How to calculate the GCD? How to evaluate the following with the help of Mobius function ?
$$\displaystyle\sum_{i=1}^n  \sum_{j=i+1}^n     \sum_{k=j+1}^n   \sum_{l=k+1}^n  {gcd(i,j,k,l)^4} .$$
In other words, we have to select all possible quadruplets from (1 ton n), and then sum up their value with power 4.
Example:-
N=4 :
(1,2,3,4) :   gcd(1,2,3,4)^4 = 1
Total sum = 1
For Second  Case, let N=5 :
(1,2,3,4) :   gcd(1,2,3,4)^4 = 1
(1,2,3,5) :   gcd(1,2,3,5)^4 = 1
(1,2,4,5) :   gcd(1,2,4,5)^4 = 1
(1,3,4,5) :   gcd(1,3,4,5)^4 = 1
(2,3,4,5) :   gcd(2,3,4,5)^4 = 1
Total sum = 1+1+1+1+1 = 5.
My approach is:
To calculate the number of quadruplets with gcd=2,3,4,...till n.
Say, the number of quadruplets with gcd=2 are x1, for gcd=3, its x2...and so on.
Now,  l  = $$\binom{n}{4}$$-(x1+x2+.......xn-1) = number of quadruplets with gcd=1. 
Now, the final answer = l^4+x1^4+x2^4+.....
The only problem I have is how to calculate the number of quadruplets with gcd=x with the help of mobius function ?
 A: Consider the sets of quadruples of integers $$Q(n)=\{(i,j,k,l) : 1\leqslant i<j<k<l\leqslant n\},\\ Q(n,d)=\{(i,j,k,l)\in Q(n) : \gcd(i,j,k,l)=d\}$$ for $1\leqslant d\leqslant n$. Clearly we have $$|Q(n)|=\binom{n}{4},\quad Q(n)=\bigcup_{d=1}^{n}Q(n,d),\quad Q(n,d)=dQ(\lfloor n/d\rfloor,1)$$
(where, for the last one, $d(i,j,k,l):=(di,dj,dk,dl)$ and $dS:=\{ds : s\in S\}$ are assumed).
This means that if $F(n)=|Q(n,1)|$, then $|Q(n,d)|=F(\lfloor n/d\rfloor)$ and $$\color{blue}{\sum_{d=1}^{n}F(\lfloor n/d\rfloor)=\binom{n}{4}}\quad\implies\quad F(n)=\sum_{d=1}^{n}\mu(d)\binom{\lfloor n/d\rfloor}{4}$$
by Möbius inversion. The sum we need is $S(n)=\sum\limits_{d=1}^{n}d^4 F(\lfloor n/d\rfloor)$. To compute it, one may avoid the computation of $\mu$, using the "blue" equation above as a recurrence for $F(n)$, and grouping terms with equal $\lfloor n/d\rfloor$ (here are my answers suggesting this too: 1, 2, 3).
A: A quadruplet with gcd a multiple of $d$ occurs iff all $i,j,k,l$ are multiples of $d$, for which there are ${\lfloor n/d\rfloor\choose d}$ possibilities.
So you'll arrive at
$${n\choose 4}\cdot 1^4+{\lfloor n/2\rfloor\choose 4}\cdot (2^4-1^4) +{\lfloor n/3\rfloor\choose 4}\cdot (3^4-1^4)+{\lfloor n/4\rfloor\choose 4}\cdot (4^4-2^4)+\cdots$$
