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

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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