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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)

 Jan 24, 2018
edited by Guest  Jan 24, 2018
edited by Guest  Jan 24, 2018
edited by Guest  Jan 24, 2018
 #1
avatar+33603 
+1

a = 6, b = 3, n = 9 results in 61965

 

9 is not prime.

 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+33603 
+2

So you did. However

 

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

 

You still need to tighten your constraints!

 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.

 Jan 24, 2018

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