find $a_n$, $a_n=\sqrt{\frac{a_{n-1}^2+a_{n+1}^2}{2}}, a_1=10(a_n\in \mathbb{N},n=2,3,4,\cdots)$ I would appreciate if somebody could help me with the following problem
Q: find $a_n$
$$a_n=\sqrt{\frac{a_{n-1}^2+a_{n+1}^2}{2}}, a_1=10(a_n\in \mathbb{N},n=2,3,4,\cdots)$$
 A: Hint : Let $b_n=a_n^2$. Then $2b_{n}=b_{n-1}+b_{n+1}$, so there are
two constants $c_1$ and $c_2$ such that $b_n=c_1+c_2n$. So
all you have to do is compute $c_1$ and $c_2$.
A: Using your equation, we get
$$
a_{n+1}^2-2a_n^2+a_{n-1}^2=0
$$
This implies that $a_{n+1}^2-a_n^2=a_n^2-a_{n-1}^2$; that is, $a_{n+1}^2-a_n^2$ is a constant, $b$.
If $b\lt0$, then for some $n$, $a_n^2\lt0$.
If $b\gt0$, then for some $n$, $2a_n+1\gt b$. However, then the next square is
$$
(a_n+1)^2=a_n^2+2a_n+1\gt a_n^2+b=a_{n+1}^2
$$
Therefore, $b=0$ and thus, $a_n=10$ for all $n$.
A: Follow Ewan's answer. 
Obviously, $c_2\ge0$, otherwise $b_n=a_n^2=c_1+c_2n<0$ for large $n$.
If $c_2>0$, then $a_{n+1}-a_{n}>0$.
However, $a_{n+1}-a_{n}=\sqrt{c_1 + c_2 n+c_2}-\sqrt{c_1 + c_2 n}=\frac{c_2}{\sqrt{c_1 + c_2 n}+\sqrt{c_1 + c_2 n+c_2}}<\frac{c_2}{2\sqrt{c_1+c_2n}}$
So for $n>c_2-c_1/c_2$, $0<a_{n+1}-a_n<\frac{c_2}{2\sqrt{c_2^2}}=1/2$, which contradicts the condition that $a_n$ and $a_{n+1}$ are both natural number.
So $c_2=0$, and $c_1=a_1-0*1=10$,
In conclusion $a_n=c_1+0*n=a_1=10$.
