Can you factorize these squares?

Here is a simple solution, without matrices and determinants:

Let's multiply the first two equations and take away the third from their product:

$(a^2 - b^2)(c^2 + d^2) - ( (ac)^2 - (bd)^2 ) = 5 × 74 - 341$

$(ac)^2 - (bc)^2 + (ad)^2 - (bd)^2 - (ac)^2 + (bd)^2 = 370 - 341$

After simplifying, we get our fourth equation with the missing value:

$(ad)^2 - (bc)^2 = \boxed {29}$

Let matrix $A = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$ and matrix $B = \begin{bmatrix} c & -a \\ d & b \\ \end{bmatrix}$ .

Then multiplying these matrices, we will obtain: $A\times B = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} \cdot \begin{bmatrix} c & -a \\ d & b \\ \end{bmatrix} = \begin{bmatrix} ac+bd & b^2 - a^2 \\ c^2 + d^2 & bd - ac \\ \end{bmatrix}$

Calculating the Determinants of these matrices, we know that the $\text{det}(AB) = \text{det}(A) \cdot \text{det}(B)$ .

Hence, $(ad-bc)(ad+bc) = (ac+bd)(bd-ac) - (b^2 - a^2)(c^2 + d^2)$ .

$(ad)^2 - (bc)^2 = [(bd)^2 - (ac)^2] + (a^2 - b^2)(c^2 + d^2) = -341 + 5\times 74 = 370 - 341 = \boxed{29}$ .

Yes, we can factorize those equations.

$a² - b² = 5$

$(a+b)(a-b) = 5 \times 1$

$a+b = 5$ ; $a-b=1$

Solving, we get $a=3$ and $b=2$ .

$(ac)² - (bd)² = 341$

$(3c)² - (2d)² = 341$

$(3c+2b)(3c-2b) = 341$

Since $3c+2b > 3c-2b$ , then we have

$3c+2b = 341$ or $31$ ; and $3c-2b = 1$ or $11$

First condition. If $3c+2b = 341$ and $3c-2b =1$ ,

then $c² + d² \neq 74$ .

Second condition; $3c +2b =31$ and $3c-2b = 11$ , solving we get $c=7$ and $d=5$ .

Then, the second equation is completed.

Now,

$(ad)² - (bc)² = (3 \times 5)² - (2 \times 7)²$

$= (15+14)(15-14) = 29 \times 1$

$= \boxed{29}$