Cryptography20785 marksNumerical: Polynomial division GF(7)Finite Fields
Divide 3x2 + 4x + 3 by 5x + 6 over GF(7).
Divide $3x^2 + 4x + 3$ by $5x + 6$ over GF(7). [5]
- Dividend: $3x^2 + 4x + 3$ - Divisor: $5x + 6$ - Field: $GF(7)$, all arithmetic mod 7. We need $5^{-1} \pmod 7$: since $5 \cdot 3 = 15 \equiv 1 \pmod 7$, we have $$5^{-1} = 3 \text{ in } GF(7)$$ $$\frac{3x^2}{5x} = 3 \cdot 5^{-1} x = 3 \cdot 3 , x = 9x \equiv 2x \pmod 7$$ Multiply back: $$2x(5x...