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.
 A: You might note that $ac-bd$ and $ad+bc$ are the real and imaginary parts of $(a+ib)(c+id)$, so the formula says $|zw|^2 = |z|^2 |w|^2$ where $z= a+ib$ and $w=c+id$.
For the second part, it might help that $25988 = 2^2 \times 73 \times 89$.  Can you write $2$, $73$, $89$ each as the sum of two squares?
A: 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.
A: You can check this directly:
$$(ac-bd)²+(ad+bc)²=(ac)²+(bd)²-2abcd+(ad)²+(bc)²+2abcd$$ $$=a²(c²+d²)+b²(c²+d²)=(a²+b²)(c²+d²).$$
$25988=2²\times 73\times 89=2²(8²+3²)(8²+5²)=2²[(8²-15)²+(40+24)²]=98²+128²$
A: The first part of the problem reduces the complexity of the second, by writing $AB$ as a sum of two squares if $A$ and $B$ are, but does not lead to a mechanical solution.  To start the process one needs to find factors of 25988 that can be written as sums of two squares.  There is a theorem of Fermat that only numbers with an even power of every prime of type $4k-1$ in their factorization can be expressed as sums of two squares, which limits the possibilities.  Beyond this a search is needed, there is no simple formula for the square summands when they exist.
brute force solution
factorization
A: A fancy approach to Lagrange's identity: differentiation.
$$\frac{\partial}{\partial a}(a^2+b^2)(c^2+d^2) = 2a(c^2+d^2),$$
$$\frac{\partial}{\partial a}\left((ac-bd)^2+(ad+bc)^2\right)=2c(ac-bd)+2d(ad+bc) $$
they match, and the same happens for $\frac{\partial}{\partial b},\frac{\partial}{\partial c},\frac{\partial}{\partial d}$, so $(a^2+b^2)(c^2+d^2)$ and $(ac-bd)^2+(ad+bc)^2$ differ by a constant, that is clearly zero by evaluating both functions at $(1,1,1,1)$. The usual approach: the norm on $\mathbb{Z}[i]$ (the Gaussian integers) defined through
$$ N(z) = N(a+ib) = z\cdot\overline{z} = (a+ib)(a-ib) = a^2+b^2$$
is multiplicative:
$$ N(zw) = zw\cdot\overline{zw} = z\overline{z}\cdot w\overline{w}  = N(z)N(w)$$
hence by taking $z=a+ib,w=b+id$ Lagrange's identity easily follows.
Lagrange's identity gives that the set $S=\{n:n=a^2+b^2,a\in\mathbb{Z},b\in\mathbb{Z}\}$ is a semigroup, i.e. if $s_1,s_2\in S$, then $s_1\cdot s_2\in S$. Using Fermat's descent, it is not difficult to show that if $m\in\mathbb{N}$ belongs to $S$ and $m=uv$ with $\gcd(u,v)=1$, then both $u$ and $v$ belong to $S$. Moreover, the only prime numbers belonging to $S$ are $2$ and the primes of the form $4k+1$. Since
$$25988=2^2\cdot 73\cdot 89 $$
and both $73$ and $89$ belong to $S$, $25988\in S$, too. To be clear, from
$$ 73=8^2+3^2,\qquad 89=8^2+5^2 $$
it follows that:
$$ 73\cdot 89 = (64-15)^2+(24+40)^2 = 49^2+64^2, $$
hence $25988 = 98^2+128^2$.
