# Rational Roots (with Lots of Square Roots!)

Mathematics Asked by Fleccerd on September 16, 2020

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.