I What are the resources for Pythagorean triples?

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What is the most motivating way in introduction to Pythagorean triples to undergraduate students? I am looking for an approach that will have an impact. Good interesting or real life examples will help. Is there any resources for this?
 

fresh_42

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What is the most motivating way in introduction to Pythagorean triples to undergraduate students? I am looking for an approach that will have an impact. Good interesting or real life examples will help. Is there any resources for this?
I have for gotten the details, but I believe to remember, that on a documentation about Stonehenge they said that the constructors must have known the theorem, or the triples. But anyway, I'm sure that they can be found in ancient constructions prior to Pythagoras.
 
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There is also the recent discovery that Ramanujan was looking at them in an effort to crack Fermats Theorem. It came from the taxicab numbers story of 1729 being the sum of two cubes two different ways (10^3 + 9^3) = (12^3 + 1^3) (edit: fixed typo )

https://plus.maths.org/content/ramanujan

You could explore the pythagorean triples and then at the end bring up Fermat's theorem as a generalization of it and then show how Ramanujan was beginning to investigate it with his collection of taxicab numbers and then on to Wiles proof.

Next, there's these resources:

 
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Thanks for this information. Should be 12^3+1^3=1729.
 

mathman

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There is a simple algorithm to get them all. Start with integers ##m## and ##n## with ##m\gt n##. Let ##a=m^2-n^2##, ##b=2mn##, then ##c=m^2+n^2##.
 

PeterO

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What is the most motivating way in introduction to Pythagorean triples to undergraduate students? I am looking for an approach that will have an impact. Good interesting or real life examples will help. Is there any resources for this?
You could refer to what square numbers mean - the number of dots in a square array.
3^2 is an array 3 x 3
4^2 is an array 4 x 4

To convert the 3 x 3 array to a 4 x 4 array, you first add 3 dots to the bottom, then 4 dots to the right hand side.

Generally, to get from one number squared to the next number squared, you add the number, then the next number. You want the sum of "the number and the next number" to be a perfect square.
Pythagorean triples occur when ever two consecutive numbers sum to a perfect square.
eg 9 = 4 + 5 , but 9 is the square of 3, so 3,4,5 are a triple
two consecutive numbers will always sum to an odd number.
The next odd square is 25 (5^2) - which is 12 + 13 - leading to the triple 5,12,13
49 = 24 + 25 leading to 7,24,25 etc.

You can also investigate triples based on two numbers 2 values apart.

To get from a 3x3 array to a 5x5 array you add 3 then 4, then 4, then 5
ie a number plus two lots of the next number plus the number after that. - which equates to 4 x the "next number". SO we are looking for squares that are a multiple of 4.

4^2 = 16 = 3+4+4+5 which leads to 4,3,5 or 3,4,5 again
8^2 = 64 = 15 + 16 + 16 + 17 which leads to 8, 15,17
144 = 35 + 36 + 36 + 37 which leads to 12, 35, 37
Set them to find the next 3 or 4 of these.

It gets harder when the difference between the two "base squares" are bigger than 2.
 

mathwonk

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the proof of mathman's algorithm is short and sweet by geometry. note that the problem is equivalent to finding all pairs of rational numbers (x,y) such that x^2+y^2 = 1, i.e. to finding allpoints of the unit circle with both coordinates rational. By projecting along a line from the north pole (0,1), rational points on the circle correspond one - one with rational points on the real axis. then just compute where the line joining the points (0,1) and (n/m,0) meets the unit circle, and you get all points with coordinates (2mn/(m^2+n^2), (m^2-n^2)/(m^2+n^2)), i.e. you get all pythagorean triples as (2mn), (m^2-n^2), (m^2+n^2). (I changed a sign which switched the order of the first two points.)
 

mathman

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the proof of mathman's algorithm is short and sweet by geometry. note that the problem is equivalent to finding all pairs of rational numbers (x,y) such that x^2+y^2 = 1, i.e. to finding allpoints of the unit circle with both coordinates rational. By projecting along a line from the north pole (0,1), rational points on the circle correspond one - one with rational points on the real axis. then just compute where the line joining the points (0,1) and (n/m,0) meets the unit circle, and you get all points with coordinates (2mn/(m^2+n^2), (m^2-n^2)/(m^2+n^2)), i.e. you get all pythagorean triples as (2mn), (m^2-n^2), (m^2+n^2). (I changed a sign which switched the order of the first two points.)
Is this a proof that all triples are given by the algorithm I gave? (Don't attribute it to me - it is well known and can easily be verified by computing ##a^2,b^2,c^2##).
 

mathwonk

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yes it is. as far as i understand your second comment, you are saying (by computing a^2 + b^2 and c^2), that it is easily verified that all triples given by your algorithm are indeed pythagorean triples. i agree that is "obvious". i am arguing the less obvious converse, that all pythagorean triples are indeed given by your algorithm.

i.e.
1) if a triple (a,b,c) is given by your algorithm then a^2+b^2 = c^2.
2) if a^2 + b^2 = c^2, then (a,b,c) is given by your algorithm.

I agree that 1) is easy by computation, and I have thus argued that 2) is also true. I.e. I have argued that if m/n is a rational point on the real line such that the corresponding point (a/c, b/c) on the circle satisfies a^2+b^2=c^2, then the number n,m satisfies your algorithm.

Moreover, I have stated but not argued, that every rational point on the circle comes from a rational point m/n on the real line. For this you have to note that a line passing through two rational points on the circle must have a rational slope and hence must meet the real line at a rational point. does this sound right? basically this is just because linear equations with rational coefficients have rational solutions.
 
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mathman

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From
we obtain
and hence
. Then
. Since
is rational, we set it equal to
in lowest terms. Thus
, being the reciprocal of
. Then solving


for
and
gives



Above is proof from Wikipedia
 

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