Concrete mathematics : a foundation for computer science

522 ANSWERS TO EXERCISES
5.82 Let ~,,(a) be the exponent by which the prime p divides a, and let
m = n - k. The identity to be proved reduces to
For brevity let’s write this as min(x,,yl,zl) =: min(xz,y2,z2). Notice that
x1 + y, + z1 = x2 + y2 + 22. The general relation
+(a) < e,(b) =+ e,,(a) = eP(/u*bl)
allows us to conclude that x.1 # x2 ==+ min(x, ,x2) = 0; the same holds also
for (~1, y.7) and (2, ,22). It’s now a simple matter to complete the proof.
5.83 If m < n, the quantity (j:“) (“‘?:iPk) is a polynomial in k of degree
less than n, for each fixed .i; hence the sum over k is zero. If m 3 n and
if r is an integer in the range n < r 6 m, the quantity (‘+kk) (m+c:iPk) is a
polynomial in j of degree less than r, for each fixed k; hence the sum over j is
zero. If m 3 n and if r = -d - 1 is an integer, for 0 6 d < n, we have
(;) = (w(qd) = (-lIq;)(;);
hence the given sum can be written
pk(i;k)(;)(:)(;I)(m+;lI:-k)
, , = pk(;) (3 (‘:“) (jy) (m+;:; -“>
= &,,+m+-l
j,kL
n
(k)(;)(‘:“)(-‘i”l’)(-“m’*,‘)
= xc-1 )k+mi-L
k,i
(;) (3 (7”) (-‘mn12).
This is zero since (I:“) is a polynomial in k of degree d < n.
If m 3 n, we have verified the identity for m different values of r. We
need consider only one more case to prove it in general. Let r = 0; then j = 0
and the sum is
pk(;) (-+;;- “) = (3
by (5.25). (Is there a substantially shorter proof?)