<s>[INST] Verify that $(a^2 + b^2)(c^2 + d^2)$ = $(ac - bd)^2 + (ad + bc)^2$ for any integer $a$,$b$,$c$,$d$ Part 1 - Verify that $(a^2 + b^2)(c^2 + d^2)$ = $(ac - bd)^2 + (ad + bc)^2$ for any integer $a$,$b$,$c$,$d$
Part 2 - Write 25988 as the sum of the two squares (of integers).  
A bit confused with this question, we covered Norm today in class, and I took the formula below from my notes because it seemed helpful with this problem.
$N((a + bi)(c + di))$ = $N((a + bi) * N(c + di)$
$(a + bi)(c + di)$ = $(ac - bd) + (ad + bc)$
$(a - bi)(c - di)$ = $(ac - bd) - (ad + bc)$
This formula goes on to multiply together, but I am not sure if this useful or as to how I would use the answer for part 2 of the problem. Any help is greatly appreciated.
 [/INST] The prime factorization of $25988$ is $2^2 \cdot 73 \cdot 89$. Observe that $73 = 3^2 + 8^2$ and $89 = 5^2 + 8^2$. Using the norm formula, we have
\begin{equation*}
\begin{aligned}
25988 &= 2^2(3^2 + 8^2)(5^2 + 8^2) \\
&= 2^2((3 \cdot 5 - 8 \cdot 8)^2 + (3 \cdot 8 + 8 \cdot 5)^2) \\
&= 2^2(49^2 + 64^2) \\
&= 98^2 + 128^2,
\end{aligned}
\end{equation*}
giving a reprsentation of $25988$ as a sum of two squares.
</s>