a) We have given the set $S = \{x^2 + 3xy + 8y^2 \;\; | \;\; x,y \in \mathbb{Z}\}$. By setting $P(x,y) = x^2 + 3xy + 8y^2$ we obtain $(1) \;\; P(x,y) = \frac{(2x + 3y)^2 + 23y^2}{4}$. Hence, if $u,v \in S$, then there exists four integers $a,b,c,d$ s.t.$u=P(a,b)$ and $v=P(c,d)$, which according to formula (1) give us $uv = \frac{(2a + 3b)^2 + 23b^2}{4} \cdot \frac{(2c + 3d)^2 + 23d^2}{4} = \frac{1}{4} \bigg[ \bigg( \frac{(2a + 3b)(2c + 3d) - 23bd}{2} \bigg)^2 + 23 \bigg( \frac{(2a + 3b)d + (2c + 3d)b}{2} \bigg)^2 \bigg] \in S$ q.e.d. b) The congruence $P(x,y) \equiv 7 \pmod {23}$ is according to formula (1) equivalent to $(2) \;\; (2x + 3y)^2 \equiv 28 \pmod{23}$. In order to solve the quadratic cngruence (2), we calculate the value of the Legendre symbol $(28/23) = (-18/23) = (3^2/23) \cdot (-1/23) \cdot (2/23) = 1 \cdot (-1)^{\frac{23-1}{2}} \cdot (-1)^{\frac{23^2-1}{8}} = 1 \cdot (-1)^{11} \cdot (-1)^{66} = (-1) \cdot 1 = -1$. Conclusion: An integer of the form $23k + 7$ does not belong to $S$.