For how many positive integers for k is the value of k^2 - 18k + 65 a positive prime number? 0, 1, 2, or 4?

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For how many positive integers for k is the value of k^2 - 18k + 65 a positive prime number? 0, 1, 2, or 4?

Mathematics
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see if you can factorise the expression:\[k^2-18k+65\]if you can then this implies it can only generate a prime number if one of the factors is equal to 1.
it factors into (x-5)(x-13) and how do you know that it can only generate a prime number if its 1? Oh! Any integer squared is automatically composite unless its 1. So we try 1 and -1 and we get 48 and 84 respectively. So, the answer is 0? But even if something is squared and it is composite, it could still be changed into prime by the -18 or the + 65
Well, not by the -18, but it could indirectly help the 65 make f(k) prime.

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Other answers:

  • phi
In general x*y is composite (by definition) unless x=1 and y is prime or y=1 and x is prime. for your case (with integer k>0) (k-5)(k-13) we ask can (k-5) = 1 and (k-13) be prime and vice versa for k=4, we get -1* -9 = 9 not prime k=6 1* -7 negative but the question asks for positive primes k= 12 7 * -1 neg k=14 9 * 1 not prime
  • phi
It looks like zero primes
How come we want to make the x-intercepts = 1? Doesnt that leave out a whole bunch of other numbers that we could try into the original equation?
the answer is 0 but i dont understand how you arrived at that
  • phi
Do you agree with this statement: x*y is composite (by definition) unless x=1 and y is prime or y=1 and x is prime.
  • phi
k^2 - 18 k + 65 = (k-5)(k-13)
Yes, because after the multiplication, that product would be divisible by both x and y
  • phi
so x=(k-5) and y= (k-13)
?
  • phi
x*y is composite unless.... (k-5)*(k-13) is composite unless....
Unless one of those equals 1! Oh. I see now. Goodness I can be a real stonehead sometimes heheh. Thanks
  • phi
I was getting nervous, cause I do not know how to say it any other way

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