$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.
 A: Assume $n=\frac pq$ with $\gcd(p,q)=1$ is a solution, i.e.
$$n^2+5=a^2,\quad n^2+10=b^2\quad\text{with }a,b\in\mathbb Q.$$
Also assume that $n$ is a solution with minimal $q$.
Then we have
$$\tag1p^2+5q^2=(qa)^2,\quad p^2 +10q^2=(qb)^2$$
that is $c:=qa$ and $d:=qb$ are integers (because the square root of an integer is either integer or irrational).
Any prime (including $5$!) dividing both $c$ and $p$ would also divide $q$, hence $\gcd(c,p)=1$. Similarly $\gcd(c,q)=\gcd(d,p)=\gcd(d,q)=1$.
From $(1)$ we find $10q^2=2(c^2-p^2)=(d^2-p^2)$, hence
$2c^2=d^2+p^2$ and finally
$$(d+p)^2+(d-p)^2=2d^2+2dp+p^2+d^2-2dp+p^2=2(d^2+p^2)=(2c)^2 $$
so that $(d-p,d+p,2c)$ is a pythagorean triple.
Note that $\gcd(d-p,d+p,2c)$ divides $\gcd(2p,2c)=2$, hence either $(d-p,d+p,2c)$ is primitive or we have $p\equiv d\pmod 2$ and 
$$\left(\frac{d-p}2,\frac{d+p}2,c\right)$$
is a primitive pythagorean triple (PPT).
The first variant is impossible because the hypothenuse in a PPT is always odd.
From $d\equiv p\pmod 2$ and $\gcd(d,p)=1$ we see that $d,p$ are both odd and either $\frac{d+p}2$ or $\frac{d-p}2$ is even.
The PPTs are  well-known: If $u,v$ are coprime natural numbers of different parity, then we have $(u^2-v^2,2uv,u^2+v^2)$ or $(2uv,u^2-v^2,u^2+v^2)$, depending on where we want the even number.
This gives us $$p=\frac{d+p}2-\frac{d-p}2=\pm(u^2-2uv-v^2)=\pm((u-v)^2-2v^2).$$
$$c=u^2+v^2$$
$$q^2=\frac15(c^2-p^2)=\frac{4uv(u+v)(u-v)}5$$
From coprimeness, we conclude that one of the following cases occurs:


*

*$(u-v,v,u)$ is a PPT and $\frac{u+v}5$ a square. But in such a PPT we have that $u+v$ is either square or twice a square, contradiction.

*$(u,v,u+v)$ is a PPT and $\frac{u-v}5$ a square. (to be continued)

*$u,u-v,u+v$ and $\frac v5$ are squares. This makes $(p,q,c,d)=(\sqrt{u-v},\sqrt{\frac v5},\sqrt u,\sqrt{u+v})$ a solution of $(1)$, contradicting the minimality of $q$.

*$v,u-v,u+v$ and $\frac u5$ are squares. (to be continued)
A: 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
$$n=\frac{31}{12}$$
A: This is an instance of the congruent number problem. We say the positive integer $m$ is a congruent number if there is a right triangle with all sides rational and area $m$. It can be shown that $m$ is a congruent number if and only if there is a rational $x$ such that $x-m,x,x+m$ are all squares of rational numbers. To see the relation to the question here, let $m=5$, $x=n^2+5$. The road to solving these problems leads through elliptic curves. An exposition is here. See also Theorem 3.1 and Example 3.3 of this. 
