in the wikipedia page about the genus of a quadratic form the wikipedia page about the genus of a quadratic form it is claimed that $Q(x,y)=x^2+82y^2$ and $P(x,y)=2x^2+41y^2$ are not equivalent (over $\mathbb{Z}$) but lie in the same genus (i.e., they are equivalent over $\mathbb{Z}_p$ for every prime $p$), but I am struggling to show this.
It seems like everything boils down to analyzing the system: $$(1) \ \ \ a^2+82c^2=2$$ $$(2) \ \ \ ac+82bd=0$$ $$(3) \ \ \ c^2+82d^2=41$$
where clearly the first equation does not have solutions over $\mathbb{Z}$, so $P$ and $Q$ are not equivalent over $\mathbb{Z}$. However, I do not know how to show that it has solutions over all $\mathbb{Z}_p$.
I have looked in the wikipedia reference, Rational quadratic forms by Cassels. It is suggested there that this follows from: $41 \in \mathbb{Q}_2^{\times}$, $2 \in \mathbb{Q}_{41}^{\times}$ and this lemma. However, I have two problems:
I do not understand the last step of the lemma. I do not understand how can we "extend" a primitive representation of 1 to an integral equivalence with $x_1^2+u_1u_2x_2^2$. I do not understand very what "extend" means.
I do not understand why the case $p=2, 41$ follows from $41 \in \mathbb{Q}_2^{\times}$, $2 \in \mathbb{Q}_{41}^{\times}$. E.g., if $p=41$, if $2 \in \mathbb{Q}_{41}^{\times}$, we can solve equation (1), but I don't see why we can solve equation (3).
If someone could help me understand these two questions of give me an alternative way to show this, I'd appreciate it.
This is Siegel's approach. It is chapter V in The Arithmetic Theory of Quadratic Forms by Burton Wadsworth Jones; pages 105-121. I will need to check how Cassels does this.....
Anyway, the two matrix identities below are a proof that $x^2 + 82 y^2$ and $2x^2 + 41 y^2$ are in the same genus. Each represents the other ``rationally without essential denominator,'' as $7$ does not divide the discriminant. Just divide everything certain matrices by $7$ to make the rationals appear, as in
$$\left( \begin{array}{rr} \frac{4}{7} & \frac{1}{7}\\ \frac{-41}{7}& \frac{2}{7} \\ \end{array} \right) $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$\left( \begin{array}{rr} 4 & 1 \\ -41 & 2 \\ \end{array} \right) \left( \begin{array}{rr} 1 & 0 \\ 0 & 82 \\ \end{array} \right) \left( \begin{array}{rr} 4 & -41 \\ 1 & 2 \\ \end{array} \right) = \left( \begin{array}{rr} 98 & 0 \\ 0 & 2009 \\ \end{array} \right) = 49 \left( \begin{array}{rr} 2 & 0 \\ 0 & 41 \\ \end{array} \right) $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$\left( \begin{array}{rr} 2 & 1 \\ -41 & 4 \\ \end{array} \right) \left( \begin{array}{rr} 2 & 0 \\ 0 & 41 \\ \end{array} \right) \left( \begin{array}{rr} 2 & -41 \\ 1 & 4 \\ \end{array} \right) = \left( \begin{array}{rr} 49 & 0 \\ 0 & 4018 \\ \end{array} \right) = 49 \left( \begin{array}{rr} 1 & 0 \\ 0 & 82 \\ \end{array} \right) $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
Let's see, for a prime $q \neq 2,3,41$ with Legendre $(-328|q) = 1,$ there is an integer expression $u^2 + 82 y^2 = q$ if and only if $x^4 + 6x^2 + 41$ factors into four distinct linear factors in the integers $\pmod q$