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.
X = (2/1, 4/3, 6/5, 8/7)
One rational multiple of X, not consisting entirely of integers, is
(1/2)*X = (1, 2/3, 3/5, 4/7)
Another rational multiple of X, not consisting entirely of integers, is
(15/2)*X = (15, 10, 9, 60/7)
Yet another rational multiple of X, this time consisting only of integers, is
(105/2)*X = (105, 70, 63, 60) = (3*5*7, 2*5*7, 3*3*7, 2*2*3*5), which consists of four integers with no common divisor
If k>1 is an integer, then (105k/2)*X will also consist only of integers, but those integers will have a common divisor.
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?