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\def\m#1;#2;#3;#4;{\ensuremath{\left[\begin{array}{cc}#1 & #2 \\ #3 & #4\end{array}\right]}}

\noindent Geoffrey Thomas \\
\noindent 18.781 pset 7 \\
\noindent Collaborators: Liz Denys

\begin{enumerate}
\item[3-5 1.]
$F = \m7;25/2;25/2;23;$. Let $M = \m1;-2;0;1;$; then $M^t F M = \m7;-3/2;-3/2;1;$, representing $7x^2 - 3xy + y^2$. Then we transform it to $x^2 + 3xy + 7y^2$, and then to $x^2 + 3xy + 5y^2$, which is reduced.

\item[3-5 3.] $\m x;y;u;v; \in \Gamma$ implies $xv - yu = 1$, which by the extended Euclidean algorithm implies $(x, y) = 1$.

\item[3-5 4.] In one direction, the quadratic form $nx^2 + bxy + cy^2$ for any $b$ and $c$ represents $n$ as $(1, 0)$. In the other direction, 

\item[3-5 5.] 

\item[3-5 9.] 

Let $n = ax^2 + bxy + cy^2$. Then $b^2 - 4ac \pmod{|d|}$, so $4an \equiv 4a^2x^2 + 4abxy + 4acy^2 \equiv 4a^2x^2 + 4abxy + b^2y^2 \equiv (2ax + by)^2 \pmod{|d|}$.

\item[3-6 1.] Consider integers less than 13. These can only be made by combining 1, 4, and 9; in particular 6, 7, 8, and 9 are not sums of squares.

\item[3-6 2.] We want a number composed of as many primes of the form $4k+1$ as possible, and the smallest such primes. $R(n)$ is 4 for each prime, plus another 4 for each unique prime. We can limit our search to about 5, 13, and 17, for anything else would be too big; by trying a few possibilities, we find that $R(5^2 \cdot 13 = 325) = 24$, and we can go no higher.

\item[3-6 4.] This problem looks silly.

\item[3-6 6.] If there are $k$ solutions $(x, y)$ to $x^2 + y^2 = n$ over the positive integers where $(x, y) = 1$, then $(x, y), (x, -y), (-x, y), (-x, -y)$ are all solutions to $x^2 + y^2 = n$ over the integers, and are distinct because neither $x$ nor $y$ can be zero. So the number of solutions $r(n) = 4k$. Therefore $k = r(n) / 4$.

\item[3-6 8.] For integers $p, q, r, s$, where $(p, q) = (r, s) = 1$, write $n = \frac{p^2}{q^2} + \frac{r^2}{s^2} = \frac{p^2s^2 + r^2q^2}{q^2s^2}$. Reduce this fracction to simplest form as follows: suppose it is not. Then there were some factor $k$ in the denominator, it must be a factor of either $q$ or $s$. WLOG let it be a factor of $q$. Then if $k | p^2 s^2 + r^2 q^2$, it is also the case that $k | p^2 s^2$. But we know $k \not | p^2$ because $(p, q) = 1$, so $k | s^2$ and $k | s$, so $k$ is a factor of both $q$ and $s$. Therefore we can cancel $k$ and rewrite using $q' = q/k$ and $s' = s/k$. Ultimately, this fraction needs to be an integer, so the denominator is 1. Therefore, $n$ is the sum of the squares $(ps)^2 + (rq)^2$.

\end{enumerate}
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