Can $\sqrt3,\sqrt5,\sqrt7$ all be elements (not necessarily consecutive) of the same arithmetic sequence? Can $\sqrt3,\sqrt5,\sqrt7$ all be elements (not necessarily consecutive) of the same arithmetic sequence?
An arithmetic sequence is a sequence such that the difference of any two consecutive elements is constant. In other words, we keep adding the same value infinitely.
I've tried making a sort of system equations between these numbers and their differences and trying to find any other connections, but I've not had any success. Can someone, perhaps with experience with these problems, help me? Thanks!
 A: Hint: If $a,b,c$ are not necessarily consecutive terms in an arithmetic sequence, then, writing the inter-term difference as $d\in\mathbb{R}$, we have $b=a+nd$ and $c=b+md$, where $n,m$ are nonnegative integers. Hence, you may aim to prove whether there is such a difference $d=\frac{\sqrt{5}-\sqrt{3}}{n}=\frac{\sqrt{7}-\sqrt{5}}{m}$. But this would rely on the integers $n,m$ satisfying $m\left(\sqrt{5}-\sqrt{3}\right)=n\left(\sqrt{7}-\sqrt{5}\right)$, which would rely on $\displaystyle \frac{\sqrt{5}-\sqrt{3}}{\sqrt{7}-\sqrt{5}}$ being rational.
A: Assume WLOG that $\sqrt{3}$ is the first term of the sequence. Let $a_r=\sqrt 5$ and $a_s=\sqrt 7$ for some integers $s>r>1$. Therefore,
$$\frac{\sqrt 5-\sqrt 3}{\sqrt 7-\sqrt 5}=\frac{r-1}{s-r}$$
This is not possible because LHS is irrational and RHS is rational.
You may need to check the irrationality of LHS.
Proof that $\dfrac{\sqrt 5-\sqrt 3}{\sqrt 7-\sqrt 5}$ is irrational
Assume that $\dfrac{\sqrt 5-\sqrt 3}{\sqrt 7-\sqrt 5}$ is rational. Then
$$\dfrac{\sqrt 5-\sqrt 3}{\sqrt 7-\sqrt 5}\cdot\dfrac{\sqrt 7+\sqrt 5}{\sqrt 7+\sqrt 5}=\dfrac{5+\sqrt{35}-\sqrt{21}-\sqrt{15}}2$$
is rational. Then
$$r=\sqrt{21}+\sqrt{15}-\sqrt{35}$$
is a rational number.
But
$$r^2+2r\sqrt{35}+35=36+6\sqrt{35}$$
If $r=3$ then $9=1$, so $r\neq 3$. Then,
$$\sqrt{35}=\frac{r^2-1}{6-2r}$$
and LHS is irrational and RHS is rational. This is a contradiction, so $\dfrac{\sqrt 5-\sqrt 3}{\sqrt 7-\sqrt 5}$ is not rational.
A: Suppose such a sequence exists and has common difference $d \ge 0$.  Then this implies $$\sqrt{5} - \sqrt{3} = kd$$ for some positive integer $k$, and similarly, $$\sqrt{7} - \sqrt{5} = md$$ for some positive integer $m$.  Note that $d$ need not be an integer.  Hence $$\frac{k}{m} = \frac{kd}{md} = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{7} - \sqrt{5}}.$$  The left-hand side, $k/m$, being the ratio of two positive integers, is a rational number.  Is the right-hand side a rational number?  If yes, then we can find such a $k$ and $m$ and in turn, an arithmetic sequence satisfying your criteria; if no, then no such sequence exists.
The answer to this question is left as an exercise for the reader.
