Finding all positive integer solutions to $(x!)(y!) = x!+y!+z!$

The equation is $(x!)(y!) = x!+y!+z!$

where $x,y,z$ are natural numbers.

How to find out them all?

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Another post about the same question: math.stackexchange.com/questions/1617225/… – Martin Sleziak Jan 18 at 20:26

Here's a start. We may assume $x\le y$. The case $x=1$ is trivial. The case $x=2$ gives $y!=z!+2$, which obviously has no solution. So assume $x\ge3$. Then $x!+z!\ge5y!$, so $z!\ge4y!$, so $z\gt y$. Suppose there is a prime number $p$ with $x\lt p\le y$. Then $p$ divides $y!$ and $z!$ but not $x!$ so we get a contradiction from $x!=x!y!-y!-z!$. So we have shown that if $x\le y$ then there can't be a prime $p$ with $x\lt p\le y$.

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I think let $p=3y!$ in your prove, all things are done. p is not necessary a prime. – Leitingok Dec 6 '11 at 8:34
I don't understand. It is not true that $3y!\le y$, and it is not true that $3y!$ divides $y!$. – Gerry Myerson Dec 6 '11 at 12:33
Hi, Gerry. I think your argument gives the whole thing. I've seen arguments of this sort, using the $\mbox{ord}_p (n)$ I defined, in contest books. – Will Jagy Dec 6 '11 at 18:44
Well, my argument (or Ross Millikan's) together with your not-the-least-bit-trivial contribution. – Gerry Myerson Dec 7 '11 at 3:19
Thank you Gerry. – Will Jagy Dec 7 '11 at 3:38

There won't be very many as the factorials are very sparse. I suspect you have already listed them all. We can assume $x \le y$. If $x!=1$ the RHS is at least $2$ greater than the left. If $x!=2, y=z$ the RHS is $2$ greater than the left, but there are no factorials with a difference of $2$. If $x! \ge 6$, divide both sides by $y!$. The LHS is an integer while the RHS is not unless $x=y$. Now you are looking for solutions to $(x!)^2=2(x!)+z!$

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You might want to take a look at the solution I just posted (prompted by the posting of a duplicate, linked to in a comment below the OP here). – Barry Cipra Jan 18 at 21:24

Gerry's argument shows, for $3 \leq x \leq y < z,$ that in fact $x=y.$ Otherwise, if we assume $y \geq x+1,$ there is some prime $p | (x+1),$ which might be $x+1$ itself or might be smaller. Suppose we define $\mbox{ord}_p (n)$ to be the largest exponent $k$ such that $p^k | n.$ We have $\mbox{ord}_p (y!) \geq 1 + \mbox{ord}_p (x!),$ also $\mbox{ord}_p (z!) \geq 1 + \mbox{ord}_p (x!).$ Indeed, we have a contradiction, since $x! = x! y! - y! - z!,$ and everything on the right hand side is divisible by $p^{1 + \mbox{ord}_p (x!)}.$

Alright, so $$x=y.$$ We do, indeed, have the solution $x=3, y=3, z=4.$

If $x \geq 4,$ then $\mbox{ord}_2 (x!) \geq 3.$ From $x=y$ we have $(x!)^2 = 2 x! + z!.$ Now, $\mbox{ord}_2 (x!)^2 \geq 3 + \mbox{ord}_2 (x!),$ while $\mbox{ord}_2 2(x!) = 1 + \mbox{ord}_2 (x!).$ As a result, $\mbox{ord}_2 (z!) = 1 + \mbox{ord}_2 (x!).$ We cannot have $z \geq x + 4$ because then there would be two extra even numbers dividing $z!.$ So $$z \leq x + 3.$$

However, Stirling's formula shows that, for $x \geq 6, \; z > x + 3$ where we regard $z$ as a real number here. Indeed, for the next few integral $x,$ we get the real solutions $(4,4,5.833), \; (5,5,7.504), \; (6,6,9.156), \; (7,7,10.814), \; (8,8,12.480), \; (9,9,14.154).$ Stirling's gives me $$z > \; 2 x \; \left( 1 - \frac{\log 2}{\log x} \right)$$

The only solution in natural numbers is $(x=3, \; y=3, \;z=4)$

ADDED: the function I have been calling $\mbox{ord}_p (n)$ is called $v_p(n)$ in Problems from the Book by Titu Andreescu and Gabriel Dospinescu, available at AOPS or XYZ

EDIT, December 8::This is sensible notation, as the function $v_p$ is the "$p$-adic valuation," see page 25, Definition 2.1.2, of $p$-adic Numbers by "Fernando Q. Gouvea," that being one of the pen-names of Mariano Suárez-Alvarez. Mariano wanted to include a letter "i" somewhere in the last name, have the full set, but was told that would seem inauthentic. P_ADIC_BOOK

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Pushing the maximum on edits. $v_p$ is good notation in English, I am not sure what Hensel wrote. Also, 'i' for the full set of vowels. About Fernando and Mariano, see math.stackexchange.com/questions/73565/… and math.stackexchange.com/questions/73832/… – Will Jagy Dec 8 '11 at 22:01

Without loss of generality, we can assume $x\le y$. We can quickly rule out any solution with $x=1$ or $2$, since the equations

$$y!=1+y!+z!\quad\text{and}\quad 2y!=2+y!+z!$$

are easily seen to have no solutions in positive integers $y$ and $z$.

Now suppose we had a solution with $x\lt y$. Then

$$x!={x!\over y!}+1+{z!\over y!}\implies{z!\over y!}=x!-1-{x!\over y!}\not\in\mathbb{N}$$

hence $z\lt y$ as well. But that implies $x!\lt1+1+1=3$, which implies $x=1$ or $2$, which have already been ruled out. Hence any solution must have $x=y\ge3$.

Now suppose we have a solution with $x=y\ge3$. The equation can now be written as

$$x!-2={z!\over x!}$$

which implies $z\ge x$. We can rule out $z=x$ since $x!$ cannot equal $3$, so we have

$$x!-2=(x+1)(x+2)\cdots(z-1)z$$

But $x\ge3$ implies $x!-2\equiv1$ mod $3$, and this implies the product $(x+1)(x+2)\cdots(z-1)z$ can only be $x+1$ (with $x\equiv0$ mod $3$) -- that is, any product of two or more consecutive numbers is either divisible by $3$ or leaves remainder $2$ mod $3$. Thus any solution $(x,y,z)$ must be of the form $(x,x,x+1)$, satisfying the equation

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

which simplifies to

$$x!=x+3$$

which has $x=3$ as its only solution. So $(x,y,z)=(3,3,4)$ is the only solution to $x!y!=x!+y!+z!$.

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Got your comment. This is very nice. It helps to know that something is likely a contest type problem; I keep asking people for the source of their question, and getting little cooperation. Today I have this bit of joy: math.stackexchange.com/questions/1617264/… – Will Jagy Jan 18 at 23:01

Observe that $x, y, z > 2$. Obviously, $y < z$. By symmetry, assume $x \leqslant y$. Then, we have $x \leqslant y < z$. Let $x = y$. Then, $$x! = 2 + z! / x!.$$ Let $k = z - x > 0$, so the last equality becomes $$x! = 2 + (x + 1) (x + 2) \cdots (x + k).$$ Since $x | x!$, we should have $x | k! + 2$. If $k \geqslant x$, then, we get $x | 2$, which is impossible. Hence $k < x$. This is only possible when $k = 1$, in which case we have $x | 3$. Hence, $x = 3$, therefore, $y = 3$ and $z = 4$.

Now suppose $x < y < z$. Let $\alpha$, $\beta$ and $\gamma$ be the largest powers of $2$ that divide $x!$, $y!$ and $z!$, respectively. Then, $$\alpha + \beta = \min (\alpha, \beta, \gamma)$$ by the original equation, since $x, y, z$ are distinct. But this is impossible since $\alpha, \beta, \gamma > 0$.

Hence, the only solution is $(x, y, z) = (3, 3, 4)$.

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