## Re: Pythagorean Sets of Consecutive Primes

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• ... [Then I screwed up, sorry: I *should* have said:] and x is odd. I am pondering the unfortunate consequences.... David (feeling more odd then even :-)
Message 1 of 7 , Sep 1, 2011
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> > Puzzle:
> > If (a, b, c) and (b, c, d) are Pythagorean triples then is a=d?
>
> Suppose the converse. Then we have 4 possibilities:
> 1) a^2 = b^2 + c^2, d^2 = b^2 - c^2
> 2) a^2 = b^2 + c^2, d^2 = c^2 - b^2
> 3) a^2 = b^2 - c^2, d^2 = b^2 + c^2
> 4) a^2 = c^2 - b^2, d^2 = b^2 + c^2
> Hence there exists a pair of coprime integers (x,y)
> such that x^2+y^2 and x^2-y^2 are both odd squares.
> But each odd square is congruent to 1 modulo 8.
> Hence 2*x^2 = 2 mod 8

[Then I screwed up, sorry: I *should* have said:]

and x is odd.

I am pondering the unfortunate consequences....

David (feeling more odd then even :-)
• ... See http://www.mathmeth.com/tom/files/pyth.pdf which claims a proof by infinite descent that x^2+y^2 and x^2-y^2 cannot both be odd squares. David (who
Message 2 of 7 , Sep 1, 2011
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> > Puzzle:
> > If (a, b, c) and (b, c, d) are Pythagorean triples then is a=d?
> >
> > Suppose the converse. Then we have 4 possibilities:
> > 1) a^2 = b^2 + c^2, d^2 = b^2 - c^2
> > 2) a^2 = b^2 + c^2, d^2 = c^2 - b^2
> > 3) a^2 = b^2 - c^2, d^2 = b^2 + c^2
> > 4) a^2 = c^2 - b^2, d^2 = b^2 + c^2
> > Hence there exists a pair of coprime integers (x,y)
> > such that x^2+y^2 and x^2-y^2 are both odd squares.
> > But each odd square is congruent to 1 modulo 8.
> > Hence 2*x^2 = 2 mod 8
>
> [Then I screwed up, sorry

See
http://www.mathmeth.com/tom/files/pyth.pdf
which claims a proof by infinite descent that
x^2+y^2 and x^2-y^2 cannot both be odd squares.

David (who miserably failed to find a simpler proof)
• ... It was apparently harder than I thought. I didn t have a proof but guessed it would be relatively easy when there is a known parametrization of all
Message 3 of 7 , Sep 1, 2011
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David wrote:
> See
> http://www.mathmeth.com/tom/files/pyth.pdf
> which claims a proof by infinite descent that
> x^2+y^2 and x^2-y^2 cannot both be odd squares.
>
> David (who miserably failed to find a simpler proof)

It was apparently harder than I thought.
I didn't have a proof but guessed it would be relatively easy when
there is a known parametrization of all Pythagorean triples.
I only did a brute force search to see there were no small solutions.

--
Jens Kruse Andersen
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