arithmetic progession question

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 $x$?

Answer 1

$$\sqrt{a-x},\sqrt{x},\sqrt{a+x}$$ 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 $x=4$.

Answer 2

GoodDeeds's answer is arguably the "right" way to go about solving this problem, but it's possibly worth noting that it could be tackled by means of brute-force search. That is, the conditions require that $a-x\lt x\lt a$, which can be re-expressed as

$$x\lt a\lt2x$$

Now we can rule out $x=1$ immediately, since there is no integer $a$ satisfying $1\lt a\lt2$.

If $x=2$, the inequality $2\lt a\lt4$ is satisfied only by $a=3$, but $1,\sqrt2,\sqrt5$ is not an arithmetic progression. (Strictly speaking, we need to show that $\sqrt5-\sqrt2\not=\sqrt2-1$, but the non-equality seems clear enough.)

Likewise, if $x=3$, the inequality $3\lt a\lt6$ is satisfied only by $a=4$ and $5$, but $1,\sqrt3,\sqrt7$ and $\sqrt2,\sqrt3,\sqrt8$ are not arithmetic progressions.

It's only at $x=4$ that we find a value of $a$, namely $a=5$, for which $\sqrt{a-4},2,\sqrt{a+4}$ is an arithmetic progression.