A set of $m$ non-zero rationals {$a_1, a_2, ... , a_m$} is called a rational Diophantine $m$-tuple if $a_i a_j+1$ is a square. It turns out an $m$-tuple can be extended to $m+2$ if it has certain properties.

I. $m=2$. Given $a,b$:

$$(ax_i+1) = \big(a\pm\sqrt{ab+1}\big)^2\tag1$$

then {$a,b,x_1,x_2$} is a quadruple if,

$$x_1 x_2+1 = 2(a^2+b^2)-(a+b)^2-3 = y^2$$

If $(a-b)^2 = 4$, then one root $x_i$ is equal to zero.

II. $m=3$. Given $a,b,c$:

$$(ax_i+1) = \big(a\sqrt{bc+1}\pm\sqrt{(ab+1)(ac+1)}\big)^2\tag2$$

then {$a,b,c,x_1,x_2$} is a quintuple if,

$$x_1 x_2+1 = 2(a^2+b^2+c^2)-(a+b+c)^2-3 = y^2\,^{\color{red}\dagger}$$

If $(a+b-c)^2 = 4(ab+1)$, then one root $x_i$ is equal to zero.

III. $m=4$. Given $a,b,c,d$:

$$\small(ax_i+1)(abcd-1)^2 = \big(a\sqrt{(bc+1)(bd+1)(cd+1)}\pm\sqrt{(ab+1)(ac+1)(ad+1)}\big)^2\tag3$$

then {$a,b,c,d,x_1,x_2$} is a sextuple if,

$$2(a^2+b^2+c^2+d^2)-(a+b+c+d)^2-3-6abcd+(abcd)^2 = y^2\,^{\color{red}\dagger}$$

If $(a+b-c-d)^2 = 4(ab+1)(cd+1)$, then one root $x_i$ is equal to zero.

$^{\color{red}\dagger}$ Note: These two can be satisfied by the parametric example by Dujella.

IV. Question:

For $m=5$, given $a,b,c,d,e$:

$$\text{LHS}? = \text{RHS}?\tag4$$

1. Can we find $(4)$ analogous to the first three? If yes, then maybe we can use known $5$-tuples or $6$-tuples to generate $7$-tuples, of which there is yet no known example.
2. Or, like quintics, is there a Galois-theoretic restriction on five variables $a,b,c,d,e$ that prevent generalization for $m>4$?