Solusi

$\large{\frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-...+\frac{1}{1319}}$

$\large{\frac{p}{q}=(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{1319})-(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1318})}$

$\large{\frac{p}{q}=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{1319})-2.(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1318})}$

$\large{\frac{p}{q}=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{1319})-(1+\frac{1}{2}+\frac{1}{3}...+\frac{1}{659})}$

$\large{\frac{p}{q}=\frac{1}{660}+\frac{1}{661}+\frac{1}{662}+...+\frac{1}{1319}}$            (1)

$\large{\frac{p}{q}=\frac{1}{1319}+\frac{1}{1318}+\frac{1}{1317}+...+\frac{1}{660}}$         (2)

Jumlahkan (1) dan (2)

$\large{\frac{2p}{q}=(\frac{1}{660}+\frac{1}{1319})+(\frac{1}{661}+\frac{1}{1318})+(\frac{1}{662}+\frac{1}{1317})+...+(\frac{1}{1319}+\frac{1}{660})}$

$\large{\frac{2p}{q}=(\frac{1}{660}+\frac{1}{1979-660})+(\frac{1}{661}+\frac{1}{1979-661})+(\frac{1}{662}+\frac{1}{1979-662})+...+(\frac{1}{1319}+\frac{1}{1979-1319})}$

$\large{\frac{2p}{q}=\sum_{k=660}^{1319}(\frac{1}{k}+\frac{1}{1979-k})}$

$\large{\frac{2p}{q}=\sum_{k=660}^{1319}\frac{1979}{k(1979-k)}}$

$\large{\frac{p}{q}=\frac{1979}{2}\sum_{k=660}^{1319}\frac{1}{k(1979-k)}}$

Misalkan $\large{\frac{a}{b}=\sum_{k=660}^{1319}\frac{1}{k(1979-k)}}$ dimana $a,b \in \mathbb{N}$ dan $gcd(a,b)=1$

Maka $\large{\frac{p}{q}=\frac{1979 \times a}{2b}}$

Misalkan $2b={x_1}^{y_1} \times {x_2}^{y_2} \times ... \times {x_n}^{y_n}$ dimana $x_i$ adalah faktor prima dari $2b$ dan $y \in \mathbb{N}$
Cukup jelas $2\le x_1 \lt x_2 \lt x_3 \lt ... \lt x_n \le 1319$

Karena $1979$ adalah bilangan prima dan $2 \le x_1 \lt x_2 \lt x_3 \lt ... \lt x_n \le 1319 \lt 1979$
Maka tidak ada satu pun $x$ yang habis membagi 1979, yaitu $gcd(2b,1979)=1$

Karena $x \nmid 1979$ maka $1979$ pastilah salah satu faktor dari $p$, yaitu $1979 \mid p$