Pythagorean Triples investigation

I have done an investigation on Pythagorean triples, concerning some of their properties, which I will give to you as a proof problem, in two parts:

  1. Suppose a,b,p a,b,p are natural numbers such that p p is prime and a2+b2=p2\large a^2 +b^2 = p^2 . Prove that if the Pythagorean triple is primitive, then one of a,b a, b is divisible by 3.

  2. If 11a 11 \mid a (11 divides a) then is it possible that 2a 2 \mid a ?

Note by Curtis Clement
4 years, 8 months ago

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Good examples for 1) include:  32+42=52\ 3^2 + 4^2 = 5 ^2 112+602=61211^2 +60^2 = 61 ^2 For 2) you need either an example or a proof (possibly by modular arithmetic) that the statement is true or false

Curtis Clement - 4 years, 8 months ago

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wow!Thanks a lot!What I realize here is it may be a or b that is divisible by 3.I also realize another thing and that is once you mix up both of these numbers and divide it by 3,it will end up with 33333 or 6666 at the back of the decimal example 3^2+4^2=25/3=8.3333333

Frankie Fook - 4 years, 8 months ago

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I am assuming that a,b,pa,b,p, with pp prime, is a primitive Pythagorean triple (in no particular order). Also, if they are in no particular order then statement 1 does not hold true, since in a (3,4,5)(3,4,5) primitive Pythagorean triangle we have the prime p=3p = 3 divisible by 33 but neither 44 nor 55 being divisible by 33. So it's unclear if you meant that one of a,b,pa,b,p is divisible by 33, (in order to cover this case), or if you meant that the prime pp is the hypotenuse in a primitive triple. I'll assume the former for the following proof.

To prove statement 1., first note that any Pythagorean triple can be found by choosing natural numbers m,nm,n with m>nm \gt n and then forming side lengths 2mn,m2n2,m2+n2.2mn, m^{2} - n^{2}, m^{2} + n^{2}. If 3m3 | m or 3n3 | n then all side lengths will be divisible by 33 and we are done. If m±1(mod3)m \equiv \pm 1 \pmod{3} and n±1(mod3)n \equiv \pm 1 \pmod{3} then m21(mod3)m^{2} \equiv 1 \pmod{3} and n21(mod3)n^{2} \equiv 1 \pmod{3}, in which case m2n20(mod3)m^{2} - n^{2} \equiv 0 \pmod{3}. This means that at least one of the side lengths is divisible by 33.

Now if m2n2=3m^{2} - n^{2} = 3 then m=2m = 2 and n=1n = 1, giving us a (3,4,5)(3,4,5) triangle, in which case neither aa nor bb is divisible by 33. However, if m2n23m^{2} - n^{2} \ne 3 then this side cannot be pp, (since it is divisible by 33 and greater than 33), and thus one of aa or bb must be given by m2n2m^{2} - n^{2}, and thus one of aa or bb is indeed divisible by 33.

I'm not clear on what you are after with for question 2. We have the triple (66,88,110)(66,88,110), for example, with 116611 | 66 as well as 2662 | 66, implying that it is possible. Or are we still working with primitive triples a,b,pa,b,p? If so, then things get a bit more interesting. We have the triple (20,99,101)(20,99,101), where p=101p = 101 and 11a=9911 | a = 99 but 22 does not divide a=99.a = 99. Looking further ..... yes, we have a winner. Look at the primitive triple (88,105,137)(88,105,137). We have p=137p = 137 and both 11a=8811 | a = 88 and 2a=882 | a = 88. So the answer is "Yes, it is possible".

Brian Charlesworth - 4 years, 8 months ago

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I am indeed assuming that a,b,pa,b,p with p{p} prime is a primitive triple with a2+b2=p2\large a^2 +b^2 = p^2 . I like your solution and also did you notice that 4|88 as well - there is a good reason for that....

Curtis Clement - 4 years, 8 months ago

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It is easy to see why it is possible for 44|a. If (a,b,c) = (2mn, m2m^{2} - n2n^{2}, m2m^{2} + n2n^{2}), then we are left to show that if 11|a then 2|mn. Now (m-n)(m+n)1\equiv1mod(2) so without loss of generality m is even and n is odd such that 4|2mn and 11|2mn \Rightarrow \therefore 44|a Q.E.D

Curtis Clement - 4 years, 8 months ago

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Excuse me,may I ask a question?

Frankie Fook - 4 years, 8 months ago

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yes of course (although technically you have already done so )

Curtis Clement - 4 years, 8 months ago

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Thanks!Can you give me some examples? I need more examples to analyze and learn it

Frankie Fook - 4 years, 8 months ago

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