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So for the first part of the question, I have to find the solutions to:

$\ x^2+4x+3=0$ over $\mathbb{Z}_{16}$ I have found these to be $X=5,7,13$ and $15$, just by using standard method to solve a quadratic over a field. However the next part of the question is to find

$\ x^{128}+4x+3=0$ over $\mathbb{Z}_{127}$ How would I go about solving this?

Bérénice
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rosiegeorge
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  • Note that every nonzero element of $\Bbb Z_{127}$ has multiplicative order dividing $126$. – Travis Willse May 17 '16 at 13:42
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    Note that $\Bbb{Z}{16}$ is NOT a field. The first equation is equivalent to $(x+2)^2=1$ though. So your set of solutions follows from knowing the four solutions of $y^2=1$ in $\Bbb{Z}{16}$. OTOH $\Bbb{Z}_{127}$ is a field as $127$ is a prime. – Jyrki Lahtonen May 17 '16 at 14:13

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Hint For the first question : Use $x^2+4x+3=(x+1)(x+3)$

Hint for the second question : In $Z_{127}$ you have $x^{127}=x$ for all $x$

Peter
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