$n ^ 2 +5$, $n ^ 2 +10$ 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.

• Since you are not new to the site, you know that, in order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, note that there is no question in your question; please consider rewriting your post. – Did Nov 24 '12 at 9:43
• Let $n$ be $p/q$, then we want $\dfrac{p^2+5q^2}{q^2}$ and $\dfrac{p^2+10q^2}{q^2}$ to be rational squares or equivalently we want $p^2+5q^2$ and $p^2+10q^2$ to be integer squares i.e. we want \begin{align}a^2 & = p^2 + 5q^2\\b^2 & = p^2 + 10q^2\end{align}Without loss of generality we could assume that $(p,q) = 1$. Hence, either $p$ or $q$ or both are odd. If $q$ is odd, then $b^2 \equiv (p^2 + 2q^2) \pmod{4} = (p^2 + 2) \pmod{4}$. No solution. Hence, $q = 2r$ is even and $p=2s+1$ is odd. Hence,\begin{align} a^2 & = (2s+1)^2 + 20r^2\\ b^2 & = (2s+1)^2 + 40r^2 \end{align} – user17762 Nov 24 '12 at 9:57

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.

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:

1. $(u-v,v,u)$ 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.
2. $(u,v,u+v)$ is a PPT and $\frac{u-v}5$ a square. (to be continued)
3. $u,u-v,u+v$ 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 $q$.
4. $v,u-v,u+v$ and $\frac u5$ are squares. (to be continued)
• Will this make it easier ? He is looking for rational solutions – Amr Nov 24 '12 at 9:55
• This is a high school students asked the question, answer $n=\frac{31}{12}$, but no answer process. I try to follow math.stackexchange.com/questions/182569/… Method to solve, If ${{n}^{2}}+5={{a}^{2}}$，${{2}^{2}}+5={{3}^{2}}$，$n-3=t(a-2)$，so a=(-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. – geometryscience Nov 25 '12 at 2:31

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 $$n=\frac{31}{12}$$