olympic mathematics For each integer $a_0 > 1$, define the sequence $a_0, a_1, a_2, 
\ldots$ by:
$$a_{n+1} = \begin{cases} \sqrt{a_n}, & \mbox{if } \sqrt{a_n}\mbox{ is an integer,} \\a_n+3, &\mbox{otherwise,}\end{cases}$$
Determine all values of $a_0$ for which there is a number $A$ such that $a_n = A$ for infinitely many values
of $n$.
I have figured that if there exists an $i$ such that $a_i 
\ \mathrm{mod}\, 3 = 2$, the elements after $a_i$ are all different.
I think there are many ways to solve this question so i would appreciate any answers.
i think the answer is $a_0\ \mathrm{mod}\, 3 = 0$ 
  but i can't prove it
examples: 7,10,13,16,4,2 doesn't work
3,6,9,3 works
please tell me if i can improve my question.
 A: That only says you have a term $a_0$ in this sequence that generates a cycle.
$a^2=a+3$ means that $a=3k$ or $a=3k+1$


*

*For the case: $a_0=3k$
Try with $a_0=6$, $a_1=a+3=9$, $a_2=3$, $a_3=6$. for infinite values of $k$, $a_k$ is cyclic.
All terms of this sort would map to this small circle, because the distance of leapback is incomparably large in relation of the incremental value which is $3$, let's denote $3k$ is the starting edge $a_0$, at some extent $a_n$ would be written as $3(3x^2)$ by incrementing repeatedly its following terms, the draw back of this sequence is evaluated to $3x(3x-1)$ for larger values of $x$ this is estimated to be very big comparingly to the incremental constant and should override the edge $3k$ where it finally leads straight to the limited 3 steps cycle.
This python code shows my conjecture:
for i in range(1000000):
    s=3*i
    while(s!=0 and s!=3 and s!=6 and s!=9):
          if (sqrt(s)==int(sqrt(s))):
               s=sqrt(s)
          else:
               s=s+3
    print i

The loop will never halt for all $3k$ terms tested up to 3 millions.


*

*For the case $3k+1$ it's trickier to prove since there is a small exception when $\sqrt{a}$ is a square itself where the oddity occurs.

A: There is no integer $a$ so that $a^2 \equiv 2 \mod 3$ (as $0^2 \equiv 0 \mod 3$, $1^2 \equiv 1 \mod 3$ and $2^2 \equiv 1 \mod 3$ so if $a_0\equiv 2 \mod 3$ then by induction $a_{i+1} = a_i + 3 \equiv a_0 \equiv 2 \mod 3$ and the sequence never repeats.
There are infinitely many squares that are multiples of $3$.  The square roots of these are multiples of $3$.  So $a_0 = 3k$ will eventually lead to a $a_n = (3m)^k$ so $a_{n+1} = 3m < a_n$ so $a_n$ will repeat infinitely.  So any multiple of $3$ will be such a starting point.
(Obviously if a sequence repeats once it will, by induction, repeat infinitely many times.
This just means we must consider $a_0 \equiv 1 \mod 3$.
If $a_0 = 1$ then $a_1 = \sqrt 1 = 1$  so $a_0$ is such a starting point.
But $a_0 = 4 \implies a_1 = 2$ so $4$ is not a valid starting point.
If $a_0 = 7; a_1 = 10;a_2 = 13; a_3 = 16; a_4 = 4$ so $7$ iss not a valid starting point.  Is any $1 + 3k$ other than $1$?
Well, it can't be less than $25$ as that will lead to $25$ which leads us to $5\equiv 2 \mod 3$.  We need to be lead to a $(3k + 1)^2 $ that leads to $3k + 1$ that does not lead to and $3k + 1 < (3m + 2)^2 < (3k+1)^2$.
Not sure any such number exists.  $49$ for example is such that $7 < 16 < 49$ is a failure.
But if $(3k+1)^2> 49$  then $k > 2$ and  $(3k-2)^2 = 9k^2 - 12k + 4 \ge 18k-12k +4 = 6k+ 4 > 3k + 1$ so $3k + 1 < (3k-2)^2 < (3k+1)^2$.
So there is no such $1 + 3k$ (other than $1$ that will work).
So $a_0 = 1$ and $a_0 = 3k$ are all such starting points.
