[ Formulas for abbreviated multiplication ]

Can the number $\left(\frac{17^{2}+19^{2}}{2}\right)^{2}$ be represented as the sum of the squares of two natural numbers?
The first method. Let 18 = $a$. Then the given number can be written and transformed as follows:

$$
\left(\frac{17^{2}+19^{2}}{2}\right)^{2}=\left(\frac{(a-1)^{2}+(a+1)^{2}}{2}\right)^{2}=\left(a^{2}+1\right)^{2}=\left(a^{2}-1\right)^{2}+(2 a)^{2}. \text{ Therefore, }
$$

$\left(\frac{17^{2}+19^{2}}{2}\right)^{2}=\left(18^{2}-1\right)^{2}+36^{2}=323^{2}+36^{2}$

The second method. $\left(\frac{17^{2}+19^{2}}{2}\right)^{2}=\left(\frac{289+361}{2}\right)^{2}=325^{2}=(13 \cdot 25)^{2}=13^{2} \cdot 25^{2}=169 \cdot 25^{2}=(144+25) \cdot 25^{2}=\left(12^{2}+\right.$

$\left.5^{2}\right) \cdot 25^{2}=12^{2} \cdot 25^{2}+5^{2} \cdot 25^{2}=300^{2}+125^{2}$

Having obtained that the given number equals 132,252, we can proceed differently: $\left(\frac{17^{2}+19^{2}}{2}\right)^{2}=13^{2} \cdot 25^{2}=13^{2}\left(20^{2}+15^{2}\right)=$ $13^{2} \cdot 20^{2}+13^{2} \cdot 15^{2}=260^{2}+195^{2}$

The third method. Since 172 ends in the digit 9, and 192 ends in the digit 1, the number in the parentheses ends in the digit 5. Therefore, the given expression has the form $(5 k)^{2}$, where $k$ is some natural number. Then the positive answer to the question of the problem follows from the equality: $(5 k)^{2}=(4 k)^{2}+(3 k)^{2}$.

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