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Thread: Questions about polynomials over Z mod 5

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    Questions about polynomials over Z mod 5

    How many irreducible quadratic polynomials are there over Z mod 5?

    Thanks!
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  2. #2
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    Quote Originally Posted by xxie View Post

    How many irreducible quadratic polynomials are there over Z mod 5?

    Thanks!
    let's solve the problem for the general case, i.e. over $\displaystyle \mathbb{Z}/p,$ where $\displaystyle p$ is any prime number. a quadratic polynomial is irreducible iff it has no root. this number is obviously $\displaystyle n=p^2(p-1) - k,$ where

    $\displaystyle k$ is the number of those polynomials of degree 2 which are reducible, i.e. they are in the form $\displaystyle a(x-b)(x-c).$ clearly $\displaystyle a$ can be any element of $\displaystyle \mathbb{Z}/p - \{0 \}.$ so there are $\displaystyle p-1$ possibilities for $\displaystyle a.$

    we also have $\displaystyle \binom{p}{2}$ distinct pairs $\displaystyle a,b$ in $\displaystyle \mathbb{Z}/p.$ it is also possible to have $\displaystyle a=b,$ which gives us $\displaystyle p$ possibilities. thus $\displaystyle k=(p-1) \left(\binom{p}{2} + p \right)=\frac{p(p^2-1)}{2}.$ thus: $\displaystyle n=p^2(p-1)-k=\frac{p(p-1)^2}{2}.$

    if you're looking for the number of monic irreducible polynomials of degree 2, then $\displaystyle a=1$ would be the only possibility for $\displaystyle a$ and thus in this case $\displaystyle n=\frac{p(p-1)}{2}.$
    Last edited by NonCommAlg; May 6th 2009 at 08:40 PM. Reason: typo!
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