Pythagorean Triples – Part 2

Slideshow 39, MathematicsMr. Richard Sasaki, Room 307

ObjectivesObjectives

• Learn another proof for Euclid’s formulae to calculate triples

• Check the points that we should have learned from the given project (relationships & rules with triples)

• Learn and use some laws in triples• Introduce triangular numbers

Euclid’s FormulaeEuclid’s FormulaeWe know all about Euclid’s formulae for generating triples now…𝒂=𝒎𝟐−𝒏𝟐 ,𝒃=𝟐𝒎𝒏 ,𝒄=𝒎𝟐+𝒏𝟐

If we know that and , we may use the following proof.Consider a triple that satisfies where where and , .

We can say Also,

∴𝑐2−𝑎2=(𝑚2+𝑛2 )2− (𝑚2−𝑛2 )2=4𝑚2𝑛2

So, As the above is in the form , the formulae above apply.

Rules and Patterns for TriplesRules and Patterns for TriplesFor the formulae , we can insert any values of and where .

𝒎>𝒏Triples always consist of three positive integers where they are either…• All even numbers (always non-primitive)• Two odd numbers and an even number

Some properties for primitive triples include…• Either or is odd, is odd• One of or has a factor of 3 and 4

• Could be either of them has both factors or has one and has one

• Either or has a factor of 5• At maximum, one of or is square

Rules and Patterns for TriplesRules and Patterns for TriplesAn interesting one is if you calculate , you always get a square number.

ExampleUsing , test whether may be a valid triple.

(65−33 ) (65−56 )2

=¿2882

=¿144The test is passed as 144 is a square number.

Note: This formula can’t be used for 100% certainty, would satisfy this for example. But all valid triples satisfy this formula.

Rules and Patterns for TriplesRules and Patterns for TriplesIf you enter any value for for a triple where , and is a prime number other than 2, and can be calculated. In other words, triples exist for where and prime. In fact they will be primitive.So how do we calculate them?ExampleConsider a triangle with its shortest leg . Find the lengths of the other two edges assuming all edges hold integer values.

13𝑐𝑚You may have found that when is prime, for the triple , .

( and are consecutive integers.)

Rules and Patterns for TriplesRules and Patterns for Triples

13𝑐𝑚Using the fact that and we are given , we can make a formula to calculate and .

As always, start with . Using , write a simpler equation about and .𝑎2+𝑏2=𝑐2

𝑎2=𝑐2−𝑏2𝑎2=(𝑏+1)2−𝑏2

𝑎2=𝑏2+2𝑏+1−𝑏2𝒂𝟐=𝟐𝒃+𝟏𝑎2=𝑏+(𝑏+1)𝒂𝟐=𝒃+𝒄

Note: Actually, is just for looking of pairs that fit and , not for calculation.

This is in terms of .

𝑎2=(𝑐−1)+𝑐𝒂𝟐=𝟐𝒄−𝟏

This is in terms of . 132=𝑏+𝑐169=𝑏+𝑐

169=84+85𝑏=84 ,𝑐=85

AnswersAnswers

(not a square number)

None have a factor of 4 or 3

𝑎=√𝑏+𝑐 ,𝑎=√2𝑏+1 ,𝑎=√2𝑐−1(3 ,4 ,5) (7 ,24 ,25)(11 ,60 ,61)

(13 ,84 ,85)(17 ,144 ,145)

(19 ,180 ,181)

(23 ,264 ,265)(29 , 420 ,421)

(31 ,480 ,481)

𝑏=𝑎2−12

,𝑐=𝑎2+12

Works but gives the wrong solution

Works but gives the wrong solution

must be prime, odd numbers work but will give the incorrect answers

Triangular NumbersTriangular NumbersYou learned about these in the Winter Homework.They follow the list…1 ,3 ,6 ,10 ,15 ,21 ,…

As you know, they are written in the form where is the number’s position, or the number of dots on an edge.What do these numbers have to do with Pythagorean triples?

Triangular NumbersTriangular NumbersIf you checked the grading sheet (as advised), you would have seen…

Your project was for , right? How many triples were there? 45For this test, .For , you’d simply ignore results where or . How many are there? ()

36In fact, as you decrease the values of , you’ll see…

1 2 3 4 5 6 7 8 9 10

36 45282115106310

This produces the triangle numbers!

Triangular NumbersTriangular NumbersSo for , is the number of valid triples.If you were asked to calculate though, how would you?𝑇 𝑥=¿1+2…+𝑥¿∑

𝑛=1

𝑥

𝑛=¿𝑥 (𝑥+1 )2

Using is the easiest way to calculate this.

𝑇 50=¿50 (50+1 )2

¿25502

¿1275Let’s prove .Consider and . All terms can be written as and there are terms.

∴2𝑇 𝑥=𝑥 (𝑥+1 )⇒𝑇 𝑥=𝑥 (𝑥+1 )2

AnswersAnswers

10 153 276

𝑇 𝑛−1

𝑛2

𝑛

𝐶2❑𝑛+1 =

(𝑛+1 )!2 (𝑛−1 )!

¿(𝑛+1 ) ∙𝑛 ∙ (𝑛−1 ) ∙…∙12 (𝑛−1) ∙…∙1

¿(𝑛+1 ) ∙𝑛2

=𝑛 (𝑛+1 )2