# Arithmetic Sequence

## Problem 1

Can you find arithmetic sequence that contains $\sqrt{2}, \sqrt{3}$ and $\sqrt{5}$ ?

Suppose we can find arithmetic sequence with different $d$ that contains $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{5}$, then there must be non-zero integers $n$ and $m$ such that $\sqrt{2}+nd=\sqrt{3}$ and $\sqrt{2}+md=\sqrt{5}$. We get $\sqrt{2}=\sqrt{3}-nd$ and $\sqrt{2}=\sqrt{5}-md$. This give us $1=\frac{\sqrt{3}-nd}{\sqrt{5}-md}$. Squaring both sides we get $1 = \frac{3-2\cdot\sqrt{3}nd +n^2d^2}{5-2\cdot\sqrt{5}md+m^2d^2}$ or equivalently $5-2\cdot\sqrt{5}md+m^2d^2=3-2\cdot\sqrt{3}nd+n^2d^2$. We get $\begin{array}[H]{lll} 2-2\cdot\sqrt{5}md+m^2d^2=-2\cdot\sqrt{3}nd+n^2d^2 \\ 2-2\cdot\sqrt{5}md+(m^2-n^2)d^2=-2\cdot\sqrt{3}nd \\ \big(2+(m^2-n^2)d^2\big)-2\cdot\sqrt{5}md=-2\cdot\sqrt{3}nd\\ \end{array}$ Once again we square both sides of the equation and get $\big(2+(m^2-n^2)d^2\big)^2-4\cdot\big(2+(m^2-n^2)d^2\big)\cdot\sqrt{5}md +20m^2d^2 = 12n^2d^2$ hence $\sqrt{5}$ must be rational. But $\sqrt{5}$ is irational, therefore no arithmetic sequence contains $\sqrt{2},\sqrt{3}$ and $\sqrt{5}$ $\square$