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matqkks

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matqkks

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jedishrfu

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This is perhaps the best presentation I've ever seen:

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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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jedishrfu

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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:

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:

http://www.teachersofindia.org/en/article/ramanujan-and-pythagoras

and this article on Pythagorean puzzle proof:

https://fossbytes.com/computer-scie...hagorean-triples-problem-with-200-tb-of-data/

and this one with the Plimpton tablet translation of a table of pythagorean triples:

https://thatsmaths.com/2014/01/23/pythagorean-triples/

and this article on Pythagorean puzzle proof:

https://fossbytes.com/computer-scie...hagorean-triples-problem-with-200-tb-of-data/

and this one with the Plimpton tablet translation of a table of pythagorean triples:

https://thatsmaths.com/2014/01/23/pythagorean-triples/

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matqkks

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

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mathman

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PeterO

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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.

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.

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mathwonk

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mathman

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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##).

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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.

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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- #11

mathman

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for

Above is proof from Wikipedia

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