A set of $m$ non-zero ***rationals*** {$a_1, a_2, ... , a_m$} is called a *[rational Diophantine $m$-tuple][1]* 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. The problem is to generalize the relations below to $m=5$. >**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, $$2(a^2+b^2)-(a+b)^2-3 = y^2$$ >**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, $$2(a^2+b^2+c^2)-(a+b+c)^2-3 = y^2\,^{\color{red}\dagger}$$ >**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, $$2(a^2+b^2+c^2+d^2)-(a+b+c+d)^2-3-6abcd+(abcd)^2 = y^2\,^{\color{red}\dagger}$$ >**IV. Notes:** $^{\color{red}\dagger}$ These two can be satisfied by the parametric [example][2] in the variable $t$ by Dujella. In general, one root $x_i$ is equal to zero if, $$(a-b)^2 = 4\\ (a+b-c)^2 = 4(ab+1)\\ (a+b-c-d)^2 = 4(ab+1)(cd+1)$$ for relations $(1), (2), (3)$, respectively. >**V. 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. The pattern is suggestive. But, like quintics, is there a Galois-theoretic restriction on five variables $a,b,c,d,e$ that prevent generalization for $m>4$? [1]: https://web.math.pmf.unizg.hr/~duje/intro.html [2]: https://web.math.pmf.unizg.hr/~duje/ratio.html