# Algebra, recurrence equation, square roots, linear or non-linear, unique solution

$$\sqrt{a_n}=\sqrt{a_{n-1}}+2\sqrt{a_{n-2}}$$

Is there a way to get rid of the square roots so i can render this equation linear? Is there another way to find the solution?

i did the following: let $b_n=\sqrt{a_n}$, thus my initial equation can be written: $b_{n+2}-b_{n+1}-2b_n=0$ to which i found the following general solution: $b_n=C_1(2)^n+C_2(-1)^n$ Now i have the initials coditions $a_0=a_1=1$...should i use those to determine the constants before or after i have set $a_n=(b_n)^2$? If i try to determine the costants after i get more than one values for them? Is the solution to my equation unique or not, how do i proceed?

hint

we have

$$b_0=\sqrt {a_0}=1=C_1+C_2$$ and

$$b_1=\sqrt {a_1}=1=2C_1-C_2$$

thus by sum $C_1=2/3$ and $C_2=1/3$.

You can rely on the convention that if $a_n\gt 0$ is a positive real number then $\sqrt{a_n}$ is a positive real number. This is a choice to make $\sqrt x$ a function for $x\in \mathbb R^+$. This is inherent in the usual definition of the $\sqrt .$ sign.

• Could be a bit more specific as to how this relates to C1, C2 being unique? Dec 4, 2017 at 22:04
• @IliasKoutroumpas Well there is a unique meaning to every symbol, so you just use arithmetic. Dec 4, 2017 at 22:07