Chapter 4.2, Problem 65E

(a)

To determine

To prove: the polynomial identity is true by showing that the simplified expressions for the left and right sides are the same.

Answer to Problem 65E

The identify is true

Explanation of Solution

Given information:

A Pythagorean triple is a set of positive integers $a,b$ and $c$ such that $a^2+b^2=c^2$ . The numbers 3, 4, and 5 forms a Pythagorean triple because $3^2+4^2=5^2$ . The polynomials identify ${\left(x^2-y^2\right)}^2+{\left(2xy\right)}^2={\left(x^2+y^2\right)}^2$ to generate other Pythagorean triples.

Calculation:

Expand and simplify the expression on the left side of equation.

  $\begin{array}{c}{\left(x^2-y^2\right)}^2+{\left(2xy\right)}^2=x^4-2x^2{y}^2+y^4+4x^2{y}^2\\ =x^4+2x^2{y}^2+y^4\end{array}$

Expand and simplify the expression on the left side of equation.

  ${\left(x^2+y^2\right)}^2=x^4+2x^2{y}^2+y^4$

Because the simplified left side is equal to the right side, so the identify is true.

(b)

To determine

To generate: the Pythagorean triple when $x=6$ and $y=5$ using the identity.

Answer to Problem 65E

The Pythagorean triplets are 11, 60 and 61.

Explanation of Solution

Given information:

A Pythagorean triple is a set of positive integers $a,b$ and $c$ such that $a^2+b^2=c^2$ . The numbers 3, 4, and 5 forms a Pythagorean triple because $3^2+4^2=5^2$ . The polynomials identify ${\left(x^2-y^2\right)}^2+{\left(2xy\right)}^2={\left(x^2+y^2\right)}^2$ to generate other Pythagorean triples.

Calculation:

Use the identity when $x=6$ and $y=5$ .

  $\begin{array}{c}{\left(x^2-y^2\right)}^2+{\left(2xy\right)}^2={\left(x^2+y^2\right)}^2\\ {\left(6^2-5^2\right)}^2+{\left(2\cdot 6\cdot 5\right)}^2={\left(6^2+5^2\right)}^2\\ {11}^2+{60}^2={61}^2\end{array}$

Therefore, the Pythagorean triplets are 11, 60 and 61.

(c)

To determine

To verify: the answer in part (b) satisfies $a^2+b^2=c^2$ .

Answer to Problem 65E

Explanation of Solution

Given information:

A Pythagorean triple is a set of positive integers $a,b$ and $c$ such that $a^2+b^2=c^2$ . The numbers 3, 4, and 5 forms a Pythagorean triple because $3^2+4^2=5^2$ . The polynomials identify ${\left(x^2-y^2\right)}^2+{\left(2xy\right)}^2={\left(x^2+y^2\right)}^2$ to generate other Pythagorean triples.

Calculation:

Using $a=11$ and $b=60$ ,

  $\begin{array}{c}{a}^2+b^2={11}^2+{60}^2\\ =121+3600\\ =3721\\ ={61}^2\end{array}$

Thus, result in part (b) satisfies $a^2+b^2=c^2$ .