1) Carlos, though I do not have an answer for you, I
do have some progress to report. I will write the function as mu(n). You
can change it to Greek if you like.

We are looking for a counterexample. However,
instead of testing mu(n) for each n, we examine a set of n such that mu(n)
= M for some fixed M, and determine whether there are consecutive numbers
in that set. The advantage of this approach is that it provides a
systematic way of testing very large numbers. No counterexamples have been
found yet. The algorithm is outlined below. We can provide a Mathematica
program to anyone interested.

Recall that the value of mu(n) is determined from
the prime factorization of n. Basically, there is a prime power p^e such
that p^e | n and mu(n) = mu(p^e). Hence, to find all the numbers n such
that mu(n) = M, we can concentrate on computing all the prime powers such
that mu(p^e) = M. The number of such prime powers is the same as the
number of prime factors of M, counting multiplicity. The primes p
appearing in the prime powers are merely the prime factors of M. Only the
exponents e must be computed. For example, when M = 60, the four prime
powers are 2^55, 2^56, 3^28, and 5^14.

If M = mu(n) = mu(n+1), then we can factor n and n+1
as

(1) n = x p^a and n+1 = y q^b

for two different prime powers p^a and q^b with
gcd(x,p) = gcd(y,q) = 1, M = mu(p^a) = mu(q^b), mu(x) <= M, and mu(y) <=
M. Eliminating n from (1) shows that we need to solve the linear
Diophantine equation

(2) y q^b - x p^a = 1

for x and y. Because p^a and q^b are known, and
gcd(p^a,q^b) = 1, the extended Euclidean algorithm can be used to compute
a basic solution x0 and y0. The family of all solutions is x = x0 + k*q^b
and y = y0 + k*p^a for any integer k. The valid values of k range from
-M!/(p^a q^b) to M!/(p^a

q^b) -- which can be huge! We used -10^6 < k < 10^6.
After computing x and y for some k, we check that gcd(x,p) = gcd(y,q) = 1,
mu(x) <= M, and

mu(y) <= M. If we find an x and y that meets all
these conditions, then a counterexample has been found. Rather than
compute mu(x) and mu(y) exactly (by factoring x and y), we merely check
that x | M! and y | M!, which is much faster than factoring.

Note that equation (2) must be solved for all pairs
of prime powers associated with M. For M = 60, there are five pairs:
(2^55,3^28), (2^55,5^14), (2^56,3^28), (2^56,5^14), and (3^28,5^14).

Observe that when M is prime, or a power of a prime,
no values of n satisfy M = mu(n) = mu(n+1).

>What is your personal opinion about the absolute
validity of the

>Conjecture? According to your approach you do not devise any theoretical

>proof on its favor, but computationally, could it be disapproved

>someday soon?

I still have an open mind about the conjecture: I'm
not sure whether it is true or false. However, if my computer does not
find a counterexample soon, then I will believe that the conjecture is
true.