Exercise 2.7.2 (When a congruence of degree 3 has three solutions?)

Question

Prove that $2 x^3 + 5x^2 + 6x + 1 \equiv 0 \pmod 7$ has three solutions by use of Theorem 2.29.

Richard Ganaye · Accepted Answer

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\begin{proof} 
Since $4(2 x^3 + 5x^2 + 6x + 1) = 8x^3 + 20 x^2 + 24 x + 4= x^3 + 6x^2 + 3x + 4 + 7(x^3 + 2x^2 +  3x)$, $2 x^3 + 5x^2 + 6x + 1 \equiv 0 \pmod 7$ is equivalent to $f(x) = x^3 + 6x^2 + 3x + 4 \equiv 0 \pmod 7$. This gives a monic polynomial $f(x)$, which allows us to apply Theorem 2.29.

Now, in $\Z[x]$,
\begin{align*}
x^7 -x &= (x^4 - 6x^3 + 33x^2 - 184x + 1029)(x^3 + 6 x^2 + 3x + 4)-5754 x^2 - 2352x - 4116\
&=(x^4 - 6x^3 + 33x^2 - 184x + 1029)(x^3 + 6 x^2 + 3x + 4) - 7(822 x^2 + 336x + 588)
\end{align*}
So $x^7 - x  = f(x) q(x) + 7 s(x)$, where $\deg(f) = 3, \deg(q) = 4, \deg(s)<n$, and $f$ is monic.

Then Theorem 2.29 shows that $f(x) \equiv 0 \pmod 7$ has three solutions, thus $2 x^3 + 5x^2 + 6x + 1 \equiv 0 \pmod 7$ has also three solutions.
\end{proof}

Note: If we work in $\F_7[x]$, (where $\F_7 = \Z/7\Z$ is the field with $7$ elements) it is sufficient to verify the equality
$$x^7 - x =  	(2 x^{3} + 5 x^{2} + 6 x + 1)(4 x^{4} + 4 x^{3} + 6 x^{2} + 6 x) \in \F_7[x].$$
So there is no need to take a monic polynomial $f$, and the expression of Theorem 2.29 is simpler:

{\it If $f(x) \in \F_p[x]$\ and $d = \deg(f) >0$, then $f$\ has $d$\ distinct roots in $\F_p$\ if and only if $f(x) \mid x^p - x$.

Exercise 2.7.2 (When a congruence of degree 3 has three solutions?)

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Exercise 2.7.2 (When a congruence of degree 3 has three solutions?)

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