$$x!+y=y^3$$ $$y=\sqrt[3]{x!+\sqrt[3]{x!+\sqrt[3]{x!+\cdots}}}$$
The only integer solutions to these identities that I have found are:
$$3!+2=2^3$$ $$4!+3=3^3$$ $$5!+5=5^3$$ $$6!+9=9^3$$
I conjecture these are all the solutions. Is that true
Asked
Active
Viewed 172 times
3
-
I think you can rephrase it as: "are $3!,4!,5!,6!$ the only faculties that can be written as a product of $3$ consecutive integers?" – drhab May 07 '16 at 09:14
-
Brute force agrees up to $x=1000$. – Ian Miller May 07 '16 at 09:17
-
Related – Kenny Lau May 07 '16 at 09:19
-
Since $x!=(y+1)y(y-1)=z(z-1)(z-2)$ (writing $z$ for $y+1$) you could maybe try and show that $\dfrac{z!}{(z-3)!}$ cannot be expressed as the factorial of a single integer for $z>10$. – George Law May 07 '16 at 10:03
-
For large $x$ it will be incredibly difficult to find a solution since $y^3 - y$ will have to be highly divisible. Consider different valutations mod different primes. 2 is a good place to start. – fretty May 08 '16 at 12:37
1 Answers
-2
Assuming the ABC conjecture the following paper shows that there should only be finitely many solutions.
https://web.math.pmf.unizg.hr/glasnik/37.2/37(2)-04.pdf
I would probably bet that the solutions you have found are all of the solutions.
fretty
- 11,156
- 1
- 26
- 37