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\noindent Geoffrey Thomas \\ 18.781 problem set 5 \\ Collaborators: Liz Denys

\begin{enumerate}
\item[2.7 2]
\item[2.7 9]
\item[1]
\item[3.1 6b] The minimum value of $\left|n^2 - 17x\left|$ is the integer of smallest absolute value in the same congruence class as $n^2 \pmod{17}$, which is either a quadratic residue of 17 or is 0, for all possible values of $n$.. The quadratic residues of 17 are $-8, -4, -2, -1, 1, 2, 4, 8$.
\item[3.1 7a] $2^{30} \equiv -1 \pmod{61} \Rightarrow$ no solutions
\item[3.1 7d] ${-2}^{29} \equiv -58 \equiv 1 \pmod {59} \Rightarrow$ two solutions
\item[3.1 7e] $x^2 \equiv 2 \pmod{61}$ has no solutions, so neither can $x^2 \equiv 2 \pmod{122}$.
\item[3.1 7h] $x^2 \equiv -2 \pmod{2}$ has one solution, so by theorem 2.20 and pard d, the original congruence has two solutions.
\item[3.1 8c] $365 = 5 \times 73$. $x^2 \equiv -1 \pmod{5}$ has 2 solutions, and $x^2 \equiv -1 \pmod{73}$ 2, so the original congruence has 4 solutions.
\item[3.1 8d] $3599 = 59 \times 61$. $x^2 \equiv -1 \pmod{59}$ has no solutions, so neither does the original congruence.
\item[3.1 8f] $244 = 4 \times 61$. $x^2 \equiv -1 \pmod{4}$ has no solutions, so neither does the original congruence.
\item[3.1 14] The sum of the quadratic residues is simply the sum of consecutive squares $\frac{1}{6}\frac{p-1}{2}\frac{p+1}{2}p + kp$ for some integer $k$; this is divisible by $p$.
\item[3.1 17]
\item[3.1 18]
\item[3.1 19] $x^8 \equiv 16 \pmod{p}$ has solutions for $x^4$ as long as $1 \equiv \left(\frac{16}{p}\right) \equiv 16^\frac{p-1}{2} \equiv 4^\frac{p-1} \pmod{p}$, which by Fermat's little theorem is true. Since a solution exists, it must be congruent to $4 \pmod{p}$. By similar reasoning, $x^4 \equiv 4 \pmod{p}$ has solutions for $x^2$ since $1 \equiv 2^{p-1} \pmod{p}$, and since a solution exists it must be congruent to $2 \pmod{p}$. $x^2 \equiv 2 \pmod{p}$ 
\item[3.1 20] If there are $m$ quadratic residues less than $p/2$
