Chapter 8.3, Problem 3E

a.

To determine

To calculate: primitive triplet with 33 as its lowest value

Answer to Problem 3E

z $33,544,545$

Explanation of Solution

Given information: Method is given to find Pythagorean triples with an odd number.

Calculation:

Let $n$ be a positive integer. A triple is given by $a=2n+1,b=2n^2+2n,c=\left(2n^2+2n\right)+1$

Put $n=16$

  $\begin{array}{l}a=2n+1=2\left(16\right)+1=32+1=33\\ b=2n^2+2n=2{\left(16\right)}^2+2\left(16\right)=512+32=544\\ c=(2n^2+2n)+1=\left[2{\left(16\right)}^2+2\left(16\right)\right]+1=544+1=545\end{array}$

b.

To determine

To explain: the method used in part a.

Explanation of Solution

According to Pythagorean theorem,

  $a^2+b^2=c^2$ where $a,b$ denote sides of the right triangle and $c$ denotes the hypotenuse.

First find $a^2+b^2$

  $\begin{array}{l}{a}^2+b^2={\left(2n+1\right)}^2+{\left(2n^2+2n\right)}^2\\ =4n^2+1+4n+4n^4+4n^2+8n^3\\ =4n^4+8n^3+8n^2+4n+1\end{array}$

Now find $c^2$

  $\begin{array}{l}{c}^2={\left[\left(2n^2+2n\right)+1\right]}^2\\ ={\left(2n^2+2n+1\right)}^2\\ =4n^4+4n^2+1+8n^3+4n+4n^2\\ =4n^4+8n^3+8n^2+4n+1\end{array}$

So,

  $a^2+b^2=c^2$

As the above numbers $a,b,c$ satisfy the Pythagorean theorem, this method can be used to find the Pythagorean triplets.