# Three coprime square root terms of an arithmetic sequence must be perfect squares.

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.

As $\sqrt{c}, \sqrt{b}$ and $\sqrt{a}$ are terms of an A.P., we can write $\sqrt{c}-\sqrt{b} = m*d$ and $\sqrt{b}-\sqrt{a} = n*d$, for some integers m and n, where d is the common difference of the A.P. Dividing the equations, we get, $$\frac{\sqrt{c}-\sqrt{b}}{\sqrt{b}-\sqrt{a}} = \frac{m}{n}, or\ (m+n)\sqrt{b} = m\sqrt{a} + n\sqrt{c}$$ Squaring, gives that $2mn\sqrt{ac} \in \mathbb{Z}$. Therfore, as a,c are coprime, they both are squares (prime factorization). Plugging it into above equation gives that $\sqrt{b}$ is also an integer.
• Do you know the $n^{th}$ term of an A.P? It is given by $a_n = a + (n-1)d$, where a is any given term, d is the common difference and n is the count of the term starting at a. You get the rational fraction from assuming $\sqrt{a}, \sqrt{b}, \sqrt{c}$ to be terms of the sequence. Note that n must be an integer as it is the number which counts how far you are from the initial term. – bat_of_doom Sep 3 '17 at 12:15
• Just checking because I tried rewriting $\sqrt{b}$ and $\sqrt{c}$ in that form and just couldn't get a nice fraction that I could show that was rational. – El Santi Sep 3 '17 at 13:25