yogi55hr

2021-11-08

Consider $A\left(x\right)={x}^{3}+{x}^{2}+1$ and $B\left(x\right)={x}^{2}+x+1$ in $GF\left(17\right)$. Also, the irreducible polynomial is given as $p\left(x\right)={x}^{3}+1$
1. Find ${A}^{2}\left(x\right)—B\left(x\right)$. (13 pts)
Hint: Use $p\left(x\right)$ if the initial result does not fall in $GF\left(17\right)$.
2. Find ${A}^{2}\left(x\right)\cdot {B}^{2}\left(x\right)$. (12 pts)
Hint: Use $p\left(x\right)$ if the initial result does not fall in $GF\left(17\right)$.

Lounctirough

1) Find ${A}^{2}\left(x\right)-B\left(x\right)$
${A}^{2}\left(x\right)={\left({x}^{3}+{x}^{2}+1\right)}^{2}$
$=\left({x}^{3}+{x}^{2}+1\right)\left({x}^{3}+{x}^{2}+1\right)$
$={x}^{6}+{x}^{5}+{x}^{3}+{x}^{5}+{x}^{4}+{x}^{2}+{x}^{3}+{x}^{2}+1$
$={x}^{6}+2{x}^{5}+{x}^{4}+2{x}^{3}+2{x}^{2}+1$
${A}^{2}\left(x\right)={x}^{6}+2{x}^{5}+{x}^{4}+2{x}^{3}+2{x}^{2}+1$
$\to {A}^{2}\left(x\right)-B\left(x\right)$
$={x}^{6}+2{x}^{5}+2{x}^{3}+{x}^{4}+2{x}^{3}+2{x}^{2}+\mathrm{¬}1-{x}^{2}-x-\mathrm{¬}1$
$={x}^{6}+2{x}^{5}+2{x}^{3}+{x}^{4}+{x}^{2}-x$
${A}^{2}\left(x\right)-B\left(x\right)={x}^{6}+2{x}^{5}+2{x}^{3}+{x}^{4}+{x}^{2}-x$
2) Find ${A}^{2}\left(x\right)\cdot {B}^{2}\left(x\right)$
${A}^{2}\left(x\right)={x}^{6}+2{x}^{5}+{x}^{4}+2{x}^{3}+2{x}^{2}+1$
${B}^{2}\left(x\right)=\left({x}^{2}+x+1\right)\left({x}^{2}+x+1\right)$
$={x}^{4}+{x}^{3}+{x}^{2}+{x}^{3}+{x}^{2}+x+{x}^{2}+x+1$

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