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\(\frac{\left(\frac{\left(a^nb-ab^n\right)}{2n}\right)}{3\left(a+b\right)}\)

 

For a and b being integers, when the result of the expression above is an integer, and n is not equal to a + b, I perdict that:

 - n is prime;

 - n isn't 2.

 

Can you prove or counter my prediction?

 

(a, b, n having different names implies that they have different values)

Guest Jan 24, 2018
edited by Guest  Jan 24, 2018
edited by Guest  Jan 24, 2018
edited by Guest  Jan 24, 2018
 #1
avatar+26716 
+1

a = 6, b = 3, n = 9 results in 61965

 

9 is not prime.

Alan  Jan 24, 2018
edited by Alan  Jan 24, 2018
 #2
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I specified n is not equal to a + b...

9 = 6+3

 

OH!

ok

actually no.

a mustn't be the square of a common multiple of a and b

Guest Jan 24, 2018
edited by Guest  Jan 24, 2018
 #3
avatar+26716 
+2

So you did. However

 

a = 6, b = 3, n = 27 results in 2105947230095214567

 

You still need to tighten your constraints!

Alan  Jan 24, 2018
 #4
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That's false...?

Guest Jan 24, 2018
 #5
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What is false? The result obtained by Alan? His result is accurate, and is NOT a prime.

Guest Jan 24, 2018

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