Determine the values of x, y, and z using one equation with three variables

Question

Solve for x, y, and z using one equation with three variables:

$xyz = x! + y! + z!$

where factorials are defined in the usual way, namely $n! = n(n-1)...2*1$

I'm not sure where to start with this question, any help would be great! Thanks!

Answer 1

So with the comments already given, two out of the three variables must be equal, and allocating the numbers 3,3,4 to the variables $x,y,z$ does it. There are 3 choices of doing that. And - that was also indicated in the comments - this is the only solution as for $x,y,z \geq 6$ no more solutions can be expected.

Answer 2

$xyz=x!+y!+z!$
$AM \ge GM$.
$\therefore$ $x^3+y^3+z^3\ge3x!+3y!+3z!$
$\Rightarrow$ $x^3+y^3+z^3-(3x!+3y!+3z!)\ge0$ -------- (1)
But we know that $k^3<k!$ (see above comment by JMoravitz) for $k>5$.

$\max(n^3-3n!)=9, n=3$. --------- (2)

That means that if we try to take value of $(x,y,z)$ greater than $4$ we won't be able to keep the above expression (expression (1), above) greater than $0$. As for $n=5$, expression(expression (2)) gives value smaller than $-18$ (which is the minimum value we can cancel out (by putting $n=3$) i.e. $18=2*9$).
Note: Expression (2) will keep on further decreasing for higher values of $n$.
Hence, it proves that $n\le4$.

Now find, value of expression (2) for $n=1,2,3,4$. ($0$ ins't possible because LHS will turn $0$).
Values of expression (2),
For $n=1$, Value $=-2$
$n=2$, Value $=2$
$n=3$, Value $=9$
$n=4$, Value $=-8$.

All possible triplets are $(1,2,2);(1,3,3);(1,2,3);(4,3,2);(4,3,3);(3,3,3);(2,2,2);(2,2,3);(3,3,2)$

Indeed a great achievement from $64$ possible triplets to $10$ (could be done faster if you have intuition and practised these steps.)

Bonus: $(x,y,z)$ aren't equal because for that $(x,y,z)$ should have $3$ as a factor and should be even. Reducing triplets to $8$.

This isn't answer you have to solve. :P