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.

Answered by Crostul on September 16, 2020

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