Question

Multiplication in GF(24 ): Compute A(x)B(x) mod P(x) in GF(24 ) using the irreducible polynomial P(x) = x 4 + x + 1, where A(x) = x2 + 1, and B(x) = x3 + x + 1

Answer 1

Answer:

Step-by-step explanation:

hello,

i advice you check the question again if it is GF($2^{4}$) or GF(24). i believe the question should rather be in this form;

multiplication in GF($2^{4}$): Compute A(x)B(x) mod P(x) = $x^{4}$ + $x$+1, where A(x)=$x^{2}$+1, and B(x)=$x^{3} + x+1$.

i will solve the above question and i believe with this you will be able to solve any related problem.

A(x)B(x)=$(x^{2} +1) (x^{3}+x+1) mod (x^{4}+x+1 ) = (x^{5} +x^{3}+x^{2} ) + (x^{3}+x+1 ) mod (x^{4} + x+1 )$

= $x^{5}+2x^{3} +x^{2} + x + 1 mod(x^{4}+x+1 )$

=$2x^{2} +1$

please note that the division by the modulus above we used

$\frac{x^{5}+2x^{3}+x^{2} +1 }{x^{4}+x+1}= x+\frac{2x^{3} +1}{x^{4}+x+1}$