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How many integers n are there such that the quantity \(\lvert 2n^2 + 23n + 11 \rvert\) is prime?

 Dec 29, 2017
 #1
avatar+107156 
+2

Not totally sure about this...but.....here's my best attempt...

 

abs [ 2n^2  + 23n   + 11 ]   factor

 

abs(2n  + 1) * abs(n + 11)

 

Note that this will  be a possible prime  if either factor  =  ±1 

 

But 2n + 1  will = 1  only when  n  = 0.....and the other factor will = abs ( 11 )  = 11

So....when  n = 0, the result will be prime, i.e, 11

 

And 2n + 1  will equal  - 1  when  n  = -1.....and the other factor will =  abs(-1  + 11) = 10......but this isn't prime

 

And n + 11  will equal  1  when n  =  -10

And the other factor will  be 2(-10) + 1  =  -19   which is prime for abs (2n + 1) = abs (2*-10 + 1) =

abs(-19)  =  19

 

And  n + 11  will =  - 1  when n  =  -12  ....so abs (-12 + 11)  = abs(-1)  =  1

And the other factor will be  abs (2(-12) + 1)  =  abs (-23)  =  23

So.....this will be prime  when n   = -12

 

So.....the  integers producing prime results  for  abs [ 2n^2  + 23n   + 11 ]  are 

n = 0 , n  = -10 and n = -12

 

EDITED ANSWER.....still don't know if it's correct, or not....!!!!!!

 

 

cool cool cool

 Dec 29, 2017
edited by CPhill  Dec 30, 2017
 #2
avatar+878 
+1

hmm, i'm getting something different

 Dec 29, 2017
 #3
avatar+191 
+2

yes, that's correct @CPhill

 Dec 30, 2017
 #4
avatar+191 
+3

We first note that \(2n^2 + 23n + 11\) factors as \((2n + 1)(n+ 11)\) . (We can find these factors using the rational root theorem.) Thus we have \( \lvert 2n^2 + 23n + 11 \rvert = \lvert 2n + 1 \rvert \cdot \lvert n + 11 \rvert . \)Now, each of the factors on the right hand side of this equation is an integer. It follows that the left hand side is a prime number if and only if one of the right hand factors is 1 and the other one is a prime number. Thus we must either have \(2n + 1 = \pm 1\) , or \(n +11 = \pm 1\). We consider these cases separately.

If 2n+1, then n=0, and n+11=11, which is prime. Thus this value of n works.

If 2n+1=-1 , then n=-1, and n+11=10 , which is not prime. Therefore we have no solution in this case.

If n+11=1 , then n=-10 , so 2n+1=-19 . Since 19 is prime, we obtain a valid solution in this case.

Finally, if n+11=-1, then n=-12 , and 2n+1=-23. Since 23 is prime, this value of n works.

Thus there are exactly \(\boxed{3}\) values of \(n\) that work: 0, \(-10\), and \(-12\) ; and these give the prime numbers 11, 19, and 23.

 Dec 30, 2017
 #5
avatar+107156 
+1

Thanks, azsun......I  hope we're correct....LOL!!!

 

 

cool cool cool

 Dec 31, 2017

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