Independence Day Problem

Number Theory Level 4

Consider all integer solutions to the following equation:

$xyz = (x - 1)(y - 1)(z + 1)$

What is the greatest even integer in powers of 2 that must divide $xyz$ ?

1 2 4 8 16 32

1 solution

X X
Jul 3, 2018

First,notice that one of the triples $(x,y,z)=(5,6,2)$ ,so we can cross out $8,16,32$

We will show that $xyz$ needs to be a multiple of 4.

Proof by contradiction. Suppose $xyz$ is not a multiple of 4, and satisfies the equation.

The only ways that $xyz$ is a multiple of 4 are:

Case 1: $x,y,z$ are all odd numbers.
Then $(x-1)(y-1)(z+1)$ is a multiple of 8, so we have a contradiction.

Case 2: One of $x,y,z$ is even (but not a multiple of 4) and the others are odd.
Then $(x-1)(y-1)(z+1)$ must be a multiple of 4, which is a contradiction.

So, $xyz$ must be a multiple of 4.

It is interesting to note that $(x,y,z)=(2^n,2^n-1,2^{n-1}-1)$ is a solution for all integers $n \ge 1$ , so any power of $2$ greater than or equal to $4$ can be a factor of $xyz$ .

Mark Hennings - 2 years, 11 months ago

I think the word "always" should be put after the word "must" because I thought the question meant if any of the solutions were divisible by powers of 2.

e r - 2 years, 11 months ago

Independence Day Problem

1 solution

1 pending report

Vote up reports you agree with