$n ^ 2 +5 $, $n ^ 2 +10 $ are rational number square Seeking a nonzero rational number $n$, 
such that $n ^ 2 +5 $,   $n ^ 2 +10 $  are rational number square。
This is a high school students asked the question, answer $n=\frac{31}{12}$, but no answer process. 
I try to follow Parametrization of a conic and rational solutions
Method to solve, If ${{n}^{2}}+5={{a}^{2}}$，${{2}^{2}}+5={{3}^{2}}$，$n-3=t(a-2)$，so $a=\frac{-3t+2t^2-\sqrt{14-12t-t^2}}{-1+t^2}$, requires radical equation is a square number. It seems, is a cycle and then proceed.

Let $p$ and $q$ integers so that $n=\frac{p}{q}$, then
$$n^2+5=\frac{p^2+5q^2}{q^2} \quad(1)$$ and $$n^2+10=\frac{p^2+10q^2}{q^2} \quad(2)$$
Let $x$ and $y$ be integers so that $x^2=p^2+5q^2$ and $y^2=p^2+10q^2$, then
$$(x-p)(x+p)=5q^2 \quad(3)$$
and
$$y^2=p^2+10q^2 \quad(4)$$
Let´s factorise $q$ as $q=f_1f_2$, then we will get from $(3)$:
$$(x-p)(x+p)=5f_1^2f_2^2 \quad(5)$$
Using equation $(5)$, let's choose the following decomposition:
$$(x-p)=5f_1 \quad(6)$$
and
$$(x+p)=f_1f_2^2 \quad(7)$$
Solving $(6)$ and $(7)$ we get:
$$p=\frac{f_1}{2}(f_2^2-5) \quad(8)$$
$$x=5f_1+p \quad(9)$$
Replacing the value of $p$ in equation $(4)$  we get:
$$y=\frac{f_1}{2}\sqrt{f_2^4+30f_2^2+25} \quad(10)$$
Now we must choose $f_1$ and $f_2$ so that $p$,$y$ and $x$ are integers.
By inspection if we choose $f_2=6$ we get from equations $(10)$ and $(8)$:
$$y=\frac{f_1}{2}49$$
$$p=\frac{f_1}{2}31$$
Now if we choose $f_1=2$, we get $p=31$, $y= 49$,$x=41$ and $q=12$.
Therefore
$$n=\frac{31}{12}$$
