Does there exist a Pythagorean triple consisting only of prime numbers?
L Loren May 2015 516 32 Arlington, VA Aug 7, 2017 #1 Does there exist a Pythagorean triple consisting only of prime numbers?
P Petek Nov 2010 174 35 Berkeley, CA Aug 8, 2017 #2 Hint: At least one term of a Pythagorean triple must be even. Reactions: 3 people
D Denis Math Team Oct 2011 14,592 1,026 Ottawa Ontario, Canada Aug 8, 2017 #3 Petek said: Hint: At least one term of a Pythagorean triple must be even. Click to expand... .....and "2" is out!
Petek said: Hint: At least one term of a Pythagorean triple must be even. Click to expand... .....and "2" is out!
J JeffM1 May 2016 1,310 551 USA Aug 8, 2017 #4 Petek said: Hint: At least one term of a Pythagorean triple must be even. Click to expand... $k,\ m,\ n \in \mathbb N^+.$ $(2k- 1)^2 + (2m - 1)^2 = (2n - 1)^2 \implies$ $4k^2 - 4k + 1 + 4m^2 - 4m + 1 = 4n^2 - 4n + 1 \implies$ $1 + 1 - 1 = 1 = 4(n^2 - n - k^2 - m^2 + k + m) \implies$ $\dfrac{1}{4} = n^2 - n - k^2 - m^2 + k + m \implies$ $k \not \in \mathbb N^+ \text { or } m \not \in \mathbb N^+ \text { or } n \not \in \mathbb N^+.$
Petek said: Hint: At least one term of a Pythagorean triple must be even. Click to expand... $k,\ m,\ n \in \mathbb N^+.$ $(2k- 1)^2 + (2m - 1)^2 = (2n - 1)^2 \implies$ $4k^2 - 4k + 1 + 4m^2 - 4m + 1 = 4n^2 - 4n + 1 \implies$ $1 + 1 - 1 = 1 = 4(n^2 - n - k^2 - m^2 + k + m) \implies$ $\dfrac{1}{4} = n^2 - n - k^2 - m^2 + k + m \implies$ $k \not \in \mathbb N^+ \text { or } m \not \in \mathbb N^+ \text { or } n \not \in \mathbb N^+.$
P Petek Nov 2010 174 35 Berkeley, CA Aug 8, 2017 #5 JeffM1 said: $k,\ m,\ n \in \mathbb N^+.$ $(2k- 1)^2 + (2m - 1)^2 = (2n - 1)^2 \implies$ $4k^2 - 4k + 1 + 4m^2 - 4m + 1 = 4n^2 - 4n + 1 \implies$ $1 + 1 - 1 = 1 = 4(n^2 - n - k^2 - m^2 + k + m) \implies$ $\dfrac{1}{4} = n^2 - n - k^2 - m^2 + k + m \implies$ $k \not \in \mathbb N^+ \text { or } m \not \in \mathbb N^+ \text { or } n \not \in \mathbb N^+.$ Click to expand... 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.
JeffM1 said: $k,\ m,\ n \in \mathbb N^+.$ $(2k- 1)^2 + (2m - 1)^2 = (2n - 1)^2 \implies$ $4k^2 - 4k + 1 + 4m^2 - 4m + 1 = 4n^2 - 4n + 1 \implies$ $1 + 1 - 1 = 1 = 4(n^2 - n - k^2 - m^2 + k + m) \implies$ $\dfrac{1}{4} = n^2 - n - k^2 - m^2 + k + m \implies$ $k \not \in \mathbb N^+ \text { or } m \not \in \mathbb N^+ \text { or } n \not \in \mathbb N^+.$ Click to expand... 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.