To generate Pythagorean triples using the identity [tex]\( (x^2-y^2)^2 + (2xy)^2 = (x^2+y^2)^2 \)[/tex], let's calculate the triples step-by-step for each given pair of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] values and find the missing values.
1. For [tex]\( x = 4 \)[/tex] and [tex]\( y = 3 \)[/tex]:
[tex]\[
a = x^2 - y^2 = 4^2 - 3^2 = 16 - 9 = 7
\][/tex]
[tex]\[
b = 2xy = 2 \cdot 4 \cdot 3 = 24
\][/tex]
[tex]\[
c = x^2 + y^2 = 4^2 + 3^2 = 16 + 9 = 25
\][/tex]
The Pythagorean triple is [tex]\( (7, 24, 25) \)[/tex].
2. For [tex]\( x = 5 \)[/tex] and the unknown [tex]\( y \)[/tex], we already know the Pythagorean triple is [tex]\( (9, 40, 41) \)[/tex]:
[tex]\[
c = 41
\][/tex]
Since [tex]\( c = x^2 + y^2 \)[/tex]:
[tex]\[
41 = 5^2 + y^2 = 25 + y^2 \implies y^2 = 41 - 25 = 16 \implies y = 4
\][/tex]
The Pythagorean triple remains [tex]\( (9, 40, 41) \)[/tex].
3. For [tex]\( y = 3 \)[/tex] and the unknown [tex]\( x \)[/tex], we have the Pythagorean triple [tex]\( (27, 36, 45) \)[/tex] already provided:
[tex]\[
c = 45
\][/tex]
Since [tex]\( c = x^2 + y^2 \)[/tex]:
[tex]\[
45 = x^2 + 3^2 = x^2 + 9 \implies x^2 = 45 - 9 = 36 \implies x = 6
\][/tex]
The Pythagorean triple remains [tex]\( (27, 36, 45) \)[/tex].
4. For [tex]\( x = 7 \)[/tex] and [tex]\( y = 5 \)[/tex]:
[tex]\[
a = x^2 - y^2 = 7^2 - 5^2 = 49 - 25 = 24
\][/tex]
[tex]\[
b = 2xy = 2 \cdot 7 \cdot 5 = 70
\][/tex]
[tex]\[
c = x^2 + y^2 = 7^2 + 5^2 = 49 + 25 = 74
\][/tex]
The Pythagorean triple is [tex]\( (24, 70, 74) \)[/tex].
Thus, filling out the table:
[tex]\[
\begin{tabular}{|c|c|c|}
\hline
$x$-value & $y$-value & Pythagorean Triple \\
\hline
4 & 3 & (7,24,25) \\
\hline
5 & 4 & (9,40,41) \\
\hline
6 & 3 & (27,36,45) \\
\hline
7 & 5 & (24,70,74) \\
\hline
\end{tabular}
\][/tex]
