 Solution

The Chinese remainder theorem says that for pairwise coprime positive integers $m_i$, the mapis a bijection. For two moduli, choose $u,v$ with $um+vn=1$; thenhas residues $a$ modulo $m$ and $b$ modulo $n$. Uniqueness follows because the difference of two solutions is divisible by both coprime moduli, hence by their product. Induction proves the general case.

The two given congruences are compatible modulo $\gcd (6,8)=2$, and checking modulo $\mathrm{lcm}(6,8)=24$ gives

Write $d=2^r{3}^{s}m$ with $(m,6)=1$. Use the Chinese remainder theorem to chooseThen $4a^2+9b^2\equiv 1$ modulo each of the pairwise coprime factors $2^r$, $3^s$, and $m$, hence modulo $d$.