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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 ?2012-11-17
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    well using the gcd algorithm for polynomials gives that answer2012-11-17
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    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 ...2012-11-17
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    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.2012-11-17
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    i do not understand why it works for regular fractions and not for polynomials...2012-11-17
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    is it because the polynomial is prime...but when we replace n with some number its not a polynomial anymore and it could be divisible by some number?2012-11-17
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    $n+1$ and $n+3$ either are both even or both odd. Their $\gcd$ is at most $2$ because any number which divides $n+1$ leaves remainder $2$ when dividing $n+3$.2012-11-17
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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 $\rm\ (k,2\!+\!k) = (\color{#C00}k,2\!+\!k\!-\!\color{#C00}k) = (k,2) = (r\!+\!2j,2) = (r\!+\!2j\!-\!\color{#C00}2j,\color{#C00}2) = (r,2)\ [ = 2\ if\ 2\mid r,\ else\ 1]$