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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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Wwhy is gcd(n+1,n+3)=1 ? –  Amr Nov 17 '12 at 20:21
    
well using the gcd algorithm for polynomials gives that answer –  phi Nov 17 '12 at 20:23
1  
If the $\gcd$ of $p_1(n) = n+1$ and $p_2(n) = n+3$ is $1$ in $\mathbb Q[n]$, this doesn't mead that the $\gcd(p_1(n), p_2(n))$ has to be 1 for every $n\in \mathbb N$, as for example $n=1$ shows ... –  martini Nov 17 '12 at 20:27
    
As another counterexample consider $\rm\:n(n\!-\!1)/2,\:$ which is never in lowest terms even though $\rm(2,n^2\!-\!n) = 1\:$ in $\rm\,\Bbb Q[n].\:$ Ditto for $\rm\:(n^p-n)/p\:$ for $\rm\,p\,$ prime, by little Fermat. –  Bill Dubuque Nov 17 '12 at 20:30
    
i do not understand why it works for regular fractions and not for polynomials... –  phi Nov 17 '12 at 20:37

1 Answer 1

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

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