Can you reduce this fraction?

$\dfrac{8n+3}{4n+2} = \dfrac{8n+4-1}{4n+2} = 2 - \dfrac{1}{4n+2}$

Note that $\forall \ n \in \mathbb{N}$ , we have $\text{gcd } (1,4n+2) = 1$ and hence the fraction is irreducible for any such $n$ .

$\therefore \$ The required answer is $\boxed{0}$ .

• For the fraction to be reducible there must be some prime p which divides both the numerator and the denominator.
• We have p | 4n + 2 and p | 8n + 3
• This means p | 8n + 4
• We have that p | 8n + 4 and p | 8n + 3
• So we have that p | 1
• Therefore there are no integers n where this is reducible
• Hence, the answer is zero.