We say that two natural numbers form a perfect pair when the sum and the product of these two numbers are perfect squares. For example, 5 and 20 form a perfect pair, since $5+20=25=5^{2}$ and $5 \times 20=100=10^{2}$. Does 122 form a perfect pair with another natural number?
Let's call $n$ the natural number that is a "candidate" to form a perfect pair with 122. Then, we must have: $122+n=A^{2}$ and $122 \times n=B^{2}$ where $A$ and $B$ are natural numbers.

Since $B^{2}=2 \times 61 \times n$, we conclude that $n$ also has the prime factors 2 and 61. Therefore, we can write $n$ as $n=2 \times 61 \times m^{2}=122 \, m^{2}$.

We then obtain $A^{2}=122+122 m^{2}=122\left(1+m^{2}\right)$. The smallest value of $\left(1+m^{2}\right)$ that satisfies this equation is $1+m^{2}=122$, that is, $m^{2}=121$. From this, it follows that $m=11$. Consequently, $n=122 \times 121$ and we have:

$$
A^{2}=122+122 \times 121=122^{2} \quad \text{and} \quad B^{2}=122 \times 122 \times 121=(122 \times 11)^{2}
$$

Therefore, 122 and $122 \times 121$ form a perfect pair.

Note. In fact, $122 \times 121$ is the smallest natural number that forms a perfect pair with 122. Are there any others?