Number of integer solutions of two similar equations

Question

Find the number of integer solutions of:

(a) $${1\over\sqrt{x}}+{1\over\sqrt{y}} = {1\over\sqrt{20}}$$

(b) $${1\over\sqrt{x}}+{1\over\sqrt{y}} = {1\over\sqrt{2014}}$$

I know the solution for (a), but not for (b).

Answer 1

More generally, consider $$ \dfrac{1}{\sqrt{x}} + \dfrac{1}{\sqrt{y}} = \dfrac{1}{a \sqrt{b}}$$ where $a$, $b$ are positive integers and $b > 1$ is squarefree, and we want $x$, $y$ to be positive integers. We can rearrange this as $$ \sqrt{x} = \dfrac{a \sqrt{by}}{\sqrt{y} - a \sqrt{b}}$$ so we must have $\sqrt{y} > a \sqrt{b}$ as well as $$ x = \dfrac{a^2by^2}{y + a^2 b - 2 a \sqrt{by}} = \dfrac{a^2 b y^2 (y + a^2 b + 2 a \sqrt{by})}{(y + a^2 b)^2 - 4 a^2by }$$ Now in order for the right side to be rational, $\sqrt{by}$ must be an integer. Thus $y = b z^2$ for some positive integer $z$.
Note that the condition $\sqrt{y} > a \sqrt{b}$ says $z > a$. Thus let $z = a + t$ where $t$ is a positive integer. Substituting $y = b (a+t)^2$ and simplifying gives us $$ x = \dfrac{a^2 (a+t)^2 b}{t^2} = \dfrac{a^4 b}{t^2} + \dfrac{2 a^3 b}{t} + a^2 b$$ Any prime that divides $t$ must divide $2$, $a$ or $b$, and can't have too high a degree.

In the first example, $a=2$, $b=5$, $80/t^2 + 80/t$ is an integer; the possible primes are $2$ and $5$, and it's not hard to show that the only possibilities are $t = 1,2,4$, corresponding to $y = 5 (2+t)^2 = 45, 80, 180$ and $x = 180, 80, 45$ respectively.

Now try $a = 1$, $b = 2014 = 2 \times 19 \times 53$.

EDIT: By the way, $2016$ will be an interesting year.

Answer 2

Let$${1\over\sqrt{x}}+{1\over\sqrt{y}} = {1\over\sqrt{20}}.$$ Then $$\sqrt {5x}+\sqrt {5y}=\frac{\sqrt{xy}}{4}=t$$ for some real number $t.$ From here we can obtain that, $$xy=16t^2,\,\,\,\,\,\ x+y=\frac{t^2}{5}-8t$$ Since $x,y$ are integers note that $t$ is also an integer. $$z^2-(\frac{t^2}{5}-8t)z+16t^2=0$$ has the roots $x$ and $y.$ $$5z^2+(40t-t^2)z+80t^2=0$$ If we consider the discriminant of this quadratic equation, there should exist a n integer $d$ such that $$d^2=(t^2-40t)^2-1600t^2=t^3(t-80).$$ Con you continue from here? Similar idea works for your (b) also.