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{\bf Corollary 17.2.}\quad {\sl Suppose $(a,p)=1.$ Then $(a↑2/p)=1$, 
$(ab/p)=(a/p)(b/p)$, and if $a≡b \mod{p} $, then $(a/p)=(b/p).$}

{\it Note.}\quad The equation $({ab\over p})=(a/p)(b/p)$ says that the product of two
residues is a residue, the product of two non-residues is a residue, and the
product of are sidue and a non--residue is a non---residue.

The problem of deciding. . . .  If $a>1$, and $(a,p)=1$, 
suppose $a=\prod q↓i↑{β↑i}$. Then $(a/p)=\prod (q↓i/p)↑{β↓i}$.
$$(a/p)=\prod↓{β↓{i}\hbox{odd}} (q↓i/p).$$
If $a<-1$, then
$$(a/p)=(-1)↑{(p-1)/2} \prod↓{β↓{i}\hbox{odd}}(q↓i/p).$$
{\sl Example.} Find (--540/7)=. . . .
$$L↓1:y= {1\over2}$$
$$L↓2:y= {q-1\over2}$$

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