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

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

Note by Curtis Clement
2 years ago

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I am assuming that \(a,b,p\), with \(p\) 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 \((3,4,5)\) primitive Pythagorean triangle we have the prime \(p = 3\) divisible by \(3\) but neither \(4\) nor \(5\) being divisible by \(3\). So it's unclear if you meant that one of \(a,b,p\) is divisible by \(3\), (in order to cover this case), or if you meant that the prime \(p\) 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 \(m,n\) with \(m \gt n\) and then forming side lengths \(2mn, m^{2} - n^{2}, m^{2} + n^{2}.\) If \(3 | m\) or \(3 | n\) then all side lengths will be divisible by \(3\) and we are done. If \(m \equiv \pm 1 \pmod{3}\) and \(n \equiv \pm 1 \pmod{3}\) then \(m^{2} \equiv 1 \pmod{3}\) and \(n^{2} \equiv 1 \pmod{3}\), in which case \(m^{2} - n^{2} \equiv 0 \pmod{3}\). This means that at least one of the side lengths is divisible by \(3\).

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

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

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@Brian Charlesworth I am indeed assuming that \(a,b,p \) with \({p}\) prime is a primitive triple with \(\large a^2 +b^2 = p^2 \). I like your solution and also did you notice that 4|88 as well - there is a good reason for that.... Curtis Clement · 2 years ago

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@Brian Charlesworth It is easy to see why it is possible for 44|a. If (a,b,c) = (2mn, \(m^{2}\) - \(n^{2}\), \(m^{2}\) + \(n^{2}\)), then we are left to show that if 11|a then 2|mn. Now (m-n)(m+n)\(\equiv1\)mod(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 · 2 years ago

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Good examples for 1) include: \[\ 3^2 + 4^2 = 5 ^2 \] \[11^2 +60^2 = 61 ^2 \] For 2) you need either an example or a proof (possibly by modular arithmetic) that the statement is true or false Curtis Clement · 2 years ago

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@Curtis Clement 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 · 2 years ago

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Excuse me,may I ask a question? Frankie Fook · 2 years ago

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@Frankie Fook yes of course (although technically you have already done so ) Curtis Clement · 2 years ago

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@Curtis Clement Thanks!Can you give me some examples? I need more examples to analyze and learn it Frankie Fook · 2 years ago

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