Rational Roots (with Lots of Square Roots!) Find the smallest positive integer $a,$ greater than 1000, such that the equation $$\sqrt{a - \sqrt{a + x}} = x$$ has a rational root.
Squaring both sides we have $a-\sqrt{a+x}=x^2.$ We shall not square again as that gives a quartic in $x.$ Rearranging, we have $$-x^2+a-\sqrt{a+x}=0 \implies x^2-a+\sqrt{a+x}=0.$$ How would we continue? Solutions? (I don't have one in my book.)

Since the quartic equation is monic, i.e. it has the form
$$x^4+ \dots =0$$
by the rational root theorem, $x$ must be an integer. Moreoever
$$x=\sqrt{a-\sqrt{a+x}}>0$$
so that $x$ must be a positive integer.
Now, consider the equation
$$\sqrt{a+x}=a-x^2$$
and call $$y=\sqrt{a+x}=a-x^2$$
Note that $y=\sqrt{a+x}>0$ is a positive integer. Then you have the two equations
$$y^2=a+x$$
$$y=a-x^2$$
Subtracting the two equations side by side you get
$$y^2-y=x+x^2$$
which can be written as
$$y(y-1)=(x+1)x$$
which has a unique positive solution: $y=x+1$.
Thus
$$a=x^2+y=x^2+x+1$$
So you have to find the smallest number $a$ which can be represented as $a=x^2+x+1$ where $x$ is a positive integer, and $x^2+x+1>1000$.
Now, $x=32$ is the smallest integer solution, and $$a=32^2+32+1=1057$$ is the number you were looking for.
