Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)

share|cite|improve this question
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
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
up vote 0 down vote accepted

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.