Solve equation $\frac{1}{x}+\frac{1}{y}=\frac{2}{101}$ in naturals My try was $$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{2}{101}\\x+y=2k,xy=101k\\x=2k-y\\y(2k-y)=101k\\2ky-y^2=101k\\y^2-2ky+101k=0\\y=k+\sqrt{k^2-101k}\\x=k-\sqrt{k^2-101k}$$
Now $\sqrt{k^2-101k}$ has to be either integer or rational,if it's an integer it has to be $k=101$ cause $gcd(k,k-101)=1\lor101$ and both $k,k-101$ can't be both squares of an integer,so $k=101t$ and $t(t-1)$ is never an square except for $t=1,0$ and $t=0$ is not possible hence $k=101$ is only possible integer solution
EDIT: So if $\gcd(k,k-101)=1$ then $k=h^2,k-101=(h-s)^2$ then $h(2h-s)=101$ which can be $s=1,h=51$ or $s=101,h=51$.$y=51^2+51\cdot50=51\cdot 101,x=51$
And since $x=2k-y$ is integer then $k=\frac{h}{2}$,if $h=2q$ then $k$ is integer otherwise if $h=2q+1$ then $$\frac{2q+1}{2}+\frac{1}{2}\sqrt{4q^2-400q-201}=\frac{2q+1}{2}+\frac{1}{2}\sqrt{(2q-10)^2-301}\\(2q-10)^2-301=r^2\\2q-10=z\\z^2-301=(z-c)^2\\c(2z-c)=301,c=1,z=151,c=7,z=25,c=43,z=25$$
The $z=151$ is impossible cause $2q-10$ is even,and $z=25$ is also impossible because $2q-10$ is even.Hence the only solutions are $(x,y)=(101,101),(51,5151),(5151,51)$ The last one clearly from symmetry

$1/x + 1/y = 2/101$
$(x+y)/xy = 2/101$
So $101|xy$.  But 101 is prime so 101|x or 101|y or both.
Wolog symmetry assume 101|x.
Let $x = 101x'$.
$(101x' + y)/101x'y = 2/101$
$(101x' + y)/x'y = 2$
So $x'|y$.  Let y = x'y'.
$(101 + y')/x'y' = 2$
So $y'|101$.
So $y' = 1$ of $y' = 101$.
So $102/x' = 2$ and $x'=51; y'=1$  or $202/x'101 = 2$ and $x'=1; y' = 101$
So $x=5151;y=51$ or $x=101;y=101$.  And removing wolog symmetry; or $x = 51; y= 5151$
