Algebra Problem Olympiad

Question

Prove that there are infinitely many pairs (n,k) of rational numbers such that $\sqrt{n+\sqrt{n}}=k$

I expressed it as $n+\sqrt{n}=\frac {a^2}{b^2}$

but cant get any further. Could someone help.

Answer 1

Let $a$, $b$ be positive integers such that $a < b$ and $(a, \: b) = 1$. If $n = \dfrac{a^4}{(b^2 - a^2)^2}$ we have $$\begin{align*} \sqrt{n + \sqrt{n}} & = \sqrt{\frac{a^4}{(b^2 - a^2)^2} + \frac{a^2}{b^2 - a^2}} = \\ & = \sqrt{\frac{a^4 + a^2(b^2 - a^2)}{(b^2 - a^2)^2}} = \frac{ab}{b^2 - a^2} \end{align*}$$ which is a rational number. So the pair $\displaystyle (n, \: k) = \left(\frac{a^4}{(b^2 - a^2)^2}, \: \frac{ab}{b^2 - a^2}\right)$ is a solution. Are these all distinct? Since $(a, \: b) = 1$, we also have $(a^2, \: (b^2 - a^2)) = 1$, whence $n$ is a fraction in the reduced form, which implies that these solutions are pairwise distinct. Since they are obviously infinite, we are done.

Answer 2

We know that $\sqrt{n}=k^2-n$ irrational, so let $m=\sqrt{n}$. Then your equation is:

$$m^2+m-k^2=0$$

There is a general technique for solving such equations, but this equation can be solved more directly by rewriting it as:

$$(2m+1)^2-(2k)^2=1$$

So if you can find infinitely many rational solutions to $u^2-v^2=1$, with $u\geq 1,v\geq 0$ you can find infinitely many solutions to your original equation by setting $n=\left(\frac{u-1}{2}\right)^2$ and $k=\frac{v}{2}$.

Writing $u^2-v^2=(u-v)(u+v)$, we then need $u-v$ and $u+v$ to be rational inverses. So $u-v=\frac{p}{q}$ and $u+v=\frac{q}{p}$, with $0<p\leq q$ integers. (These requirements ensure $u\geq 1$ and $v\geq 0.$)

You get $$\begin{align}u&=\frac{q^2+p^2}{2pq}\\v&=\frac{q^2-p^2}{2pq}.\end{align}$$

Then $$\begin{align}k&=\frac{q^2-p^2}{4pq}\\n&=\frac{(q-p)^4}{16p^2q^2}.\end{align}$$

If $p\neq q$, let $r=\frac{p}{q-p}$ then you get $r+1=\frac{q}{q-p}$ and thus $$n=\frac{1}{16r^2(r+1)^2},\quad k=\frac{2r+1}{4r(r+1)}.$$ The value for $n$ is decreasing as $r$ increases, so we have a distinct solutions for any rational $r>0$. In particular, there are infinitely many solutions.

That gives all solutions except when $p=q$, which gives the values $n=0,k=0$.

If $s=2r$ then we get the slightly simpler formula:

$$n=\frac{1}{s^2(s+2)^2}, \quad k=\frac{s+1}{s(s+2)}$$

Letting $t=s+1$, we get:

$$n=\frac{1}{(t^2-1)^2},\quad k=\frac{t}{t^2-1},\quad t> 1$$

Answer 3

Let $X=\{3, 8, 15, \ldots, m^2-1, \ldots \}$. $X$ has infinite elements. Then, for $c \in X$, $c+1$ is a perfect square. It follows that

$$\sqrt{\frac{1}{c^2}+\sqrt{\frac{1}{c^2}}}= \sqrt{\frac{c+1}{c^2}}$$

The right-hand side is rational because $c \in X$, so the infinite set

$$\left\{\left(\frac{1}{c^2},\sqrt{\frac{c+1}{c^2}}\right), c \in X\right\}$$

of pairs of rational numbers is contained in the set of rational solutions.