[Guest]

A primitive Pythagorean triangle is defined as any triangle with integral sides such that one of the angles is 90º, and also that there is no factor which divides all sides leaving an integer solution. Hence the (3,4,5) triangle is a PPT while (6,8,10) and (9,12,15) are not.

If one googles the Wikipedia entry on the topic, one finds that it mentions an intriguing property of the length of the hypotenuse, namely that it has no factor which leaves a remainder 3 when divided by 4. It is easy enough to show that c = 1 (mod 4), but unfortunately this does not itself prevent say 7² being a factor.

Can you find a simple argument to establish the claim, using modulo arithmetic only?

Edited October 5, 2009 by jerbil

[Guest]

A primitive Pythagorean triangle is defined as any triangle with integral sides such that one of the angles is 90º, and also that there is no factor which divides all sides leaving an integer solution. Hence the (3,4,5) triangle is a PPT while (6,8,10) and (9,12,15) are not.

If one googles the Wikipedia entry on the topic, one finds that it mentions an intriguing property of the length of the hypotenuse, namely that it has no factor which leaves a remainder 3 when divided by 4. It is easy enough to show that c = 1 (mod 4), but unfortunately this does not itself prevent say 7² being a factor.

Can you find a simple argument to establish the claim, using modulo arithmetic only?

I didn't quite understand the question after the "..but unfortunately" part.

Sure its easy enough to show c = 1 mod(4) using mod maths but i guess that is not the question. Mind elaborating what exactly is your question?

[Guest]

From first asppearances any two prime factors of the form (4n+3), when multiplied together, form a composite factor of the form (4n+1), which on the surface seems to make them an eligible candidate for c. But in fact this is not the case. No factor of c has the form (4n+3). One way of generating PPTs is to choose two mutually prime odd numbers u and v, with u > v, and calculating the values of

a = ((u²-v²)/2

b = uv

c = (u²+v²)/2.

For example, u = 3, v = 1 give us the (3,4,5) triangle. u = 5, v = 3 give us the (8,15,17) triangle.

Edited October 5, 2009 by jerbil

[Guest]

Any square of an integer, which is then divided by 4 will have a remainder of 1 or 0, never a 2 or 3. And it is in a patern! Remainders for:

1/4 = r of 1

4/4 = r of 0

9/4 = r of 1

16/4 = r of 0

25/4 = r of 1

36/4 = r of 0

and so on. The patern is 10101010101010101010101.........

[Guest]

Any square of an integer, which is then divided by 4 will have a remainder of 1 or 0, never a 2 or 3. And it is in a patern! Remainders for:

1/4 = r of 1

4/4 = r of 0

9/4 = r of 1

16/4 = r of 0

25/4 = r of 1

36/4 = r of 0

and so on. The patern is 10101010101010101010101.........

show the definition of even and odd.. just a little more complicated that's all... o-0

an odd number is defined as such when dividing the number by 2 produces a remainder of 1...

what did you do?

n²/2²...

the squares basically cancel each other out

not the answer though.

[Guest]

Any square of an integer, which is then divided by 4 will have a remainder of 1 or 0, never a 2 or 3. And it is in a patern! Remainders for:

1/4 = r of 1

4/4 = r of 0

9/4 = r of 1

16/4 = r of 0

25/4 = r of 1

36/4 = r of 0

and so on. The patern is 10101010101010101010101.........

From jerbil's post, we know

One way of generating PPTs is to choose two mutually prime odd numbers u and v, with u > v, and calculating the values of

a = ((u2-v2)/2

b = uv

c = (u2+v2)/2.

Now, u2+v2 for two mutually prime odd numbers can be re-written as 2*u2+8*x. Dividing by 2, we have c = u2+4*x. Therefore, dividing c by 4 will have the same remainder as dividing u2 by 4, which must be 0 or 1 as psycho has brought to our attention.

[Guest]

Please ignore an irrelevant commemt.

Edited October 5, 2009 by jerbil

[Guest]

From jerbil's post, we know

Now, u2+v2 for two mutually prime odd numbers can be re-written as 2*u2+8*x. Dividing by 2, we have c = u2+4*x. Therefore, dividing c by 4 will have the same remainder as dividing u2 by 4, which must be 0 or 1 as psycho has brought to our attention.

I mistakingly switched u and v, but I believe the logic still holds, simply replacing my u above with v.

[Guest]

One way of generating PPTs is to choose two mutually prime odd numbers u and v, with u > v, and calculating the values of

a = ((u2-v2)/2

b = uv

c = (u2+v2)/2.

Now, u2+v2 for two mutually prime odd numbers can be re-written as 2*v2+8*x. Dividing by 2, we have c = v2+4*x. Therefore, dividing c by 4 will have the same remainder as dividing v2 by 4, which must be 0 or 1 as explained below.

We can express the odd prime number v as either 4n+1 or 4n+3. When we square both of these, we get a number with mod4 = 1. Therefore, C should have mod4 = 1.

[Guest]

I understand your last post, which, in my view is a dead end. I have to say that, in my approaches, I certainly came to many dead ends myself.

In your future posts, may I suggest you utilize the toolbar to allow you use it to make your equations more simple to read? For example, using y2 to represent y² is unnecessary. Just place in "y2" then backspace on the "2", and then press the relevant button on the toolbar. The same goes for suffices.

[Guest]

I understand your last post, which, in my view is a dead end. I have to say that, in my approaches, I certainly came to many dead ends myself.

In your future posts, may I suggest you utilize the toolbar to allow you use it to make your equations more simple to read? For example, using y2 to represent y² is unnecessary. Just place in "y2" then backspace on the "2", and then press the relevant button on the toolbar. The same goes for suffices.

Sorry. I simply copied the quote, and expected the formatting to not change.

[Guest]

Perhaps I can make the claim I am making somewhat more vivid. I am stating that the hypotenuse of a PPT can never be a multiple of such prime numbers as 3, 7, 11, 19, 23, 31 and 43 (and so on.)

[Guest]

Perhaps I can make the claim I am making somewhat more vivid. I am stating that the hypotenuse of a PPT can never be a multiple of such prime numbers as 3, 7, 11, 19, 23, 31 and 43 (and so on.)

I understand the question, can you post a spoiler hint? I don't see any way of going about it

[Guest]

Definitely not, methinks. I do not see how a hint, as such, can be of any benefit. The solution, unfortunately not originating from yours truly (I wish I were that intelligent) will be posted shortly.

Edited October 9, 2009 by jerbil

[Guest]

I immediately felt guilty about my previous post. Here is a hint, underneath a spoiler.

Consider the modular equation a² + b² = 0 (mod 4). Under what circumstances might this occur?

[Guest]

I immediately felt guilty about my previous post. Here is a hint, underneath a spoiler.

Consider the modular equation a² + b² = 0 (mod 4). Under what circumstances might this occur?

this can only be true when a² AND b² are both multiples of 4. This also implies that c² is a multiple of 4 and that the equation can therefore be divided by 4 and is not a PPT. For a PPT, c² can not have mod4 of 0. It can only have mod4 of 1.

I'm still missing the part about how the factors of c² can not include a number with mod4 of 3.

[Guest]

I apologise for an earlier typo - maybe this will compensate.

1. There is no solution to the equation j² = -1 (mod p) if p = 3 (mod 4). Even this takes a bit of proving.

2. If a prime number p = 3 (mod 4) divides a number n = a² + b², then this prime number also divides both a and b.

The idea is that if the hypotenuse c is divisible by a prime number of the form 3 (mod 4), then such a number also divides the two legs a and b, so that the triad (a,b,c) cannot be a PPT.

[Guest]

I apologise for an earlier typo - maybe this will compensate.

1. There is no solution to the equation j² = -1 (mod p) if p = 3 (mod 4). Even this takes a bit of proving.

2. If a prime number p = 3 (mod 4) divides a number n = a² + b², then this prime number also divides both a and b.

The idea is that if the hypotenuse c is divisible by a prime number of the form 3 (mod 4), then such a number also divides the two legs a and b, so that the triad (a,b,c) cannot be a PPT.

I believe the only thing remaining is to show that c^{2[/sup can not have a factor p=3 (mod4) that is NOT prime.}

[Guest]

Is it not the case that any number which is 3 (mod 4) must have a prime factor 3 (mod 4)?