Examples of Pythagorean Triangle Questions

Before we look at some examples of Pythagorean Triangle questions, let's first recall some important aspects of Pythagorean triangles:

• 1) If $(a , b) = 1$, and $a^2 + b^2 = c^2$ for $a, b, c \in \mathbb{Z}$, then (a, b, c) is a fundamental solution.

• 2) In a fundamental solution, one of a or b is odd and c is always odd.

• 3) If (a, b, c) = (x, y, z) is a fundamental solution to $x^2 + y^2 = z^2$ and a is even and b is odd, then for some $m, n \in \mathbb{Z}$ where m > n, $(m, n) = 1$, and one of m or n is even, then $a = 2mn$, $b = m^2 - n^2$, and $c = m^2 + n^2$.

We will now apply this to the following questions.

Example 1

Let (a, b, c) be a fundamental solutions. Find a value of a, b, and c that satisfy $a + b + c = 3290$.

We first note that $a = 2mn$, $b = m^2 - n^2$, and $c = m^2 + n^2$ for some positive integers m and n. Hence we can substitute these values into the linear equation given to get that:

(1)

\begin{align} a + b + c = 3290 \\ (2mn) + (m^2 - n^2) + (m^2 + n^2) = 3290 \\ 2mn + 2m^2 = 3290 \\ 2m(n + m) = 3290 \end{align}

We note that m is a factor of 3290. Hence let's choose m = 35. Hence it follows that n = 12. We note that 35 > 12, (35, 12) = 1, and one of 35 and 12 is odd. Hence it follows that:

(2)

\begin{align} a = 2mn = 840 \\ b = m^2 - n^2 = 1081 \\ c = m^2 + n^2 = 1369 \end{align}

Hence (a, b, c) = (840, 1081, 1369) is a fundamental solution such that $a + b + c = 3290$.

Example 2

Prove that if n is an odd number, then $(n, (n^2 - 1)/2, (n^2 + 1)/2$ is always a Pythagorean triple.

First let's substitute this into Pythagoras' theorem to get that:

(3)

\begin{align} n^2 + \left ( \frac{n^2 - 1}{2} \right ) ^2 = \left ( \frac{n^2 + 1}{2} \right ) ^2 \\ n^2 + \frac{(n^2 - 1)^2}{4} = \frac{(n^2 + 1)^2}{4} \\ n^2 + \frac{n^4 - 2n^2 + 1}{4} = \frac{n^4 + 2n^2 + 1}{4} \\ n^2 = \frac{n^4 +2n^2 + 1}{4} - \frac{n^4 - 2n^2 + 1}{4} \\ n^2 = \frac{n^4 +2n^2 + 1}{4} - \frac{n^4 - 2n^2 + 1}{4} \\ n^2 = \frac{4n^2}{4} \\ n^2 = n^2 \end{align}

We note that n cannot be even. If n is even, then $n^2 - 1$ is odd, and cannot be divided by 2. Hence the proof is complete.

Example 3

Prove that $(2n, n^2 - 1, n^2 + 1)$ is a Pythagorean triple for any $n \in \mathbf{Z^+}$. Show that it is a fundamental solution only if n is even.

We will first prove this this set is always a Pythagorean triple by plugging it into Pythagoras' theorem:

(4)

\begin{align} (2n)^2 + (n^2 - 1)^2 = (n^2 + 1) \\ \quad 4n^2 + n^4 - 2n^2 + 1 = n^4 + 2n^2 + 1 \\ 4n^2 - 2n^2 = 2n^2 \\ n^2 = n^2 \end{align}

We now need to show that it is only a fundamental solution if n is even. First, suppose that n is odd. Hence $n = 2k + 1$. Substituting that in, we get the Pythagorean triple $(2(2k + 1), (2k + 1)^2 - 1, (2k + 1)^2 + 1)$. We will now expand the second and third terms as follows:

(5)

\begin{align} (2k+1)^2 - 1 \\ = 4k^2 + 4k + 1 - 1 \\ = 4k^2 + 4k \\ = 2(2k^2 + 2k) \\ \end{align}

And for the hypotenuse it follows that:

(6)

\begin{align} (2k+1)^2 + 1\\ = 4k^2 + 4k + 1 + 1\\ = 4k^2 +4k + 2 \\ = 2(2k^2 + 2k + 1) \end{align}

We thus note that our Pythagorean triple can be written as $( 2(2k + 1), 2(2k^2 +2k), 2(2k^2 + 2k + 1) )$. All terms have a common factor of 2, hence when n is odd, $(2n, n^2 - 1, n^2 + 1)$ is not a fundamental solution.

Furthermore, suppose that n is even. Hence $n = 2k$ for some k. Substituting this into our Pythagorean triple as we get $(4k, 4k^2 - 1, 4k^2 + 1)$. We note that $(4k, 4k^2 - 1) = 1$ since the divisors of 4k are 1, 2, 4, and k, none of which are divisors of 4k² - 1 EXCEPT 1. Hence when n is even, $(2n, n^2 - 1, n^2 + 1)$ yields a fundamental solution.

Example 4

Find a Pythagorean triple where one of a, b, or c is 11.

We first note that 11 is an odd number. Hence let's let $a = 11$. It thus follows that $b = \frac{11^2 - 1}{2} = 60$, and $c = \frac{11^2 + 1}{2} = 61$. Hence (x, y, z) = (11, 60, 61) is a Pythagorean triple containing 11.

We can verify this as $11^2 + 60^2 = 61^2$ as 3721=3721.

Example 5

Find a Pythagorean triple where one of a, b, or c is 14.

We note that c CANNOT be 14 since c must always be odd. Let's let a be 14. It thus follows that $14 = 2mn$, or rather $7 = mn$. Let's choose m = 7, and n = 1. Hence m > n, and (m, n) = 1. It thus follows that $b = 7^2 - 1^2 = 48$, and $c = 7^2 + 1^2 = 50$. Hence (x, y, z) = (14, 48, 50) is a Pythagorean triple containing 14.

Example 6

Find a Pythagorean triple where a = 52.

We will use the proof in example 3 to easily find a Pythagorean triple with this property. Since a = 52 is even, we get that our n = 26, and thus we have a Pythagorean triple in the form $( 2(26), (26)^2 - 1, (26)^2 + 1)$. When we simplify this, we get $(52, 675, 677)$.

We can verify this to be a valid Pythagorean triple. That is:

(7)

\begin{align} 52^2 + 675^2 = 677^2 \\ 2704 + 455625 = 458329 \end{align}