Michael's prime pairs

$104$ is even, so pairs of even numbers add up to it. These can be eliminated from the $103$ possible pairs of positive integers that would work. There are $51$ even pairs, from $2+102$ to $102+2$ . Other pairs that would not be coprime are odd multiples of $13$ . These are $13+91$ , $39+65$ , $65+39$ , and $91+13$ . This is $4$ pairs. $103-51-4=48$ .

If $\gcd(x, y) = 1$ , then we must have $\gcd(x, x + y) = \gcd(x, 104) = 1$ . And any positive $x < 104$ with $\gcd(x, 104)$ gives rise to a unique ordered pair $(x, 104-x)$ . The number of pairs must be the number of $x$ so that $\gcd(x, 104) = 1$ , which is $\phi(104) = 48$

By Euclidean algorithm $(x, 104 - x) = (104 - y,y) = (104,y) = (x,104) = (x,y) = 1$ Hence $x,y$ are odd and not divisible by $13$ , principle of inclusion-exclusion gives us $104 - \frac{104}{13} - \frac{104}{2} + \frac{104}{2 \cdot 13} = \boxed{48}$ such pairs

1. $x+y=104$
2. ${gcd(x,y)=gcd(x,x+y)=gcd(x,x+y)}$
3. If $gcd(x,y)=1 \Rightarrow gcd(x,x+y)=1 \Rightarrow gcd(x,104)=1$
4. $\phi(104) = \phi(2^3.13) = (2^3-2^2) \cdot (13-1) = 48$ pairs

Note: $\phi(n)$ - Euler's function

As 104 is the main character, we analyse it

$104=2^3\times13$

First, we find all possibility of $(x,y)$ , which is

$(1,103)\ldots(103,1)$ total of 103 pairs

Then, we remove the pairs which contains both $x$ and $y$ are a multiple of 2

which is $\big(2(1),2(51)\big)\ldots\big(2(51),2(1)\big)$ total of 51 pairs

Next, remove the pairs which contains both $x$ and $y$ are a multiple of 13

which is $\big(13(1),13(7)\big)\ldots\big(13(7),13(1)\big)$ total of 7 pairs

However, we have removed more as some pairs contains both $x$ and $y$ are multiple of 2 and 13

which is $\big(26(1),26(3)\big)\ldots\big(26(3),26(1)\big)$ total of 3 pairs

In conclusion, the number of ordered pairs of positive coprime integers are

$103-51-7+3=48$