Determine whether the following integers are quadratic residues of 47:
(a) 46
(b) 95
(c) 90
(d) 58
(e) 90 × 58
(a) Using Proposition (7.9) we have
$$ (\frac{46}{47}) = (\frac{-1}{47}) = (-1)^{\frac{47-1}{2}} = (-1)^{23} = -1 $$
By Definition (7.7) of the Legendre Symbol, 46 is not a quadratic residue of 47.
(b) Using Proposition (7.9) we have
$$ (\frac{95}{47}) = (\frac{1}{47}) = 1 $$
And so 95 is a quadratic residue of 47.
(c) Using Proposition (7.9) we have
$$ (\frac{90}{47}) = (\frac{-1 \times 4}{47}) = (\frac{-1}{47}) \times (\frac{2^2}{47}) = -1 \times 1 = -1 $$
And so 90 is not a quadratic residue of 47.
(d) Using Proposition (7.9) we have
$$ (\frac{58}{47}) = (\frac{11}{47}) \equiv (11)^{23} \equiv (11^4)^5 \times 11^3 \equiv (24)^5 \times 15 \equiv 25 \times 15 \equiv -1 \pmod {47}$$
And so 58 is not a quadratic residue of 47.
(e) Using Proposition (7.9) we have
$$ (\frac{90 \times 58}{47}) = (\frac{90}{47}) \times (\frac{58}{47}) = -1 \times -1 = 1$$
And so $90 \times 58$ is a quadratic residue of 47.