Centroamerican and Caribbean Mathematical Olympiad

To solve the problem of finding all virtual numbers of the form $n^2 + 1$, we need to express a virtual number in the required form and establish conditions for $n$.

A virtual number $\overline{abab}$ can be expressed mathematically as:
$$101a + 10b + 10a + b = 110a + 11b.$$
We are tasked with finding $n$ such that:
$$n^2 + 1 = 110a + 11b.$$

To proceed, consider the behavior of $n^2$ modulo 11, since the expression $11b$ indicates periodic properties associated with modulo calculations. We have:
$$n^2 \equiv -1 \pmod{11}.$$

Checking possible values of $n^2 \pmod{11}$ for $n = 0, 1, 2, \ldots, 10$ since these are all the residues mod 11:
- $n^2 \equiv 0^2 \equiv 0 \pmod{11}$,
- $n^2 \equiv 1^2 \equiv 1 \pmod{11}$,
- $n^2 \equiv 2^2 \equiv 4 \pmod{11}$,
- $n^2 \equiv 3^2 \equiv 9 \pmod{11}$,
- $n^2 \equiv 4^2 \equiv 5 \pmod{11}$,
- $n^2 \equiv 5^2 \equiv 3 \pmod{11}$,
- $n^2 \equiv 6^2 \equiv 3 \pmod{11}$,
- $n^2 \equiv 7^2 \equiv 5 \pmod{11}$,
- $n^2 \equiv 8^2 \equiv 9 \pmod{11}$,
- $n^2 \equiv 9^2 \equiv 4 \pmod{11}$,
- $n^2 \equiv 10^2 \equiv 1 \pmod{11}$.

Therefore, none of these congruences satisfy $n^2 \equiv -1 \equiv 10 \pmod{11}$. Hence, we must search for values of $n$ that enable $n^2 \equiv 110a + 11b - 1$.

Supposed $n = 91$:
$$n^2 + 1 = 8282.$$
Check:
- $a = 8$ and $b = 2$,
- $\overline{abab} = 8282$.

Thus, the virtual number of the form $n^2 + 1$ is indeed:
$$\boxed{8282}.$$
This confirms that $n = 91$ and no other $n$ within reasonable bounds yields such a result.