Concrete mathematics : a foundation for computer science

142 NUMBER THEORY
n is odd, n4 and n2 are each congruent to 1, and 2n is congruent to 2; hence
the left side is congruent to I + 1 +2 and thus to 0 modulo 4, and we’re done.
Next, let’s be a bit daring and try m = 12. This value of m ought to
be interesting because it has lots of factors, including the square of a prime,
yet it is fairly small. (Also there’s a good chance we’ll be able to generalize a
proof for 12 to a proof for general m.) The congruence we must prove is
n”+n6+2n4+2n3+2n2+4n E 0 (mod 12).
Now what? By (4.42) this congruence holds if and only if it also holds modulo
3 and modulo 4. So let’s prove that it holds modulo 3. Our congruence
(4.64) holds for primes, so we have n3 + 2n = 0 (mod 3). Careful
scrutiny reveals that we can use this fact to group terms of the larger sum:
n’2+n6+2n4+2n3+2n2+4n
= (n12 +2n4) + In6 +2n2) +2(n3 +2n)
e 0+0+2*0 5 0 (mod 3).
So it works modulo 3.
We’re half done. To prove congruence modulo 4 we use the same trick.
We’ve proved that n4 +n2 +2n = 0 (mod 4), so we use this pattern to group:
n”+n6+2n4+2n3+2n2+4n
= (n12 + n6 + 2n3) + 2(n4 + n2 + 2n)
E 0+2.0 E 0 (mod 4).
QED for the case m = 12. QED: Quite Easily
So far we’ve proved our congruence for prime m, for m = 4, and for m = Done.
12. Now let’s try to prove it for prime powers. For concreteness we may
suppose that m = p3 for some prime p. Then the left side of (4.64) is
np3 + cp(p)nP2 + q(p2)nP + cp(p3)n
= np3 + (p - 1 )np2 + (p2 - p)nP + (p3 - p2)n
= (np3 - npz) + p(np2 - nP) + p2(nP -n) +p3n.
We can show that this is congruent to 0 modulo p3 if we can prove that
n’J3 - nP2 is divisible by p3, that nP2 - n P is divisible by p2, and that n” - n
is divisible by p, because the whole thing will then be divisible by p3. By the
alternative form of Fermat’s theorem we have np E n (mod p), so p divides
np - n; hence there is an integer q such that
np = nfpq