3. Find any four pairwise distinct natural numbers $a, b, c, d$, for which the numbers $a^{2}+2 c d+ b^{2}$ and $c^{2}+2 a b+d^{2}$ are perfect squares.
3. If it suddenly turned out that $a b=c d$, then

$$
\begin{aligned}
& a^{2}+2 c d+b^{2}=a^{2}+2 a b+b^{2}=(a+b)^{2} \\
& c^{2}+2 a b+d^{2}=c^{2}+2 c d+d^{2}=(c+d)^{2}
\end{aligned}
$$

Thus, it is sufficient to find four different natural numbers $a, b, c$, and $d$ for which $a b=c d$. Or, equivalently, to find a number $n$ that can be represented as the product of two different factors in two different ways: $n=a b=c d$. An example of such a number: $6=1 \cdot 6=2 \cdot 3$.