1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I understand the rational fraction; however, I do not see how to get the rational fraction from the given information. I know from the halfway point to the end but not seeing how to get to the halfway point. $\endgroup$ – El Santi Sep 3 '17 at 10:54
  • $\begingroup$ 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. $\endgroup$ – bat_of_doom Sep 3 '17 at 12:15
  • $\begingroup$ 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. $\endgroup$ – El Santi Sep 3 '17 at 13:25
  • $\begingroup$ I have edited my answer to include that step too. Hope it's clear now. $\endgroup$ – bat_of_doom Sep 3 '17 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.