Does this system of Diophantine equations have a solution?

Question

Are there natural numbers $a,b,c,d,e,f$ such that we have $a \neq b$ and $a \neq c$ and $b \neq c$ and that they are solution of this system of equations:

$9ab-3a-3b+1=d^2$

$9ac-3a-3c+1=e^2$

$9bc-3b-3c+1=f^2$

Answer 1

Yes, let $3a-1,3b-1,3c-1$ be $2^3,2^5,2^7$.

Answer 2

There are infinitely many. Let,

$$a = (2p+1)^2+2p^2\\ b = (2q+1)^2+2q^2\\ c = (2r+1)^2+2r^2$$

then,

$$d =4(3p+1)^2(3q+1)^2\\ e =4(3p+1)^2(3r+1)^2\\ f =4(3q+1)^2(3r+1)^2$$

Answer 3

Yes, infinitely many. All the solutions can be obtained from the conditions $$\begin{cases} 3a-1=(3t-1)x^2\\ 3b-1=(3t-1)y^2\\ 3c-1=(3t-1)z^2\\ 3\not|\ xyz,\\ x,y,z,t\in\mathbb Z \end{cases}$$