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Problem 34 (Why Are Zero Divisors Such A Problem)
Consider the polynomial $p(x)=x^2-5x+6 = (x-2)(x-3)$. We can view this as a polynomial in $\mathbb{Z}_n[x]$ for any given $n$.
- Find all solutions to the equation $x^2-5x+6=0$ as a polynomial in $\mathbb{Z}_5[x]$.
- Show that the equation $x^2-5x+6=0$ as a polynomial in $\mathbb{Z}_6[x]$ has 4 solutions. Then show that $x^2-5x+6$ can be factored as both $(x+4)(x+3)$ and $(x)(x+1)$. As a suggestion, since there are only 6 numbers in $\mathbb{Z}_6$, you can quickly determine which is a zero by just plugging it into the equation and then seeing if you get 0 mod 6.
- Find all solutions to $x^2-5x+6=0$ as a polynomial in $\mathbb{Z}_8[x]$. Then factor this polynomial using only coefficients in $\mathbb{Z}_8$. Again, there are only 8 numbers to test to see if they are zeros. Just plug them all in. You might want to use Excel or write a computer program to make this fast.
- Find all solutions to $x^2-5x+6=0$ as a polynomial in $\mathbb{Z}_{14}[x]$. Factor this polynomial using only coefficients in $\mathbb{Z}_{14}$. Then factor it in a different way.
- Find all solutions to $x^2-5x+6=0$ as a polynomial in $\mathbb{Z}_{30}[x]$. (You should get more than 4 answers.) One way to factor the polynomial is of course $(x-2)(x-3)=(x+28)(x+27)$. Find one other way.
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