The Euler function phi(n) is defined to be
phi(n) = #{ 1 < i < n | gcd(i, n) = 1 }
the number of natural numbers less than n and prime to it; equivalently,
it's the size of the multiplicative group (Z/nZ)^*.
If n = p q is the product of two primes then phi(n) = (p - 1)(q - 1).
But phi(n) is not (if n is composite) the exponent of (Z/nZ)^*. It's
certainly true that
a^{phi(n)} = 1
for all a in (Z/nZ)^*; but the exponent of a group G is the /smallest/
positive integer e such that
a^e == 1
for all a in G. This quantity is denoted lambda(n); in our simple case
where n = p q is the product of two primes it's true that
lambda(n) = lcm(p - 1, q - 1)
Since p and q are large primes, both p - 1 and q - 1 are even, so
lambda(n) is at least a factor of 2 smaller than phi(n).
In fact, lambda(2) = 1, lambda(2^f) = 2^{f-2} for f >= 1, and
lambda(p^f) = p^{f-1} (p - 1) for prime p > 2; and, in general, if n =
p_1^{f_1} ... p_m^{f_m} is the prime factorization of n then
lambda(n) = lcm(lambda(p_1^{f_1}), ... lambda(p_m^{f_m}))
Signed-off-by: Mark Wooding <mdw@distorted.org.uk>
/*
* Verify that d*e is congruent to 1 mod (p-1), and mod
* (q-1). This is equivalent to it being congruent to 1 mod
- * lcm(p-1,q-1), i.e. congruent to 1 mod phi(n). Note that
- * phi(n) is _not_ simply (p-1)*(q-1).
+ * lambda(n) = lcm(p-1,q-1). The usual `textbook' condition,
+ * that d e == 1 (mod (p-1)(q-1)) is sufficient, but not
+ * actually necessary.
*/
mpz_mul(&tmp, &d, &e);
mpz_sub_ui(&tmp2, &st->p, 1);