Solving the two quadratics for $a, c$ we find \[a = 2b \pm \sqrt{3(b^2 - 1)}\] \[c = 2d \pm \sqrt{3(d^2 - 1)}\] For $a, c$ to be integers, $(b, p), (d, q)$ for some $p, q$ must be solutions to the Pell equation \[x^2 - 3y^2 = 1\] Take $(x_n, y_n)$ to be the $n^{th}$ smallest solution to this equation (for example, $x_1 = 2, y_1 = 1$) and consider the quadruple \[(2x_t + 3y_t, x_t, 2x_{3t} + 3y_{3t}, x_{3t})\] From our work above, the first and second equations are satisfied. But since all solutions to the Pell equation can be explicitly calculated as \[x_k = \frac{(2 + \sqrt{3})^k + (2 - \sqrt{3})^k}{2}\] \[y_k = \frac{(2 + \sqrt{3})^k - (2 - \sqrt{3})^k}{2}\] some quick algebra reveals that the third equation is also satisfied, and we generate an infinitude of solutions by letting $t$ vary.

consider $c^2+d^2+3=4cd$....(*). then write it as a quadratic in c .$ c^2-4cd+(d^2+3)=0 $, then $D=16d^2-4d^2-12=12(d^2-1)$ =perfect square = $y^2$ , now clearly 6 divides y , put y=6m. then it reduces to $d^2-3m^2=1.$ a solution is (2,1). now using the well known Pell equation recursion , we can enumerate all (d,n) , hence (c,d) recursion is ${d_k+ 3^(1/2).[m_k]}{d_k-3^(1/2).[m_k]}= [(2+ sqrt.3)^k.][(2-sqrt.3]^k$. similarly , we can find all (a,b) from the 1st eqn.. now using the explicit forms of c,d ,a, b ,it is not tough to see that there are infinitely many solutions where $a=4c^3-3c.$[because,actually we can show that there are infinitely many c such that if (c,d) is a solution , then also $(4c^3-3c, D)$ is a solution for (*) , so also a solution for the 1st eqn.with some algebric manipulation & our seeked solution is reached.].hence done.