Exercise 5.5.6 (Reformulation of Legendre's Theorem)

Question

Show that in the proof of Theorem 5.11 we have established more than the theorem stated, that the following stronger result is implied. Let $a,b,c$ be nonzero integers not of the same sign such that the product $abc$ is square-free. Then the following three conditions are equivalent.
\begin{enumerate}
\item[(a)] $ax^2 + by^2 + cz^2 = 0$ has a solution $x,y,z$ not all zero;
\item[(b)] $ax^2 + by^2 + cz^2 = 0$ factors into linear factors modulo $abc$;
\item[(c)] $-bc, -ac, -ab$ are quadratic residues modulo $a,b,c$, respectively.
\end{enumerate}

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $a,b,c$ be nonzero integers not of the same sign such that the product $abc$ is square-free.  We show the equivalence $(a) \iff (b) \iff (c)$.
\begin{enumerate}
\item[$(a) \Rightarrow (c)$]  This is established at the beginning of the proof of Theorem 5.11.

\item[$(c) \Rightarrow (b)$] The assertion (b) in proved in 5.39 under the hypothesis (c).

\item[ $(b) \Rightarrow (a)$] Starting from 5.39, the end of the proof shows that $ax^2 + by^2 + cz^2 = 0$ has a solution $x,y,z$ not all zero.
\end{enumerate}
There is nothing new to prove.
\end{proof}

Exercise 5.5.6 (Reformulation of Legendre's Theorem)

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Exercise 5.5.6 (Reformulation of Legendre's Theorem)

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