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interests / rec.puzzles / Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2

SubjectAuthor
* Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 =henh...@gmail.com
`* Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2Eric Sosman
 `* Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2Richard Heathfield
  +* Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2Mike Terry
  |`- Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2Gareth Taylor
  +* Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2Eric Sosman
  |`- Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2Richard Heathfield
  `* Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2Richard Tobin
   `- Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2Richard Heathfield

1
Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2

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Subject: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 =
(integer) ^ 2
From: henha...@gmail.com (henh...@gmail.com)
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 by: henh...@gmail.com - Sat, 16 Jul 2022 14:10 UTC

(Sum of 2 squares)

Is there a Pythagorean triple of the following form ?

(odd) ^ 2 + (odd) ^ 2 === (integer) ^ 2

------- pls PLEASE wait a few days before posting answers or hints.

>>> There are 16 primitive Pythagorean triples of numbers up to 100:

(3, 4, 5) (5, 12, 13) (8, 15, 17) (7, 24, 25)
(20, 21, 29) (12, 35, 37) (9, 40, 41) (28, 45, 53)
(11, 60, 61) (16, 63, 65) (33, 56, 65) (48, 55, 73)
(13, 84, 85) (36, 77, 85) (39, 80, 89) (65, 72, 97)

Each of these points forms a radiating line in the scatter plot. Other small Pythagorean triples such as (6, 8, 10) are not listed because they are not primitive; for instance (6, 8, 10) is a multiple of (3, 4, 5).

---------------------- Can this passage above revised for better clarity or accuracy ?

Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2

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From: esos...@comcast-dot-net.invalid (Eric Sosman)
Newsgroups: rec.puzzles
Subject: Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2
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 by: Eric Sosman - Sat, 16 Jul 2022 14:27 UTC

On 7/16/2022 10:10 AM, henh...@gmail.com wrote:
> (Sum of 2 squares)
>
> Is there a Pythagorean triple of the following form ?
>
> (odd) ^ 2 + (odd) ^ 2 === (integer) ^ 2

Obviously not.

--
esosman@comcast-dot-net.invalid
Look on my code, ye Hackers, and guffaw!

Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2

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From: rjh...@cpax.org.uk (Richard Heathfield)
Newsgroups: rec.puzzles
Subject: Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2
Date: Sat, 16 Jul 2022 16:04:56 +0100
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 by: Richard Heathfield - Sat, 16 Jul 2022 15:04 UTC

On 16/07/2022 3:27 pm, Eric Sosman wrote:
> On 7/16/2022 10:10 AM, henh...@gmail.com wrote:
>> (Sum   of 2 squares)
>>
>> Is there  a Pythagorean triple   of the following  form ?
>>
>>                              (odd) ^ 2   +   (odd) ^ 2
>> ===   (integer)  ^ 2
>
> Obviously not.

Being no more mathematically sophisticated than Piglet, I ask
myself why it's obvious to you but not to me. The obvious reason
is of course that I am no more mathematically sophisticated than
Piglet, but is there a deeper reason? There must be.

It is obvious to me that a solution must be bigger than the
smallest Pythagorean triple 3, 4, 5.

Clearly the longest side must be the hypotenuse. Let us define a
and b as odd and c as even (a little thought about the sum of two
odd squares was enough to convince me that c is even).

So we have a^2 + b^2 = c^2

We know c must be > 4, because the smallest possible Pythagorean
triple isn't a solution.

That's as far as I got before wasting a good 30 minutes in
algebra and diagrams and tearing my hair out trying to see the
obvious... and failing.

Please understand, Eric, I don't doubt you for a moment. I'm sure
you're right. I'm just too dense to see the obvious.

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2

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From: news.dea...@darjeeling.plus.com (Mike Terry)
Date: Sat, 16 Jul 2022 16:28:07 +0100
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 by: Mike Terry - Sat, 16 Jul 2022 15:28 UTC

On 16/07/2022 16:04, Richard Heathfield wrote:
> On 16/07/2022 3:27 pm, Eric Sosman wrote:
>> On 7/16/2022 10:10 AM, henh...@gmail.com wrote:
>>> (Sum   of 2 squares)
>>>
>>> Is there  a Pythagorean triple   of the following  form ?
>>>
>>>                              (odd) ^ 2   +   (odd) ^ 2 ===   (integer)  ^ 2
>>
>> Obviously not.
>
> Being no more mathematically sophisticated than Piglet, I ask myself why it's obvious to you but not
> to me. The obvious reason is of course that I am no more mathematically sophisticated than Piglet,
> but is there a deeper reason? There must be.
>
> It is obvious to me that a solution must be bigger than the smallest Pythagorean triple 3, 4, 5.
>
> Clearly the longest side must be the hypotenuse. Let us define a and b as odd and c as even (a
> little thought about the sum of two odd squares was enough to convince me that c is even).
>
> So we have a^2 + b^2 = c^2
>
> We know c must be > 4, because the smallest possible Pythagorean triple isn't a solution.
>
> That's as far as I got before wasting a good 30 minutes in algebra and diagrams and tearing my hair
> out trying to see the obvious... and failing.
>
> Please understand, Eric, I don't doubt you for a moment. I'm sure you're right. I'm just too dense
> to see the obvious.
>
..
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Try looking at potential solutions modulo 4.

(Not that I'd say this is "obvious"...)

Regards,
Mike.

Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2

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From: esos...@comcast-dot-net.invalid (Eric Sosman)
Newsgroups: rec.puzzles
Subject: Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2
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 by: Eric Sosman - Sun, 17 Jul 2022 03:24 UTC

On 7/16/2022 11:04 AM, Richard Heathfield wrote:
> On 16/07/2022 3:27 pm, Eric Sosman wrote:
>> On 7/16/2022 10:10 AM, henh...@gmail.com wrote:
>>> (Sum   of 2 squares)
>>>
>>> Is there  a Pythagorean triple   of the following  form ?
>>>
>>>                              (odd) ^ 2   +   (odd) ^ 2 ===
>>> (integer)  ^ 2
>>
>> Obviously not.
>
> Being no more mathematically sophisticated than Piglet, I ask myself why
> it's obvious to you but not to me. The obvious reason is of course that
> I am no more mathematically sophisticated than Piglet, but is there a
> deeper reason? There must be.

Yes! The deeper reason is that my "obviously" sprang from an
argument that was both wrong and too embarrassing to be contained
in the margin of this Usenet. Here's a better (I hope) proof:

We want C^2 = A^2 + B^2, with A and B odd and A,B,C all positive
integers. Writing A=2a+1 and B=2b+1 (a,b integers), we get

C^2 = (2a+1)^2 + (2b+1)^2
= (4a^2 + 4a + 1) + (4b^2 + 4b + 1)
= 4(a^2 + a + b^2 + b) + 2
= 4(integer) + 2

C^2 is even so C is also even, so we write C=2c (c an integer):

C^2 = (2c)^2 = 4(c^2) = 4(integer) + 2

Dividing by 4, we find

c^2 = (integer) + 0.5

.... showing that c^2 is not an integer, hence neither is c,
hence C=2c cannot be even as required.

Now I'll return (still smarting from my original error) to helping
Piglet carry a balloon over to Eeyore's place. Hope I don't trip
and fall (again)!

--
esosman@comcast-dot-net.invalid
Look on my code, ye Hackers, and guffaw!

Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2

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From: rjh...@cpax.org.uk (Richard Heathfield)
Newsgroups: rec.puzzles
Subject: Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2
Date: Sun, 17 Jul 2022 08:15:55 +0100
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 by: Richard Heathfield - Sun, 17 Jul 2022 07:15 UTC

On 17/07/2022 4:24 am, Eric Sosman wrote:
> On 7/16/2022 11:04 AM, Richard Heathfield wrote:
>> On 16/07/2022 3:27 pm, Eric Sosman wrote:
>>> On 7/16/2022 10:10 AM, henh...@gmail.com wrote:
>>>> (Sum   of 2 squares)
>>>>
>>>> Is there  a Pythagorean triple   of the following  form ?
>>>>
>>>>                              (odd) ^ 2   +   (odd) ^ 2 ===
>>>> (integer)  ^ 2
>>>
>>> Obviously not.
>>
>> Being no more mathematically sophisticated than Piglet, I ask
>> myself why it's obvious to you but not to me. The obvious
>> reason is of course that I am no more mathematically
>> sophisticated than Piglet, but is there a deeper reason? There
>> must be.
>
> Yes!  The deeper reason is that my "obviously" sprang from an
> argument that was both wrong and too embarrassing to be contained
> in the margin of this Usenet.  Here's a better (I hope) proof:

It looks a lot like Wol's pencil. Hmm. Be that as it may,
although there's something terribly embarrassing about 'fessing
up there's something rather noble about it, too.

> Now I'll return (still smarting from my original error) to helping
> Piglet carry a balloon over to Eeyore's place.  Hope I don't trip
> and fall (again)!

At least you didn't try to initialise an array with = 0 when
writing a program to count the results of modulo 4 on perfect
squares. I thought by now gcc would have become used to my little
foibles and learned a little forgiveness.

(The result is always 0 for even squares and 1 for odd squares -
Richard's Law Of Bloody Obviousness.)

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2

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From: rich...@cogsci.ed.ac.uk (Richard Tobin)
Newsgroups: rec.puzzles
Subject: Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2
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 by: Richard Tobin - Sun, 17 Jul 2022 10:35 UTC

In article <taur4p$3e1ql$1@dont-email.me>,
Richard Heathfield <rjh@cpax.org.uk> wrote:

>> If you're familiar with the proof that the square root of 2 is
>> irrational, it may give you a hint.

>Familiar? No. We have met and shared a beer, but that is all.

The observation common to solving both problems is that an even square
is not merely even, but divisible by 4. And as you have now seen, the
sum of two odd squares has a remainder of 2 when divided by 4.

-- Richard

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Newsgroups: rec.puzzles
Subject: Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2
Date: Sun, 17 Jul 2022 12:11:52 +0100
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 by: Richard Heathfield - Sun, 17 Jul 2022 11:11 UTC

On 17/07/2022 11:35 am, Richard Tobin wrote:
> In article <taur4p$3e1ql$1@dont-email.me>,
> Richard Heathfield <rjh@cpax.org.uk> wrote:
>
>>> If you're familiar with the proof that the square root of 2 is
>>> irrational, it may give you a hint.
>
>> Familiar? No. We have met and shared a beer, but that is all.
>
> The observation common to solving both problems is that an even square
> is not merely even, but divisible by 4. And as you have now seen, the
> sum of two odd squares has a remainder of 2 when divided by 4.

But surely that's obvious?

;-)

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2

<Tgk*LQlTy@news.chiark.greenend.org.uk>

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https://www.novabbs.com/interests/article-flat.php?id=338&group=rec.puzzles#338

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From: gtay...@chiark.greenend.org.uk (Gareth Taylor)
Newsgroups: rec.puzzles
Subject: Re: Pythagorean triple : (odd) ^ 2 + (odd) ^ 2 = (integer) ^ 2
Date: 16 Jul 2022 17:43:01 +0100 (BST)
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References: <c50a7ae6-8e48-4542-9727-5d1a148b2e8en@googlegroups.com> <tauhsf$3d68m$1@dont-email.me> <tauk2p$3dcgm$1@dont-email.me> <CtCdnV_TS5qaRE__nZ2dnUU7-LPNnZ2d@brightview.co.uk>
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 by: Gareth Taylor - Sat, 16 Jul 2022 16:43 UTC

In article <CtCdnV_TS5qaRE__nZ2dnUU7-LPNnZ2d@brightview.co.uk>,
Mike Terry <news.dead.person.stones@darjeeling.plus.com> wrote:

> Try looking at potential solutions modulo 4.
>
> (Not that I'd say this is "obvious"...)

I don't think it's much of a leap to write the odd numbers as "2n+1" and
even numbers as "2n". So let's call our odd numbers 2a+1 and 2b+1. The
sum of their squares is immediately even, so is a square of an even
number, which we'll call 2c. We get

(2a+1)^2 + (2b+1)^2 = (2c)^2

(4a^2+4a+1) + (4b^2+4b+1) = 4c^2

4(a^2+a+b^2+b-c^2) = -2

which is silly because the right-hand side isn't a multiple of 4.

(Or: if we're thinking of our numbers modulo 2, let's think of their
squares modulo 2 squared.)

Gareth

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