I'm reasonably certain that it is true, but it's one of those things where at first glance, you think "That's intuitive", but when you try to write down a proof, you realize that there are messy details to wade through.

Am I missing a clever approach by which you can prove it instantaneously? Or does it require a bit of wading through details?

The desired lemma:

(w,x,y,z) is a fixed 4-tuple of rational numbers. We consider all rational multiples of this 4-tuple, defined in the obvious way:

t(w,x,y,z) = (tw,tx,ty,tz) for any rational number t.

Some rational multiples of this rational 4-tuple will consist entirely of integers. Others will not.

My claim is that at most one rational multiple of the 4-tuple will consist of integers with no common divisor.

Example:

**Spoiler:**

Maybe you can sort of see why I thought this might be "intuitive". However, writing a proof might be messy, and it feels like there should be an easier way.

Rough idea behind proof:

Write X = (a1/b1, a2/b2, a3/b3, a4/b4) where each of those fractions is in lowest terms

Let G = gcd(a1,a2,a3,a4) and let L = lcm(b1,b2,b3,b4)

Show that multiplying X by L/G would give you all integers

Show, very roughly speaking, that multiplying by "anything else" would either give you non-integers, or integers all having a common factor.

It seems like that would probably work, but it also seems like writing it all out would be a pain.

Am I missing a clever shortcut?

EDITED TO ADD:

Maybe the more "elegant" trick is to do something like this:

If X = (a1/b1, a2/b2, a3/b3, a4/b4), then certainly there exist some rational numbers t with the property that t*X consists only of integers (because for example we can choose t = b1b2b3b4).

Then maybe choose the infimum of all rational t such that t*X consists only of integers?