$$\sum_{i=1}^n\sum_{j=1}^m[gcd(x,y)==1]$$

$$=\sum_{i=1}^n\sum_{j=1}^m\sum_{d|gcd(i,j)}\mu(d)$$

$$=\sum_{p}\sum_{d=1}^{\lfloor \frac{min(n,m)}{p}\rfloor}\mu(d)\lfloor\frac{n}{pd}\rfloor\lfloor\frac{m}{pd}\rfloor$$

$$2^n=\sum_{i=0}^{n}\binom{n}{i}$$

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