Longrightarrow (x-2)!(x-1)^2x=48x(x^y-1) \Longrightarrow (x-2)!(x-1)^2=48(x^y-1)$$ Maybe you can find something if you investigate last equation, but I am not sure about where it goes. $\endgroup$ – Mathelogician Mar 1 '17 at 20:42

The smallest value for which $\frac{\large x! }{\large 48 }$ is integer is $x=6$, when $\frac{\large x! }{\large 48 } = 15$. Clearly this is not $1$ less than a power of $7$. Then $x=7$ gives $\frac{\large x! }{\large 48 } = 105$, which again is not $1$ less than a power of $8$.

For $x>8$, $\frac{\large x! }{\large 48 }$ is a multiple of $8$ , so would need $x$ even. However $\frac{\large x! }{\large 48 }$ is also divisible by all odd primes less than or equal to $x$, so none of these can divide $x{+}1$.

This leaves only $x=2^k$ where $x{+}1$ is prime as possible solutions to consider, which also gives us that $k$ is a power of two, see OEIS A092605.

$17^k{-}1$, $257^k{-}1$ and $65537^k{-}1$ (and any future discoveries) cannot work to generate all the needed small factors, because among the odd primes less than $x$ there must be some which are primitive roots of $x{+}1$. (In fact $3$ will be a primitive root). Take one of these, $p$, and then the smallest power $k$ for which $(x+1)^k\equiv 1 \bmod p$ is $x$, and we know that $(x+1)^x > x!$ .

Thus since we know that $p$ divides the LHS, and we cannot have a solution in range for which $p \mid (x+1)^k {-} 1$, there are no solutions.

  • $\begingroup$ The Diophantine equation $x!=(x+1)^y-1$ has been studied by Liouville. One can use this, and treat the factor $48$. $\endgroup$ – Dietrich Burde Mar 1 '17 at 19:51
  • $\begingroup$ @DietrichBurde updated, what do you think? $\endgroup$ – Joffan Mar 1 '17 at 20:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.