Chapter 8.3, Problem 2E

To determine

To prove: The following statement is true or not.

Answer to Problem 2E

The equation given is true.

Explanation of Solution

Given:

Euclid’s method of Pythagorean triplet

  ${\left(2xy\right)}^2+{\left(x^2-y^2\right)}^2={\left(x^2+y^2\right)}^2$

Formula used:

  $\begin{array}{l}{\left(a+b\right)}^2=a^2+b^2+2ab\\ {\left(a-b\right)}^2=a^2+b^2-2ab\end{array}$

Calculation:

Given expression is,

  $\Rightarrow {\left(2xy\right)}^2+{\left(x^2-y^2\right)}^2={\left(x^2+y^2\right)}^2$

Now taking left hand side,

  $\Rightarrow {\left(2xy\right)}^2+{\left(x^2-y^2\right)}^2$

  $\Rightarrow 4x^2{y}^2+x^4+y^4-2x^2{y}^2$$\left[\because {\left(a-b\right)}^2=a^2+b^2-2ab\right]$

  $\Rightarrow 2x^2{y}^2+x^4+y^4$

Now taking right hand side,

  $\Rightarrow {\left(x^2+y^2\right)}^2$

  $\Rightarrow x^4+y^4+2x^2{y}^2$$\left[\because {\left(a+b\right)}^2=a^2+b^2+2ab\right]$

  $\Rightarrow x^4+y^4+2x^2{y}^2$

Thus, left hand side $=$ right hand side.

Conclusion:

Therefore, the following statement is true that is Euclid’s method is correct.