How to solve $\sqrt{x!y!}=xy$ for $(x,y)\in\mathbb{Z}_{\geq0}\times\mathbb{Z}_{\geq0}$?

Question

How to solve $\sqrt{x!y!}=xy$ for $(x,y)\in\mathbb{Z}_{\geq0}\times\mathbb{Z}_{\geq0}$?

In this task they are asking to find ordered pair couple in $\mathbb{Z}$ that satisfies the above equation, So that, i started to take squaring both sides to get $x!y!=xy\times xy$

then i tired to use the definition of factorial $x!=x(x-1)(x-2)....2\times 1$

Finally i obtain that $(x-1)!(y-1)!=xy \text{ or } xy=0 $

Now is there any shortcut to kill this problem easily ?

Answer 1

You can easily show that $i^2 / i! \leq 2$ for $i\in\mathbb N$. So $x,y < 6$. From that is just a matter of checking $36$ pairs.

>>> for i in range(7):
...     f = math.factorial(i)
...     print(i, i*i, f, i*i/f)
... 
0 0 1 0.0
1 1 1 1.0
2 4 2 2.0
3 9 6 1.5
4 16 24 0.6666666666666666
5 25 120 0.20833333333333334
6 36 720 0.05

So the solutions are $(1,1), (3,4), (4,3)$.

Answer 2

Note that $(x-1)! > x$ for $x > 4$. So for $x,y > 4$ there are no solutions. Suppose without loss of generality that $x \leq y$. Then we can check each possible value of $x$ separately.

If $x = 1$ then we get $(y-1)! = y$ which has solution $y=1$. If $x = 2$ then $(y-1)! = 2y$ which has no solutions. For $x = 3$, $2(y-1)! = 3y$ which gives $y = 4$. $x = 4$ also yields no solutions so we are left with the three solutions $(x,y) \in \{(1,1),(3,4),(4,3)\}$

Answer 3

$$\frac{(x-1)!}{x}=\frac{y}{(y-1)!}.$$

The possible values of the $LHS$ are

$$1,\frac12,\frac23,\frac32,\frac{24}5\cdots$$

There is no need to go further, as the next inverses are below $\dfrac12$. The only solutions correspond to

$$1\leftrightarrow1$$ and $$\frac23\leftrightarrow \frac32.$$