Exercise 2.3.16 (Solve $x^3 -9x^2 + 23 x - 15 \equiv 0 \pmod {503}$)

Question

Solve the congruence $x^3 - 9 x^2 + 23x - 15 \equiv 0 \pmod {503}$ by observing that $503$ is a prime and that the polynomial factors into $(x-1)(x-3)(x-5)$.

Richard Ganaye · Accepted Answer

\begin{proof} All is said in the sentence.
\begin{align*}
(x-1)(x-3)(x-5) &= x^3 -\sigma_1 x^2 + \sigma_2 x - \sigma_3\
&=x^3 - 9x^2 +23 x - 15.\
\end{align*}
Since $503$ is prime, the congruence $x^3 - 9 x^2 + 23x - 15 \equiv 0 \pmod {503}$ is equivalent to
$$x \equiv 1 	ext{ or } x \equiv 3 	ext{ or } x \equiv 5 \pmod{503}.$$
\end{proof}

With Sage:
\begin{verbatim}
sage:  p = 503
sage: p.is_prime()
True
sage: R.<x> = GF(p)[]
sage: p = x^3 - 9*x^2 + 23*x - 15
sage: p.factor()
(x + 498) * (x + 500) * (x + 502)
sage: p == (x-1)*(x-3)*(x-5)
True
\end{verbatim}

Exercise 2.3.16 (Solve $x^3 -9x^2 + 23 x - 15 \equiv 0 \pmod {503}$)

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Exercise 2.3.16 (Solve $x^3 -9x^2 + 23 x - 15 \equiv 0 \pmod {503}$)

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