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Does the following diophantine equation have no positive integer solutions?

$$x^3-y^3=z!$$

Many problems involving diophantine equations are hard. Is it an open problem? I hope someone can give some references for this question.

- Find all integer solutions of $1+x+x^2+x^3=y^2$
- When the $a^2+b+c+d,b^2+a+c+d,c^2+a+b+d,d^2+a+b+c$ are all perfect squares?
- Find all integer solutions to $7595x + 1023y=124$
- Can n! be a perfect square when n is an integer greater than 1?
- Do there exist an infinite number of integer-solutions $(x,y,z)$ of $x^x\cdot y^y=z^z$ where $1\lt x\le y$?
- $y^2 = \frac{x^5 - 1}{x-1}$ & $x,y \in \mathbb{Z}$

- Diophantine Quintuple?
- Infinite Sum Calculation
- Factorial of 0 - a convenience?
- Given $XX^\top=A$, solving for $X$
- Closed form for $(p-n)!\pmod{p}$ where $p$ is prime
- Solve $y^2= x^3 − 33$ in integers
- Find All $x$ values where $f(x)$ is Perfect Square
- How to find solutions of linear Diophantine ax + by = c?
- Factorial of infinity
- Generating Pythagorean triples for $a^2+b^2=5c^2$?

See abc-conjecture. Probably this problem has some easy tricky solution, but most likely it doesn’t. However, abc-conjecture implies it has only finitely many solutions (see below), and if any explicit bounds will be proven (which is not the case today AFAIK), then one may probably brute-force $z$’s to prove it has no solutions (for fixed $z$ with known factorization this doesn’t seem hard).

Denote $c=x^3, a=y^3, b=z!$, so we have $a+b=c$. Then $$rad(abc)=rad(z!xy)\le xy\cdot rad(z!)$$

$rad(z!)$ is the product of primes up to $z$ and you can prove that this is smaller than $z!^{\delta}$ for any $\delta>0$ for large enough $z$, because the density of primes among all positive integers is zero. But then $$rad(abc)^{1+\varepsilon}\le x^{1+\varepsilon}y^{1+\varepsilon}(z!)^{\delta(1+varepsilon)}\le x^{2+2\varepsilon+\delta(1+\varepsilon)}< x^3=c$$

for large enough $z$ and small enough $\delta,\varepsilon>0$. Then abc-conjecture asserts that there exist only finitely many such $a,b,c$.

I wrote a routine “cubes” in PARI/GP. For $n>1$ it checkes whether $n$ is the difference of two cubes :

```
cubes(n)={gef=0;w=factor(n);u=component(w,1)~;v=component(w,2)~;forvec(z=vector(length(v),m,[0,v[m]]),t=prod(j=1,length(v),u[j]^z[j]);if((gef==0)*(t^3<n),if(issquare((12*n-3*t^3)/t)==1,gef=1)));gef}
```

The routine is based on the fact that with $a:=x-y$ , $k:=3(x+y)$ , $N:=x^3-y^3$ we have $$3a^3+ak^2=12N$$

So, $N>1$ is the difference between two cubes if and only if for some $a|N$, the number $\frac{12N-3a^3}{a}$ is a perfect square. Since I required $a^3<n$ in the program, the program returns “$0$” if $N$ is a cube.

This is because $N=x^3-y^3$ and $y=0$ is ruled out and a solution with $y>0$ cannot exist because we would have a counterexample for Fermat’s last theorem for exponent $3$. Anyway, a factorial (besides $1$) cannot be a cube, so this case plays no role.

The following program (I copied the output after some minutes, so the program is not finished!) shows that no factorial $z!$ with $2\le z\le 35$ is the difference between two cubes. I will continue the search.

```
? gefunden=0;j=0;while(gefunden==0,j=j+1;print1(j," ");if(cubes(j!)==1,gefunden=
1))
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36
```

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- Is $\sqrt{2}$ contained in $\mathbb{Q}(\zeta_n)$?
- Multivariable integral over a simplex
- “Binomial theorem”-like identities
- The meaning of $\lambda$ in Lagrange Multipliers