Concrete mathematics : a foundation for computer science

4.9 PHI AND MU 143
Now we raise both sides to the pth power, expand the right side according to
the binomial theorem (which we’ll meet in Chapter 5), and regroup, giving
TIP2 = (n + pq)p = np + (pq)‘nPm’ y + (pq)2nPP2 i +
0 0
= np + p2Q
for some other integer Q. We’re able to pull out a factor of p2 here because
($ = p in the second term, and because a factor of (pq)’ appears in all the
terms that follow. So we find that p2 divides npz - np.
Again we raise both sides to the pth power, expand, and regroup, to get
np3 = (nP + P~Q)~
= nP2 + (p2Q)‘nP’Pp’l y + (p2Q)2nP’P-2’ 1 + . .
= np2 + p3Q
0
for yet another integer Q. So p3 divides nP3- np’. This finishes the proof for
m = p3, because we’ve shown that p3 divides the left-hand side of (4.64).
Moreover we can prove by induction that
n~k = n~km’ + pkD
for some final integer rl (final because we’re running out of fonts); hence
nPk E nPk-’
Thus the left side of (4.64), which is
(mod ~~1, for k > 0. (4.65)
(nPk-nPkm’) + p(nPkm’-nPkmZ) + . . . + pkpl(nP-,) + pkn,
is divisible by pk and so is congruent to 0 modulo pk.
We’re almost there. Now that we’ve proved (4.64) for prime powers, all
that remains is to prove it when m = m’ m2, where m’ I ml, assuming that
the congruence is true for m’ and m2. Our examination of the case m = 12,
which factored into instances of m = 3 and m = 4, encourages us to think
that this approach will work.
We know that the cp function is multiplicative, so we can write
x q(d)nm’d = x (P(d’d2)nm1mz’d1d2
d\m dl \ml> dr\mz
= t oldl)( x
di\ml dz\mz
0