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August 7th, 2017, 09:15 PM   #1
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All prime Pythagorean triple

Does there exist a Pythagorean triple consisting only of prime numbers?
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August 8th, 2017, 12:27 AM   #2
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Hint: At least one term of a Pythagorean triple must be even.
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August 8th, 2017, 07:12 AM   #3
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Quote:
Originally Posted by Petek View Post
Hint: At least one term of a Pythagorean triple must be even.
.....and "2" is out!
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August 8th, 2017, 09:11 AM   #4
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Hint: At least one term of a Pythagorean triple must be even.
$k,\ m,\ n \in \mathbb N^+.$

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

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

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

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

$k \not \in \mathbb N^+ \text { or } m \not \in \mathbb N^+ \text { or } n \not \in \mathbb N^+.$
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August 8th, 2017, 07:23 PM   #5
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Quote:
Originally Posted by JeffM1 View Post
$k,\ m,\ n \in \mathbb N^+.$

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

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

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

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

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