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.

The general solution of $b^2-a^2=5$ has
$$b=\frac12\left(t+\frac 5t\right)$$
for $t\in\Bbb Q^*$. Then
$$c^2=b^2+5=\frac{t^4+10t^2+25}{4t^2}+5=\frac{t^4+30t^2+25}{4t^2}.$$
The problem boils down to whether the genus-one curve
$$y^2=x^4+30x^2+25$$
has rational points.
