Question: Let $a, b,$ and $c$ be relatively prime positive integers. Show that if $\sqrt{a},\sqrt{b}, $ and $\sqrt{c}$ are terms of the same arithmetic sequence, then $a,b,$ and $c$ must be perfect squares.
When I first read the question I thought it was saying that $\sqrt{a},\sqrt{b},$ and $\sqrt{c}$ are consecutive. My work for that problem:
$$\sqrt{a}+\sqrt{c} = 2\sqrt{b} $$ $$a+c+2\sqrt{ac} = 4b$$ $$ 2\sqrt{ac}=4b-(a+c) \mbox{ which is an integer}$$ thus $\sqrt{ac}$ is an integer.
As $a$ and $c$ are relatively prime, then both $a$ and $c$ are perfect squares, and then $b$ must be a perfect square.
When I read the problem again, and it doesn't have the requirement that $\sqrt{a},\sqrt{b}$ and $\sqrt{c}$ are consecutive in the arithmetic progression.
With this understanding of the problem, I can get that if $\frac{\sqrt{c}-\sqrt{a}}{\sqrt{b}-\sqrt{a}}$ is rational then it should lead to the variable being perfect squares; however, the requirement of being terms of an arithmetic progression lacking in that method of attack.
Could you please help me find a solution?