Find three rational numbers $a,b,c$ s.t. $b^2-a^2=c^2-b^2=5$. A rational numbers cannot have irrational as$\exists n \in \mathbb{Z}, \, \sqrt[n]{\frac xy}$, but the two equalities give: $b^2=\frac{(a^2+c^2)}{2} \implies b = \sqrt[2]{\frac {(a^2+c^2)}{2}}$.
To avoid this, need $4\mid a$, & $a=c$; so that if $\exists t=a/2, \, b = \sqrt[2]{\frac {(a^2+c^2)}{2}} => 2t$.
As an example, $a=c=8, t=4, b=8$.
I am giving an idea for a particular case only, & am unable to pursue.

We can rewrite these numbers with common denominator $D$:
$a=\dfrac{A}{D}$, $\;b=\dfrac{B}{D}$, $\;c=\dfrac{C}{D}$. Then we'll get diophantine equation (system of diophantine equations):
$$
B^2-A^2=C^2-B^2=5D^2.\tag{1}
$$
The smallest solution of $(1)$ is $(A,B,C,D)=(31,41,49,12)$.
So, example of such three rational numbers $a,b,c$ is: $$a=\dfrac{31}{12},\quad b=\dfrac{41}{12}, \quad c=\dfrac{49}{12}.$$

Short algorithm (to find $A,B,C,D$) description:


*

*for each $C$ (from some range) consider $B = 1,2,\ldots,C-1$;


*

*if $2B^2-C^2$ is square number, then denote $A=\sqrt{2B^2-C^2}$;


*

*if $\dfrac{B^2-A^2}{5}$ is square number, then denote $D=\sqrt{\frac{B^2-A^2}{5}}$ and output $(A,B,C,D)$.



