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Ansewrs should be done in excell to improve readability. Any answer submitted in written form will receive a thumbs down as I can never read them fully. thank you/

part A. Determine the multiplicative inverse of (x3 + x + 1) in GF(24) with m(x) = x4 + x + 1

Part B. Develop a set of tables similar to Table 5.3 from the textbook for GF(4) with Modulo (x2 + x + 1).

001 010 011 100 101 110 r2+x+ 1 xx1 12 +x r2 r +1 001 010 011 100 101 110 0 2x x + 1 2 +x r+ 1 r + 1 x +1 x2 + x +x1 r 1 (a) Addition 101 011 x +1 0 r +1 001 010 100 110 0 0 xx1 001 010 01 1 100 101 110 x t 1 2x 2x1 12+x x2x1 2 +x +1 r+x+ 1 r21 (b) Multiplication Table 5.3 Polynomial Arithmetic Modulo z+1)

Explanation / Answer

PART A )

GF(16) is isomorphic to the field of all polynomials of degree, less than
4 = deg(m(x)) with coefficients, belonging to GF(2) = {0, 1}, i.e. 0 or 1.

The multiplicative inverse of P(x) = x³ + x + 1 is a polynomial
Q(x) = ax³ + bx² + cx + d /a, b, c, d = 0 or 1/ such that
P(x) * Q(x) 1 (mod x + x + 1)
Since m(x) = x + x + 1 is irreducible over GF(2), unique multiplicative inverse does exist. Using the Undetermined Coefficients Method we can find that
a = c = 0, b = d = 1, so Q(x) = x² + 1, indeed
(x³ + x + 1)(x² + 1) x + (1 + 1)x³ + x² + x + 1
x + x² + x + 1 x(x + x + 1) + 1 1 (mod m(x))
/remember, 1 + 1 = 0 over GF(2)/, or
(x³ + x + 1)(x² + 1) = 1 over GF(16).

Or you can find it as

PART B)

doing the modulo x^2+x+1: Since all your elements of degree 2 or greater can be reduced considering         x^n=x^2x^(n2) =(x+1)x^(n2), you only will have elements of degree 1 or 0, and this will be the table:

For the case of the multiplication, you can do it

0 1 x x+1 0 0 1 x x+1 1 1 0 x+1 x x x x 0 1 x+1 x+1 x 1 0

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