$(\Rightarrow)$$$(\frac{7}{p})=1\;\; \text{olsun}.$$
$(i)$ Eğer $$p \equiv 1 \mod 4 \;\; \text{ ise} \;\; (\frac{7}{p})=(\frac{p}{7})$$ $$1\equiv p^{\frac{7-1}{2}} \equiv p^3 \mod 7$$
$$p\equiv 1,2,3,4,5,6\;\; \text{sayarken}$$$$p\equiv 1,2 \; \text{ veya} \;4 \mod7$$ olur.
$$p \equiv 1\;\; \mod 4 \; \text{ve} \; p\equiv 1,2 \; \text{veya} \; 4 \mod7 \;\;\; \text{olduğundan}$$ $$p\equiv 1,9 \; \text{veya} \; 25\;\;\mod28$$
$(ii)$ Eğer $$p \equiv 3\;\; \mod 4 \;\; \text{ise} \;\;(\frac{7}{p})=-(\frac{p}{7})$$ $$-1\equiv p^{\frac{7-1}{2}} \equiv p^3 \; \mod 7$$ $$p\equiv 3,5 veya 6 (mod7 \;\; \text{ve} \;\;p \equiv 3 \mod 4 \;\; \text{olduğundan}$$ $$p\equiv 3, 19veya 27 \mod28$$
$$(\frac{7}{p})=1\;\; \text{ ise} \;\; p\equiv 1,3,9,19,25 \;\text{veya} \; 27 \mod28$$ elde edilir.
$(\Leftarrow)$ $$p\equiv 1,3,9,19,25 \;\text{veya} 27 \;\; \mod28 \;\; \text{ olsun}$$
$$(\frac{7}{p})\;\; \text{ tanımı gereği p tek asal.}$$
$(i)$Eğer $$p\equiv 3, 19 \; \text{veya}\;27 \;\mod28 \;\; \text{ise} $$ $$p \equiv 3 \;\; \mod 4 \;\text{,} \;p\equiv 3, 19\; \text{veya} \; 27 \;\; \mod7 \;\; \text{için}\;\; (\frac{7}{p})=-(\frac{p}{7})$$
$$p\equiv 3, 19 \;\text{veya}\; 27 \; \;\mod7 \Rightarrow p\equiv 3, 5\; \text{veya} \;6 \;\; \mod7 $$
$$(\frac{p}{7})=(\frac{3}{7}),(\frac{5}{7}) \; \text{veya} \; (\frac{6}{7})$$
$$(\frac{3}{7})=-(\frac{7}{3})=-(\frac{1}{3})=-1$$
$$(\frac{5}{7})=(\frac{7}{5})=(\frac{2}{5})=-1$$
$$(\frac{6}{7})=(\frac{3}{7})(\frac{2}{7})=(-1)(1)=-1$$
O halde
$$(\frac{7}{p})=-(\frac{p}{7})=1$$
$(ii)$ $$p\equiv 1, 9\;\text{veya}\; 25 \;\;\mod28 \;\; \text{ise} \;\;p \equiv 1\;\; \mod 4$$ , $$p\equiv 1, 9\;\text{veya}\; 25\;\; \mod7\;\;\text{için}\;\;(\frac{7}{p})=(\frac{p}{7})$$
$$p\equiv 1, 9\;\text{veya} \; 25 \;\;\mod7 \Rightarrow p\equiv 1, 2\;\text{veya}\; 4 \;\;\mod7$$
$$(\frac{p}{7})=(\frac{1}{7}),(\frac{2}{7})\;\text{ veya}\; (\frac{4}{7})\;\;\text{=}\;\;1$$
$$ (\frac{7}{p})=(\frac{p}{7})=1$$
$$p\equiv 1,3,9,19,25\;\text{veya}\; 27\;\;\mod28 \;\;\text{ise} (\frac{7}{p})=1$$ elde edilir.