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neat_maths
Member since Aug2203

May1906, 06:46 AM (EST) 

"Sum of 3 mutually prime numbers"

It'seems possible to prove that the sum of 3 mutually prime numbers x, y and z cannot contain all the primes contained in each of the pairs (x+y), (y+z) and (x+z). For example x=3, y=4 and z=5 gives (x+y+z)=12 which contains the prime factors in (x+z)=8, (y+z)=9 but not (x+y)=7Can anyone find a counterexample? take care 

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sfwc
Member since Jun1903

May2206, 08:07 AM (EST) 

3. "RE: Sum of 3 mutually prime numbers"
In response to message #2

x = 2 y = 3 z = 5>Good, >Now lets exclude 1, 0 and 1 This seems like rather arbitrary condition. You have said it 'seems possible to prove' this result. Does this mean that you have an idea for how to prove it? If so, please post that idea here. If not, it is unclear why you imposed this condition; if you wish to impose further conditions please give your reasoning. Thankyou sfwc <>< 

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sfwc
Member since Jun1903

May2206, 06:10 PM (EST) 

5. "RE: Sum of 3 mutually prime numbers"
In response to message #4

x = 3 y = 5 z = 22>if we exclude 1, then the choice of x=2 is not allowed, >because a prime number is defined as a positive integer > 1 I am not sure exactly what you mean by this. Please explain what you understand the condition 'exclude 1, 0 and 1' to mean. Thankyou sfwc <>< 

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neat_maths
Member since Aug2203

May2206, 08:24 PM (EST) 

6. "RE: Sum of 3 mutually prime numbers"
In response to message #5

Thank you sfwc, x=3, y=5 and z=22 is a good counter example. Back to the drawing board!Can we change the question to the sum of 3 mutually prime integers greater than 1 containing all the primes in each of the 3 numbers? For example x=3, y=5 and z=22 their sum (30) contains 2, 3 and 5 but not 11 Take care


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