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 August 7th, 2017, 10:15 PM #1 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 281 Thanks: 24 Math Focus: Number theory All prime Pythagorean triple Does there exist a Pythagorean triple consisting only of prime numbers?
 August 8th, 2017, 01:27 AM #2 Senior Member   Joined: Nov 2010 From: Berkeley, CA Posts: 174 Thanks: 35 Math Focus: Elementary Number Theory, Algebraic NT, Analytic NT Hint: At least one term of a Pythagorean triple must be even. Thanks from Denis, Loren and JeffM1
August 8th, 2017, 08:12 AM   #3
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Quote:
 Originally Posted by Petek Hint: At least one term of a Pythagorean triple must be even.
.....and "2" is out!

August 8th, 2017, 10:11 AM   #4
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Quote:
 Originally Posted by Petek Hint: At least one term of a Pythagorean triple must be even.
$k,\ m,\ n \in \mathbb N^+.$

$(2k- 1)^2 + (2m - 1)^2 = (2n - 1)^2 \implies$

$4k^2 - 4k + 1 + 4m^2 - 4m + 1 = 4n^2 - 4n + 1 \implies$

$1 + 1 - 1 = 1 = 4(n^2 - n - k^2 - m^2 + k + m) \implies$

$\dfrac{1}{4} = n^2 - n - k^2 - m^2 + k + m \implies$

$k \not \in \mathbb N^+ \text { or } m \not \in \mathbb N^+ \text { or } n \not \in \mathbb N^+.$

August 8th, 2017, 08:23 PM   #5
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Quote:
 Originally Posted by JeffM1 $k,\ m,\ n \in \mathbb N^+.$ $(2k- 1)^2 + (2m - 1)^2 = (2n - 1)^2 \implies$ $4k^2 - 4k + 1 + 4m^2 - 4m + 1 = 4n^2 - 4n + 1 \implies$ $1 + 1 - 1 = 1 = 4(n^2 - n - k^2 - m^2 + k + m) \implies$ $\dfrac{1}{4} = n^2 - n - k^2 - m^2 + k + m \implies$ $k \not \in \mathbb N^+ \text { or } m \not \in \mathbb N^+ \text { or } n \not \in \mathbb N^+.$
This argument appears to be correct. However, more simply, if {a, b, c} is a Pythagorean triple all of whose terms are odd, then $a^2+b^2 = c^2$ implies that odd + odd = odd, an obvious contradiction.

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