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\noindent Geoffrey Thomas \\
\noindent 18.781 problem set 4

\begin{enumerate}
\item[2.6 2.] The single solution of $x^5 + x^4 + 1 \equiv 0 \pmod{3}$ is $x \equiv 1 \pmod{3}$. Note that $f'(1) = 9 \equiv 0 \pmod{3}$, but $1^5 + 1^4 + 1 \not\equiv 0 \pmod{9}$, so the root cannot be lifted and there are no solutions.

\item[2.6 3.] The single solution of $x^3 + x + 57 \equiv 0 \pmod{5}$ is $x \equiv 4 \pmod{5}$. We lift this root to $4 - 125 \cdot 4 \equiv 4 \pmod{5^2}$, which then lifts to $4 - 125 \cdot 4 \equiv 4 \pmod{5^3}$.

\item[2.6 7.] The single solution of $x^3 + x^2 - 5 \equiv 0 \pmod{7}$ is $x \equiv 2 \pmod{7}$. We lift this root to $2 - 7 \cdot 4 \equiv 23 \pmod{7^2}$, which then lifts to $23 - 12691 \cdot 4 \equiv 23 \pmod{7^3}$.

\item[2.6 9.]
\item[2.6 10.] Hensel's lemma tells us that if there is a solution $x_0$ to $x^2 \equiv a \pmod{p^j}$, then as long as $2x_0 \not\equiv 0 \pmod{p}$, then there is a solution to $x^2 \equiv a \pmod{p^{j+1}}$. Our constraint is true, because if $2x_0 \equiv 0 \pmod{p}$, then $0 \equiv 4x_0^2 \equiv a \pmod{p}$, but we know that $0 \not\equiv a \pmod{p}$. Therefore, by mathematial induction, as long as $x^2 \equiv a \pmod{p^j}$ has a solution for $j=1$, it has a solution for all positive integers $j$.

\item[2.7 3.] As $(x^{13} + 12x)x \equiv 0 \pmod{13}$, the congruence is true if $x \equiv 0$ or $x^{13} + 12x \equiv 0 \pmod{13}$. By Fermat's Little Theorem, the latter is equivalent to $x + 12x \equiv 0 \pmod{13}$, which is identical.

\item[2.7 8.]
\item[2.7 10.] The unreduced numerator is $\sigma_{p-2}$ from the discussion at the end of the section, so by Wolstenholme's congruence, $p^2$ divides it. The reduced numerator $a$ is the same as the unreduced one
\item[2.7 11.]

\item[2.8 2.] By guessing numbers until they work, 5.
\item[2.8 9.] $3^4 \equiv -4 \pmod{17}$, so $3^8 \equiv (-4)^2 \equiv -1 \pmod{17}$. Therefore $3^16 \equiv (-1)^2 \equiv 1 \pmod{17}$, so the order of 3 divides 16. But the order of 3 does not divide 8, so the order must be exactly 16.
\item[2.8 12.]
\item[2.8 13.]
\item[2.8 16.]
\item[2.8 24.] Under these conditions, the order of $a$ modulo $n$ is $n-1$, so $\phi(n) \ge n-1$, so $n$ must be prime.
\end{enumerate}

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