Suppose $a_n \in \mathbb{N}$ are natural numbers such that: $$a_n^2=\frac{a_{n-1}^2+a_{n+1}^2}{2}\quad (n=2,3,\ldots), \quad a_1=10 $$ Find $a_n$
Progress: So far I had come up with $a_n^2=100+(n-1)(a_2^2-100)$ result.
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I had come up with $a_n^2=100+(n-1)(a_2^2-100)$ result – Young Dec 05 '13 at 12:49
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$a_n = 10$, what you need to show is this is the only solution such that $a_n \in \mathbb{N}$ for all $n \in \mathbb{N}$. – achille hui Dec 05 '13 at 12:57
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Hint : If $b_{n} = a_{n}^{2}$, then you have $$b_{n+1} - 2\,b_{n} + b_{n-1} = 0$$ which is a second order recurrence
Tristan
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Well, let's denote $d=a_2^2-100$. Thus our $a_n^2$ is an arithmetic progression with common difference of $d$. Can it be infinite? Obviously, $d$ can't be negative, otherwise at some point the terms will become negative, which is bad for squares. Can it be positive, then? Suppose it can; this would lead to unbounded growth of $a_n^2$. At some point it will exceed $d^2$. But wait; with numbers that big, the difference of two squares is at least $2d+1$ (that is, if they are adjacent; otherwise even more). Can we still continue our sequence with squares which are exactly $d$ apart? No, we can't.
This leaves the only possibility of $d=0$ and $a_n=10$ for every $n$.
Ivan Neretin
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I would suggest considering your equation as a recurrence relation and try substituting, $a_n=Ar^n$. Then you will get, $r=\pm 1$ and therefore you have the general solution as,
$$a_n=A+B(-1)^{n}$$
for $n\geq 2$.
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We can't do that with $a_n$, because we don't have the linear recurrence relation for $a_n$. We have one for $a_n^2$, but its characteristic equation roots are not $\pm1$, just $+1$ (twice). – Ivan Neretin Oct 10 '15 at 20:24