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

    From henhanna@gmail.com@21:1/5 to All on Sat Jul 16 07:10:47 2022
    (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 ?

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  • From Eric Sosman@21:1/5 to henh...@gmail.com on Sat Jul 16 10:27:29 2022
    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!

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Heathfield@21:1/5 to Eric Sosman on Sat Jul 16 16:04:56 2022
    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

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  • From Mike Terry@21:1/5 to Richard Heathfield on Sat Jul 16 16:28:07 2022
    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.

    .
    spoiler space
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    Try looking at potential solutions modulo 4.

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


    Regards,
    Mike.

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  • From Richard Heathfield@21:1/5 to Richard Tobin on Sat Jul 16 18:05:29 2022
    On 16/07/2022 5:13 pm, Richard Tobin wrote:
    In article <tauk2p$3dcgm$1@dont-email.me>,
    Richard Heathfield <rjh@cpax.org.uk> wrote:

    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).

    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 mod 4 hint from Mike made me wonder.

    Firstly, if c is even, then c = 2d + 1

    (2d+1)^2 = (2d+1)(2d+1) = 2d(2d+1) + 1(2d+1)

    = 4d^2+2d+2d+1=4d^2+4d+1 = 4(d^2+1) + 1.

    We can cast out d^2+1 4s, leaving 1 for all even c.

    Eeyore (who has never trusted the modulo operation) was
    sceptical, so I asked Winnie Thur Pooh to write a C program to
    prove it to him, which he did, compiling somewhat embarrassingly
    on the second try.

    Since a^2 must be odd, (a^2)%4 must be 1 or 3.

    Since b^2 must be odd, (b^2)%4 must be 1 or 3.

    Spelling it out for Eeyore:

    1+1 mod 4 is 2
    1+3 mod 4 is 0
    3+1 mod 4 is 0
    3+3 mod 4 is 2

    When Christopher Robin saw that none of these can be 1, he yelled
    "QED", which, as Wol explained to me, means "I know more Latin
    than you".

    I showed the proof to Piglet, and he sniffed and said "Well yeah;
    it's obvious, innit?"

    --
    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

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Tobin@21:1/5 to rjh@cpax.org.uk on Sat Jul 16 16:13:01 2022
    In article <tauk2p$3dcgm$1@dont-email.me>,
    Richard Heathfield <rjh@cpax.org.uk> wrote:

    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).

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

    -- Richard

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Mike Terry@21:1/5 to Richard Heathfield on Sat Jul 16 18:45:20 2022
    On 16/07/2022 18:05, Richard Heathfield wrote:
    On 16/07/2022 5:13 pm, Richard Tobin wrote:
    In article <tauk2p$3dcgm$1@dont-email.me>,
    Richard Heathfield <rjh@cpax.org.uk> wrote:

    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).

    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 mod 4 hint from Mike made me wonder.

    Firstly, if c is even, then c = 2d + 1

    (2d+1)^2 = (2d+1)(2d+1) = 2d(2d+1) + 1(2d+1)

    = 4d^2+2d+2d+1=4d^2+4d+1 = 4(d^2+1) + 1.

    We can cast out d^2+1 4s, leaving 1 for all even c.

    Eeyore (who has never trusted the modulo operation) was sceptical, so I asked Winnie Thur Pooh to
    write a C program to prove it to him, which he did, compiling somewhat embarrassingly on the second
    try.

    Since a^2 must be odd, (a^2)%4 must be 1 or 3.

    Since b^2 must be odd, (b^2)%4 must be 1 or 3.

    Spelling it out for Eeyore:

    1+1 mod 4 is 2
    1+3 mod 4 is 0
    3+1 mod 4 is 0
    3+3 mod 4 is 2

    When Christopher Robin saw that none of these can be 1, he yelled "QED", which, as Wol explained to
    me, means "I know more Latin than you".

    I showed the proof to Piglet, and he sniffed and said "Well yeah; it's obvious, innit?"


    Well, if I had to grade your effort above, I'd have to say "You're playing ALL the right notes, but
    not necessarily in the right order!"

    <https://www.dailymotion.com/video/x4pl0gr>

    (The first problem is c = 2d + 1 is ODD not even... but seriously, if you shuffled the lines around,
    and switched some c's with a's or b's, maybe swap some odd's and even's you've pretty much got it! :) )


    Mike.

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  • From Richard Heathfield@21:1/5 to Mike Terry on Sat Jul 16 18:55:17 2022
    On 16/07/2022 6:45 pm, Mike Terry wrote:
    On 16/07/2022 18:05, Richard Heathfield wrote:

    <snip>

    I showed the proof to Piglet, and he sniffed and said "Well
    yeah; it's obvious, innit?"


    Well, if I had to grade your effort above, I'd have to say
    "You're playing ALL the right notes, but not necessarily in the
    right order!"

    Sirrah, I was not expecting to laugh out loud tonight, but you
    have confounded my expectations.

    How can I break this to Piglet?

    --
    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

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Eric Sosman@21:1/5 to Richard Heathfield on Sat Jul 16 23:24:09 2022
    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!

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Heathfield@21:1/5 to Eric Sosman on Sun Jul 17 08:15:55 2022
    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

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Tobin@21:1/5 to rjh@cpax.org.uk on Sun Jul 17 10:35:31 2022
    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

    --- SoupGate-Win32 v1.05
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  • From Richard Heathfield@21:1/5 to Richard Tobin on Sun Jul 17 12:11:52 2022
    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

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Gareth Taylor@21:1/5 to news.dead.person.stones@darjeeling. on Sat Jul 16 17:43:01 2022
    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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