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Lets assume $d$ is a natural number which makes $(n+1)/(n+3)$ reducible, then $d|n+1$ and $d|n+3$.

$d|[n+3-(n+1)] = d|2$ which means $d=1$ or $d=2$.

$n+1$ and $n+3$ must be divisible by $2$ so all natural numbers of the form $2n+1$ will work.

Now $\text{gcd}(n+1,n+3) = 1$ so shouldn't this fraction be irreducible for any $n$? ($n$ natural number)

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    You should say "Any number of the form $n=2k+1$ will work." That way you will not get confused between your two different uses of $n$. You get $\frac{n+1}{n+3}=\frac{2k+2}{2k+4}$, obviously reducible to $\frac{k+1}{k+2}$.2012-11-17

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Hint $\rm\ (k,2\!+\!k) = (\color{#C00}k,2\!+\!k\!-\!\color{#C00}k) = (k,2) = (r\!+\!2j,2) = (r\!+\!2j\!-\!\color{#C00}2j,\color{#C00}2) = (r,2)\ [ = 2\ if\ 2\mid r,\ else\ 1]$