What is the product of the two integer values for x for which |x^2 - 16| is a prime number?

cormier123 Jun 13, 2018

#1**0 **

The two numbers can be +3 and -5. Also -9 and -11 will work. There maybe an infinite number of them.

abs[3^2 - 16] = 7 which is a prime number.

abs[-5^2 - 16] =41 which is a prime number.

+3 x -5 =-15

Guest Jun 13, 2018

edited by
Guest
Jun 13, 2018

#2**0 **

No, there is only a finite number of those numbers.

proof: suppose |x^{2}-16| is a prime number. |x^{2}-16|=|x-4|*|x+4|, so one of the multiplied numbers=1 (because the product is a prime number).

|x-4|=1 means x=5 or x=3, and |x+4|=1 means x=-5 or x=-3. if x=5 then |x+4|=9, and because 9 is not a prime number (9=3*3) x cannot be equal to 5.

if x=-5 then |x-4|=9, so x cannot be equal to -5.

so if |x^{2}-16| is a prime number, then x=3 or x=-3. the product of the solutions is 3*-3=-9, and that's the answer.

Guest Jun 13, 2018

#3**0 **

Do you mean: abs[-5^2 - 16] =9? If so, why is W/A giving 41 as the answer? This is how they come to that answer:

Simplify the following:

abs(-5^2 - 16)

Evaluate 5^2.

5^2 = 25:

abs(-25 - 16)

Evaluate -25 - 16.

-25 - 16 = -41:

abs(-41)

Absolute value gives the distance to the origin.

Since -41<=0, then abs(-41) = 41:

**=41**

**https://www.wolframalpha.com/input/?i=abs%5B-5%5E2+-+16%5D+%3D%3F**

Guest Jun 13, 2018

edited by
Guest
Jun 13, 2018