Show that the following integers are quadratic residues of 31:
(a) 35
(b) 71
(c) 56
(d) 94
(e) 47
Let's remind ourselves of Proposition (7.9).
Let $p$ be an odd prime and $a, b$ be integers such that $p \not \mid a$ and $p \not \mid b$. We have
(a) If $a \equiv b \pmod p$ then $(\frac{a}{p}) = (\frac{b}{p})$.
(b) $(\frac{a^2}{p})=1$.
(c) $(\frac{a \times b}{p}) = (\frac{a}{p}) \times (\frac{b}{p})$, multiplicative property.
(a) Using Proposition (7.9)(a) and (b) we have
$$ (\frac{35}{31}) = (\frac{4}{31})= (\frac{2^2}{31}) = 1 $$
And so by Definition (7.7) of a Legendre Symbol, 35 is a quadratic residue of 31.
(b) Similarly
$$ (\frac{71}{31}) = (\frac{9}{31}) = (\frac{3^2}{31}) = 1 $$
And so 71 is a quadratic root of 31.
(c) Similarly
$$ (\frac{56}{31}) = (\frac{25}{31}) = (\frac{5^2}{31}) = 1$$
And so 56 is a quadratic residue of 31.
(d) Similarly
$$ (\frac{94}{31}) = (\frac{1}{31}) = (\frac{1^2}{31})= 1$$
And so 94 is a quadratic residue of 31.
(e) Similarly
$$ (\frac{47}{31}) = (\frac{16}{31}) = (\frac{4^2}{31}) = 1$$
And so 47 is a quadratic residue of 31.