Does this system of Diophantine equations have a solution? 
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:
$9ab-3a-3b+1=d^2$
$9ac-3a-3c+1=e^2$
$9bc-3b-3c+1=f^2$


There are infinitely many. Let,
$$a = (2p+1)^2+2p^2\\
b = (2q+1)^2+2q^2\\
c = (2r+1)^2+2r^2$$
then,
$$d =4(3p+1)^2(3q+1)^2\\
e =4(3p+1)^2(3r+1)^2\\
f =4(3q+1)^2(3r+1)^2$$
