beautiful riddle dealing with pascal triangle

I don't quite understand.

Are you saying that the GCF (greatest common factor) of all the numbers greater than 1 in a given row is greater than 1? This would be false.

Or are you saying that for any pair of numbers greater than 1 in a given row, the GCF is greater than 1?

Or are you saying something else?

Part of the confusion is due to the example ("4 with 1 (1)." The problem is trivial if you accept 1 as the GCF. Every set of positive integers has a GCF that is greater than or equal to 1.

1 6 15 20 15 6 1

I think the point is to prove that every number bigger than 1 has a GCD with every other number bigger than 1.
So 6 and 15 have 3
15 and 20 have 5
6 and 20 have 2

It seems to be true, but I have no clue why this is so.

Prime numbered rows are easy. All terms (other than the 1s) are multiples of the prime. This is because when you compute n!/[(n-r)!*r!], n doesn't appear in the denominator, and being prime, can't be cancelled out by other terms.

Here's a site with the answer (small font)
http://www.qbyte.org/puzzles/p093s.html