Find at least one example of five rational numbers $x_1, \; x_2, …, \; x_5$ such that
i) $x_k > 0$ for all $k=1,2,…,5$;
ii) $x_k$ is not an integer for all $k=1,2,…,5$;
iii) if $k \ne m$ then $x_k \ne x_m$ (for all $k=1,2,…,5; \; m=1,2,…,5$);
iv) if $k \ne m$ then $x_k x_m+1$ is a square of some rational number (for all $k=1,2,…,5; \; m=1,2,…,5$).
My work. I found infinitely many examples of four rational numbers. Let $a$ and $b$ are rational numbers. Let $$x_1=a \\ x_2=ab^2+2b \\ x_3=a(b+1)^2+2(b+1) \\ x_4=a \left( 2b(b+1)a+4b+2 \right)^2+2\left( 2b(b+1)a+4b+2 \right).$$ Then $$x_1x_2+1=\left( ab+1\right)^2 \\ x_1x_3+1=\left( a(b+1)+1\right)^2 \\ x_1x_4+1=\left(a \left( 2b(b+1)a+4b+2 \right)+1\right)^2 \\ x_2x_3+1=\left( ab(b+1)+2b+1\right)^2 \\ x_2x_4+1=\left( 2a^2b^2(b+1)+2ab(3b+1)+4b+1\right)^2 \\ x_3x_4+1=\left( 2a^2b(b+1)^2+2a(b+1)(3b+1)+6b+3\right)^2. $$
I also have one idea. Can see that $x_4=x_2 \left(2a(b+1)+2 \right)^2+2 \left(2a(b+1)+2 \right)$. Let $x_5= x_2 \left(2a(b+1)+2 \pm 1\right)^2+2 \left(2a(b+1)+2 \pm 1\right)$. Then $$x_4x_5+1=\left(x_2 \left(2a(b+1)+2 \right) \left(2a(b+1)+2 \pm 1\right)+2\left(2a(b+1)+2 \right) \pm 1 \right)^2 \\ x_2x_5+1=\left( x_2\left(2a(b+1)+2 \pm 1 \right)+1\right)^2.$$
It remains to find the rational numbers $a$ and $b$ such that the numbers $x_1x_5+1$ and $x_3x_5+1$ are squares of some rational numbers.
Best Answer
See OEIS A192629 for $$1,3,8,120,\frac {777480}{8288641}$$ and OEIS A030063 (which adds $0$ and deletes the fraction) for more information.