Pythagorean Triples

Pythagorean triples are sets of three integers which satisfy the property that they are the side lengths of a right-angled triangle (with the third number being the hypotenuse).

$(3, 4, 5)$ is the most popular example of a Pythagorean triple. If you multiply each of the numbers in this triple by an integer, the result will also be a Pythagorean triple.

If $(a, 35, 37)$ is a Pythagorean triple, what is $a?$

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Since $(a, 35, 37)$ is a Pythagorean triple, the following must hold:

$$\begin{aligned} a^2+35^2&=37^2\\ a^2&=37^2-35^2\\ &=144. \end{aligned}$$

Since only a positive integer counts, the answer is $a=12.$ $_\square$

Which of the following is not a Pythagorean triple?

$$\begin{array}{c}&&\textbf{A. }(5, 12, 13) &&\textbf{B. }(7, 24, 25) &&\textbf{C. }(9, 40, 41) &&\textbf{D. }(9, 60, 61) \end{array}$$

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We have

$$\begin{aligned} 5^2+12^2&=25+144=169=13^2\\ 7^2+24^2&=49+576=625=25^2\\ 9^2+40^2&=81+1600=1681=41^2\\ 9^2+60^2&=81+3600=3681 \ne 3721=61^2. \end{aligned}$$

Hence, the answer is $\textbf{D}.$ $_\square$

If the largest number of a Pythagorean triple is $17,$ what is the smallest number in that triple?

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First, if the second largest number is $16,$ we need $$17^2-16^2=289-256=34$$ to be a square, which it is not.

Next, if the second largest number is $15,$ we need $$17^2-15^2=289-225=64$$ to be a square, which it is because $64=8^2.$ Hence, we have $8^2+15^2=17^2.$

Therefore, the smallest number in the Pythagorean triple is $8.$ $_\square$

Generating triples has always interested mathematicians, and Euclid came up with a formula for generating Pythagorean triples. Primitive Pythagorean triples are Pythagorean triples $a, b$ and $c$ such that $a, b$ and $c$ are coprime. Note however that this formula generates all primitive triples but not all non-primitive triples. Euclid gives us the following formula:

$$\begin{aligned} a& = m^2 - n^2 \\ b& = 2mn \\ c& = m^2 + n^2, \end{aligned}$$

where $m$ and $n$ are any two positive integers that are usually called parameters. If we wanted to generate all the Pythagorean triples, we could then introduce a third parameter $k$ to our formula:

$$\begin{aligned} a&= k\big(m^2 - n^2\big) \\ b&=k(2mn) \\ c&= k\big(m^2 + n^2\big). \end{aligned}$$

You can also generate the triples by using the formula

$$\begin{aligned} a& = m^2 -1 \\ b& = 2m\\ c& = m^2 + 1, \end{aligned}$$

where

$$\begin{aligned} a^2 + b^2 = c^2. \end{aligned}$$

The primitive Pythagorean triple $(3, 4, 5)$ is arguably the most famous and commonly used. Find the parameters $m$ and $n$ used in Euclid's formula to generate this triple.

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Start with $$\begin{aligned} a &= 3 = m^2 - n^2 \\ b &= 4 = 2mn \\ c&= 5 = m^2 + n ^2. \end{aligned}$$ If we add equation 1 and 3, we get $$\begin{aligned} a + c = 8 &= 2m^2 \\ m &= 2, \end{aligned}$$ and substituting we get $$n = \frac{4}{2(2)} = 1 .$$

The parameters $m$ and $n$ are thus 2 and 1, respectively. $_\square$

Another famous but non-primitive Pythagorean triple is $(6,8,10)$. What is the value of the parameter $k$ required to generate this triple?

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Notice that $(6,8,10)$ is just a multiple of $(3,4,5)$. All terms are just multiplied by two. Thus $k=2$ and $m$ and $n$ are still $2$ and $1,$ respectively. $_\square$

What are the primitive triples such that the smallest of the triple is $a=20?$

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Observe that $a=20=2mn$ has two options remembering that $m>n.$

Then if $n=1, m=10,$ $$a=2mn=20, b=m^2-n^2=99, c=m^2-n^2=101 \implies 20^2+99^2=101^2.$$ If $n=2, m=5,$ $$a=2mn=20, b=m^2-n^2=21, c=m^2-n^2=29 \implies 20^2+21^2=29^2.$$

Therefore, the triples are $(20, 99, 101)$ and $(20, 21, 29).$ $_\square$

One might wonder why this parameterization of Pythagorean triples is so useful. One could easily think of infinitely many parameterizations for different number of variables. The answer is because this one allows us to generate primitive Pythagorean triples using only $2$ variables. This comes in handy when trying to generate large amounts of triples. One would not have to check every possible integer $a, b$ and $c$ to obtain all Pythagorean triples, but instead could check every possible $m$ and $n$ and then multiply by the triples by required values of $k$.