I did the following experiment:

Let $p$ be a prime number. Then a necessary condition for the sequence to remain in $\mathbb{Z}$ is that $x_{p-1} \equiv \pm 1 \mod p$.

So for every starting value $x_1$, I calculated $x_{p-1} \mod p$ for the first several prime numbers $p$ to see if there are obstructions. It turns out that for every choice of $x_1$ between $2$ and $100000$, there are always obstructions. The first obstruction (i.e. the smallest prime $p$ such that $x_{p-1}$ is not congruent to $\pm 1$ modulo $p$) for $2 \leq x_1 \leq 100$ is listed below.

```
x[1] = 2: obstruction at 2
x[1] = 3: obstruction at 5
x[1] = 4: obstruction at 2
x[1] = 5: obstruction at 5
x[1] = 6: obstruction at 2
x[1] = 7: obstruction at 5
x[1] = 8: obstruction at 2
x[1] = 9: obstruction at 23
x[1] = 10: obstruction at 2
x[1] = 11: obstruction at 7
x[1] = 12: obstruction at 2
x[1] = 13: obstruction at 5
x[1] = 14: obstruction at 2
x[1] = 15: obstruction at 5
x[1] = 16: obstruction at 2
x[1] = 17: obstruction at 5
x[1] = 18: obstruction at 2
x[1] = 19: obstruction at 11
x[1] = 20: obstruction at 2
x[1] = 21: obstruction at 7
x[1] = 22: obstruction at 2
x[1] = 23: obstruction at 5
x[1] = 24: obstruction at 2
x[1] = 25: obstruction at 5
x[1] = 26: obstruction at 2
x[1] = 27: obstruction at 5
x[1] = 28: obstruction at 2
x[1] = 29: obstruction at 13
x[1] = 30: obstruction at 2
x[1] = 31: obstruction at 7
x[1] = 32: obstruction at 2
x[1] = 33: obstruction at 5
x[1] = 34: obstruction at 2
x[1] = 35: obstruction at 5
x[1] = 36: obstruction at 2
x[1] = 37: obstruction at 5
x[1] = 38: obstruction at 2
x[1] = 39: obstruction at 7
x[1] = 40: obstruction at 2
x[1] = 41: obstruction at 11
x[1] = 42: obstruction at 2
x[1] = 43: obstruction at 5
x[1] = 44: obstruction at 2
x[1] = 45: obstruction at 5
x[1] = 46: obstruction at 2
x[1] = 47: obstruction at 5
x[1] = 48: obstruction at 2
x[1] = 49: obstruction at 7
x[1] = 50: obstruction at 2
x[1] = 51: obstruction at 19
x[1] = 52: obstruction at 2
x[1] = 53: obstruction at 5
x[1] = 54: obstruction at 2
x[1] = 55: obstruction at 5
x[1] = 56: obstruction at 2
x[1] = 57: obstruction at 5
x[1] = 58: obstruction at 2
x[1] = 59: obstruction at 7
x[1] = 60: obstruction at 2
x[1] = 61: obstruction at 11
x[1] = 62: obstruction at 2
x[1] = 63: obstruction at 5
x[1] = 64: obstruction at 2
x[1] = 65: obstruction at 5
x[1] = 66: obstruction at 2
x[1] = 67: obstruction at 5
x[1] = 68: obstruction at 2
x[1] = 69: obstruction at 11
x[1] = 70: obstruction at 2
x[1] = 71: obstruction at 11
x[1] = 72: obstruction at 2
x[1] = 73: obstruction at 5
x[1] = 74: obstruction at 2
x[1] = 75: obstruction at 5
x[1] = 76: obstruction at 2
x[1] = 77: obstruction at 5
x[1] = 78: obstruction at 2
x[1] = 79: obstruction at 29
x[1] = 80: obstruction at 2
x[1] = 81: obstruction at 7
x[1] = 82: obstruction at 2
x[1] = 83: obstruction at 5
x[1] = 84: obstruction at 2
x[1] = 85: obstruction at 5
x[1] = 86: obstruction at 2
x[1] = 87: obstruction at 5
x[1] = 88: obstruction at 2
x[1] = 89: obstruction at 13
x[1] = 90: obstruction at 2
x[1] = 91: obstruction at 7
x[1] = 92: obstruction at 2
x[1] = 93: obstruction at 5
x[1] = 94: obstruction at 2
x[1] = 95: obstruction at 5
x[1] = 96: obstruction at 2
x[1] = 97: obstruction at 5
x[1] = 98: obstruction at 2
x[1] = 99: obstruction at 11
x[1] = 100: obstruction at 2
```

Up to $x_1 = 100000$, the biggest "first obstruction" appears at:

```
x[1] = 13589: obstruction at 103
```

Even if one allows $x_1$ to be $\sqrt{k}$ for some integer $k$, the results are similar - one just starts from $x_2$, and for $2 \leq x_2 \leq 10000$ there are always obstructions at (small) prime numbers.

These results seem to support the original conjecture.

**EDIT**

Following this idea, I further calculated, for a given prime $p$, the residue classes of $x_1 \mod p$ that will lead to an obstruction at $p$. Let us call them "bad" residues. The result seems to be interesting for its own sake.

```
p bad residues x[1] mod p
mod 2: 0
mod 3:
mod 5: 0 2 3
mod 7: 0 3 4
mod 11: 0 3 5 6 8
mod 13: 0 2 3 5 8 10 11
mod 17: 2 4 5 12 13 15
mod 19: 0 6 8 11 13
mod 23: 3 7 8 9 11 12 14 15 16 20
mod 29: 0 3 5 6 7 8 10 11 13 14 15 16 18 19 21 22 23 24 26
mod 31: 0 2 5 8 10 13 15 16 18 21 23 26 29
mod 37:
mod 41:
mod 43: 0 2 3 4 5 6 7 8 9 10 11 14 16 18 19 20 21 22 23 24 25 27 29 32 33 34 35 36 37 38 39 40 41
mod 47: 0 7 8 9 10 11 12 13 14 15 18 21 23 24 26 29 32 33 34 35 36 37 38 39 40
mod 53: 0 9 12 15 17 18 20 23 30 33 35 36 38 41 44
mod 59: 3 5 8 13 14 15 16 18 21 26 33 38 41 43 44 45 46 51 54 56
mod 61: 0 2 5 8 11 12 15 17 19 20 21 22 23 24 26 29 32 35 37 38 39 40 41 42 44 46 49 50 53 56 59
mod 67:
mod 71: 5 6 15 19 20 24 25 31 33 35 36 38 40 46 47 51 52 56 65 66
mod 73: 9 23 27 28 29 44 45 46 50 64
mod 79:
mod 83:
mod 89: 0 2 3 4 5 16 17 22 23 24 27 30 31 32 35 40 49 54 57 58 59 62 65 66 67 72 73 84 85 86 87
mod 97: 0 2 3 8 11 14 15 17 21 23 24 28 29 30 35 38 39 44 47 50 53 58 59 62 67 68 69 73 74 76 80 82 83 86 89 94 95
```

And here is the table which counts the number of bad residues modulo $p$:

```
p number of bad residues x[1] mod p
mod 2: 1
mod 3: 0
mod 5: 3
mod 7: 3
mod 11: 5
mod 13: 7
mod 17: 6
mod 19: 5
mod 23: 10
mod 29: 19
mod 31: 13
mod 37: 0
mod 41: 0
mod 43: 33
mod 47: 25
mod 53: 15
mod 59: 20
mod 61: 31
mod 67: 0
mod 71: 20
mod 73: 10
mod 79: 0
mod 83: 0
mod 89: 31
mod 97: 37
mod 101: 50
mod 103: 35
mod 107: 29
mod 109: 20
mod 113: 30
mod 127: 22
mod 131: 93
mod 137: 33
mod 139: 115
mod 149: 121
mod 151: 59
mod 157: 6
mod 163: 111
mod 167: 85
mod 173: 111
mod 179: 98
mod 181: 127
mod 191: 0
mod 193: 83
mod 197: 4
mod 199: 130
mod 211: 85
mod 223: 34
mod 227: 77
mod 229: 57
mod 233: 85
mod 239: 137
mod 241: 56
mod 251: 140
mod 257: 79
mod 263: 0
mod 269: 44
mod 271: 129
mod 277: 20
mod 281: 26
mod 283: 231
mod 293: 171
```

The most noticeable thing, to me, is those primes with "$0$" bad residues. Here are they:

```
3, 37, 41, 67, 79, 83, 191, 263, 347, 353, 373, 379, 421, 449, 463, 509, 557, 619, 647, 661, 673, 719, 733, 757, 787, 823, 839, 911
```

This sequence is not found in OEIS.

Let us call those primes "exceptional". If one excludes those exceptional primes, then the **proportion** of bad residues (i.e. [number of bad residues mod $p$] divided by $p$) seems to distribute uniformly on the interval $[0, 1)$. This suggests that the exceptional primes may of particular interest.

**EDIT**

To illustrate the distribution of the proportion of bad residues, here I add the statistical data (for primes $p < 5000$):

```
"bad proportion" number of primes
0 77 (i.e. number of exceptional primes)
(0.00, 0.02] 10
(0.02, 0.04] 15
(0.04, 0.06] 13
(0.06, 0.08] 18
(0.08, 0.10] 7
(0.10, 0.12] 11
(0.12, 0.14] 8
(0.14, 0.16] 14
(0.16, 0.18] 21
(0.18, 0.20] 14
(0.20, 0.22] 16
(0.22, 0.24] 11
(0.24, 0.26] 17
(0.26, 0.28] 15
(0.28, 0.30] 13
(0.30, 0.32] 11
(0.32, 0.34] 15
(0.34, 0.36] 15
(0.36, 0.38] 17
(0.38, 0.40] 17
(0.40, 0.42] 19
(0.42, 0.44] 13
(0.44, 0.46] 15
(0.46, 0.48] 23
(0.48, 0.50] 20
(0.50, 0.52] 16
(0.52, 0.54] 11
(0.54, 0.56] 15
(0.56, 0.58] 16
(0.58, 0.60] 14
(0.60, 0.62] 12
(0.62, 0.64] 6
(0.64, 0.66] 13
(0.66, 0.68] 20
(0.68, 0.70] 14
(0.70, 0.72] 8
(0.72, 0.74] 9
(0.74, 0.76] 9
(0.76, 0.78] 6
(0.78, 0.80] 11
(0.80, 0.82] 12
(0.82, 0.84] 7
(0.84, 0.86] 6
(0.86, 0.88] 5
(0.88, 0.90] 6
(0.90, 0.92] 3
(0.92, 0.94] 3
(0.94, 0.96] 2
(0.96, 0.98] 0
(0.98, 1.00] 0
```

There is clearly a concentration on $0$, i.e. on the exceptional primes.

The "average proportion", calculated as $\frac{\sum_p proportion_p}{\sum_p 1}$, is about $0.37551$.