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 any other natural number?
Let $n$ be the natural number "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 $B^{2}=2 \times 61 \times n$, the prime factors 2 and 61 of $B^{2}$ must appear an even number of times, which ensures that $n$ has the prime factors 2 and 61, i.e., $n=2 \times 61 \times m^{2}=122 m^{2}$, for some positive natural number $m$. It follows that

$$
A^{2}=122+122 m^{2}=122\left(1+m^{2}\right)
$$

The smallest value of $\left(1+m^{2}\right)$ that satisfies this equality is $1+m^{2}=122$, i.e., $m^{2}=121$ and $m=11$. Consequently, $n=122 \times 121$ and we conclude that $A^{2}=122+122 \times 121=122^{2}$ and $B^{2}=122 \times 122 \times 121=(122 \times 11)^{2}$. Thus, 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?