<s>[INST] arithmetic progession question if $\sqrt{a-x}, \sqrt x, \sqrt{a+x}$ are in AP provided $a>x$ and $a,x$ are positive integers then what is the least possible value of $x$?
 [/INST] $$\sqrt{a-x},\sqrt{x},\sqrt{a+x}$$ are in an AP. By the definition of an arithmetic progression,
$$\sqrt{a+x}-\sqrt{x}=\sqrt{x}-\sqrt{a-x}$$
Thus
$$2\sqrt{x}=\sqrt{a+x}+\sqrt{a-x}$$
Squaring both sides,
$$4x=a+x+a-x+2\sqrt{a^2-x^2}$$
$$2x-a=\sqrt{a^2-x^2}$$
Squaring both sides,
$$4x^2+a^2-4ax=a^2-x^2$$
$$5x^2-4ax=0$$
$$x(5x-4a)=0$$
Since $x\ne0$,
$$x=\frac{4a}{5}$$
Since $a,x$ are positive integers, for the least value, $a$ must be the smallest positive integer divisible by $5$. Thus, $a=5$.
Correspondingly, the smallest value of $x$ is $x=4$.
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