Relatively Prime in 2015

WLOG, let $n > m.$ Since $\gcd(m, n) = 1,$ the fraction $\frac{m}{n}$ is in lowest terms. Thus, we find the number of fractions from $\frac{1}{2014}$ to $\frac{1007}{1008}$ such that $m + n = 2015$ and $\frac{m}{n}$ is in lowest terms. Note that

$\frac{m}{n} = \frac{2015 - n}{n} = \frac{2015}{n} - 1.$

We reduce our problem to finding the number of integers $1008 \leq n \leq 2014$ that are relatively prime to $2015.$ We do this by counting the number of integers in the range that share a common factor with $2015$ :

The prime factorization of $2015$ is $5 \cdot 13 \cdot 21.$ There are $1007$ integers between $1008$ and $2014$ inclusive. Of these,

• $201$ have a factor of $5$ .
• $77$ have a factor of $13$ .
• $32$ have a factor of $31$ .
• $15$ have a factor of $5 \cdot 13$ .
• $6$ have a factor of $5 \cdot 21$ .
• $2$ have a factor of $13 \cdot 31$ .
• $0$ have a factor of $5 \cdot 13 \cdot 31$ .

Using PIE, we find that the number of integers in the range that share at least one common factor with $2015$ is $201 + 77 + 32 - 15 - 6 - 2 + 0 = 287,$ and the number of integers that are relatively prime to $2015$ is $1007 - 287 = 720.$

Since this covers only half of the possible pairs $(m, n)$ , our total number of ordered pairs is $720 \times 2 = \boxed{1440}.$

If d = gcd(m, 2015) = gcd(m, 5 * 13 * 31) > 1, then d | n, which is not allowed. Similar reason holds if gcd(n, 2015) > 1. Therefore both m and n are relatively prime to 2015. Clearly there are 2014 positive integral solutions to the given equation, namely (m, n) = (1, 2014), (2, 2013), ..., and (2014, 1). Moreover, 2015 is not relatively prime to 2015. Therefore the number of solutions is the number of positive integers <= 2015 relatively prime to 2015, i.e. phi(2015) = phi(5 * 13 * 31) = (5-1)(13-1)(31-1) = 1440.