Product of two primes

The number of positive integers less than $n$ that are relatively primes or coprime numbers to $n$ is given by Euler's totient function .

$\begin{aligned} \phi(n) & = n \prod_{k=1}^m \frac {p_k-1}{p_k} & \small \color{#3D99F6} \text{where }p_k \text{ is the prime factor of }n \end{aligned}$

$\implies \phi (110179) = \dfrac {110179(p-1)(q-1)}{pq} = \dfrac {110179(p-1)(q-1)}{110179} = (p-1)(q-1)$

$\begin{aligned} \implies (p-1)(q-1) & = 109480 \\ pq - p - q + 1 & = 109480 \\ 110179 - p - q + 1 & = 109480 \\ \implies p+q & = 700 \\ (p+q)^2 & = 700^2 \\ p^2 + 2pq + q^2 & = 490000 \\ p^2 + q^2 & = 490000 - 2(110179) = 269642 \\ (p-q)^2 & = p^2 - 2pq + q^2 = 269642 - 2(110179) = 49284 \\ \implies p-q & = \sqrt {49284} = \boxed{222} \end{aligned}$

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Note: Since $p+q = 700$ and $p-q=222$ , $\implies p = 461$ and $q = 239$ both primes.

The number of positive integers less than n = 110178. Let's assume that $\frac{n}{p}$ = y + 1 and $\frac{n}{q}$ = z +1. So y = $\frac{n}{p}$ - 1 and z = $\frac{n}{q}$ - 1. Therefore, p and q have y and z positive multiple respectively less than n. Hence y + z = (The number of positive integers less than n ) - (The number of positive integers less than n that are relatively prime to n ). So y + z = 110178 - 109480 = 698 . Putting the values of y and z ,we get: $\frac{n}{p}$ - 1 + $\frac{n}{q}$ - 1 = 698 . $\frac{n}{p}$ + $\frac{n}{q}$ = 700 . $\frac{n(p + q)}{pq}$ = 700 . So p + q = 700 [ n = pq]. pq = 110179. Calculation gives us p - q = 222