Determine all tuples of positive integers $(x, y, z, t)$ such that: $$ xyz = t!$$$$ (x+1)(y+1)(z+1) = (t+1)!$$holds simultaneously.
Problem
Source: Latvian TST for Baltic Way 2019 Problem 16
Tags: algebra, system of equations, factorial
Kimchiks926
29.05.2020 19:08
Note that:
$$(x+1)(y+1)(z+1) = (t+1)! = t!(t+1) = xyz(t+1) \qquad (1)$$Expanding $LHS$ of $(1)$ yields:
$$xyz +xy+yz+zx+x+y+z+1=xyz(t+1)$$Or equivalently:
$$x+y+z+xy+yz+zx+1 =xyzt \implies t = \frac{1}{x} + \frac{1}{y} +\frac{1}{z} + \frac{1}{xy}+ \frac{1}{xz}+\frac{1}{zy}+\frac{1}{xyz}$$Note that:
$$ t = \frac{1}{x} + \frac{1}{y} +\frac{1}{z} + \frac{1}{xy}+ \frac{1}{xz}+\frac{1}{zy}+\frac{1}{xyz} \le 1 + 1 + 1 + 1 + 1 +1 +1 = 7$$Checking cases $t \le 7$ yields to solutions $(1,2,3,3)$, where we can permutate $x,y,z$
skrublord420
29.05.2020 19:26
Let $f(x) = 1 + \frac{1}{x} \in (1, 2].$ Now we suppose for a moment $x \ge y \ge z$ and perform casework on $z, t.$ If $z \ge 4$ then $1 < f(x)f(y)f(z) \le 1.25^3 < 2,$ so no solutions. If $z = 3$ then $1 < f(x)f(y)f(z) \le (4/3)^3 < 3$ so we must have $t = 1, f(x)f(y) = 2/f(3) = 3/2,$ for which we find $(x,y)=(8,3), (5,4), $ If $z=2,$ then $1 < f(x)f(y)f(z) \le 1.5^3 < 4$ so $f(x)f(y) \in \{3/f(2), 2/f(2)\} = \{2, 4/3\},$ for which we find $(x,y) = (3,2), (7,6), (9,5), (15,4).$ Finally, $z=1$ means $2<2f(x)f(y) \le 8,$ so $f(x)f(y) \in \{4, 3.5, 3, 2.5, 2, 1.5\},$ giving $(x,y) = (1,1), (2,1), (3,2), (8,3), (5,4).$ Thus, the tuples are any permutation of the first $3$ entries of $(8,3,3,1), (5,4,3,1), (3,2,2,2), (7,6,2,1), (9,5,2,1), (15,4,2,1), (1,1,1,7), (2,1,1,5), (3,2,1,3), (8,3,1,2), (5,4,1,2).$ It is easy to make a mistake in the casework. In fact, this problem is all casework once you bound $z.$