There are 16 primitive Pythagorean triples of numbers up to 100:
(Sum of 2 squares)
Is there a Pythagorean triple of the following form ?
(odd) ^ 2 + (odd) ^ 2 === (integer) ^ 2
On 7/16/2022 10:10 AM, henh...@gmail.com wrote:
(SumÂ Â of 2 squares)
Is thereÂ a Pythagorean tripleÂ Â of the followingÂ form ?
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (odd) ^ 2Â Â +Â Â (odd) ^ 2
===Â Â (integer)Â ^ 2
Obviously not.
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)Obviously not.
Is there a Pythagorean triple of the following form ?
(odd) ^ 2 + (odd) ^ 2 === (integer) ^ 2 >>
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.
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.
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).
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?"
On 16/07/2022 18:05, Richard Heathfield wrote:
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!"
On 16/07/2022 3:27 pm, Eric Sosman wrote:
On 7/16/2022 10:10 AM, henh...@gmail.com wrote:
(SumÂ Â of 2 squares)
Is thereÂ a Pythagorean tripleÂ Â of the followingÂ form ?
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (odd) ^ 2Â Â +Â Â (odd) ^ 2 ===
(integer)Â ^ 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.
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:
(SumÂ Â of 2 squares)
Is thereÂ a Pythagorean tripleÂ Â of the followingÂ form ?
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (odd) ^ 2Â Â +Â Â (odd) ^ 2 ===
(integer)Â ^ 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!Â 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.Â Here's a better (I hope) proof:
Now I'll return (still smarting from my original error) to helping
Piglet carry a balloon over to Eeyore's place.Â Hope I don't trip
and fall (again)!
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.
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.
Try looking at potential solutions modulo 4.
(Not that I'd say this is "obvious"...)
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