94 II Divisibility
Similarly, sinceam+ 1 Am+ 1 ,n=am+n+ 1 Am,n, am+ 1 Am+ 1 ,n− 1 =anAm,n,am+ 1 divides (an,am+n+ 1 )Am,nand, since (an,am+n+ 1 )=(am+ 1 ,an), it follows that
am+ 1 /(am+ 1 ,an)dividesAm,n. In the same way, since
an+ 1 Am,n+ 1 =am+n+ 1 Am,n, an+ 1 Am− 1 ,n+ 1 =amAm,n,an+ 1 divides (am,am+n+ 1 )Am,n and hencean+ 1 /(am,an+ 1 )divides Am,n.This
proves (iii).
By multiplying bya 1 ···an+ 1 /am+ 2 ···am+n− 1 , we see that (iv) is equivalent to
(anan+ 1 am+ 1 ,an+ 1 am+nam+n+ 1 ,amam+ 1 am+n)
=(an+ 1 amam+ 1 ,anan+ 1 am+n,am+ 1 am+nam+n+ 1 ).Since here the two sides are interchanged whenmandnare interchanged, it is suf-
ficient to show that any common divisoreof the three terms on the right is also a
common divisor of the three terms on the left. We have
(an+ 1 amam+ 1 ,anan+ 1 am+ 1 )=an+ 1 am+ 1 (am,an)=an+ 1 am+ 1 (am,am+n)
=(an+ 1 amam+ 1 ,am+ 1 an+ 1 am+n),and similarly
(anan+ 1 am+n,an+ 1 am+nam+n+ 1 )=(anan+ 1 am+n,am+ 1 an+ 1 am+n),
(am+ 1 am+nam+n+ 1 ,amam+ 1 am+n)=(am+ 1 am+nam+n+ 1 ,am+ 1 an+ 1 am+n).Hence if we putg=am+ 1 an+ 1 am+n,then
(e,g)=(e,anan+ 1 am+ 1 )=(e,an+ 1 am+nam+n+ 1 )=(e,amam+ 1 am+n)and if we putf=(e,g),then
1 =(e/f,anan+ 1 am+ 1 /f)=(e/f,an+ 1 am+nam+n+ 1 /f)=(e/f,amam+ 1 am+n/f).Hence(e/f,P/f^3 )=1, where
P=anan+ 1 am+ 1 ·an+ 1 am+nam+n+ 1 ·amam+ 1 am+n.ButPis divisible bye^3 , since we can also write
P=an+ 1 amam+ 1 ·anan+ 1 am+n·am+ 1 am+nam+n+ 1.Hence the previous relation impliese/f =1. Thuse= fis a common divisor of
anan+ 1 am+ 1 ,an+ 1 am+nam+n+ 1 andamam+ 1 am+n,aswewishedtoshow.
For the binomial coefficient case, i.e.an = n, the property (iv) of Proposi-
tion 12 was discovered empirically by Gould (1972) and then proved by Hillman and
Hoggatt (1972). It states that if in thePascal triangleone picks out the hexagon sur-
rounding a particular element, then the greatest common divisor of three alternately