1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

[solved] Is x^2+16 prime?

  1. Aug 3, 2015 #1

    JR Sauerland

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    x2+16


    2. Relevant equations
    ???

    3. The attempt at a solution
    If you attempt to solve it with (x+4)(x+4), it results in x2+8x+16, which is not equivalent. I believe it may be prime. I am looking for the formula (if there is any) to explain this. Allow me to give an example:
    a2+2ab+b2 <<< This is the square of a binomial formula. Does my type of problem have a formula that explains its nature? If it's prime, how would I phrase it? axn+b=???

    edit: back of the book states it is prime. Oops :p
     
    Last edited: Aug 3, 2015
  2. jcsd
  3. Aug 3, 2015 #2
    Well if x = 2k, with k some natural number then we get (2k)^2 +16. Is that prime?
     
  4. Aug 3, 2015 #3

    JR Sauerland

    User Avatar
    Gold Member

    I honestly have no idea :oops: 2k2+16... It seems like we can factor out a 2. 2(k2+8)?
     
  5. Aug 3, 2015 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Your question is unclear. I can see two possible meanings: (1) is ##x^2 + 16## a prime number when ##x## is an integer?
    (2) Is the polynomial ##x^2 + 16## prime (= "irreducible?") in the algebraic sense?

    In case (1) the answer is NO: ##p(x) = x^2 + 16## is the prime number 17 when ##x = 1##, but for ##x = 0, 2, 3, 4, \ldots## it is not prime. I suppose a valid question is whether ##p(x)## can ever be prime again for some integer ##x \geq 2##. I don't have the answer to that.

    In case (2) you need to specify the number field or ring over which you polynomials taken; that is, what types of numbers will you allow in a factorization? Your polynomial ##p(x)## is irreducible (="prime"?) over the integers, rationals or reals, but is reducible (not prime?) over the complex numbers.
     
  6. Aug 10, 2015 #5
    In my experience, case (2) is what this means. I much prefer the terminology irreducible over . . . rather than prime, but many textbooks, esp. high school level books prefer the word prime.

    Following back to JR Sauerland's original question, this is an interesting problem to consider tactics for proof. What techniques or methods can you use to argue that this cannot be reduced over the reals? There are several arguments that could be made. Feel free to demonstrate that this is not reducible.
     
  7. Aug 11, 2015 #6

    Mentallic

    User Avatar
    Homework Helper

    What are the roots of [itex]x^2+16[/itex]? In other words, what are the values of x (if there are any) such that [itex]x^2+16=0[/itex].

    If x=2k, then [itex]x^2=(2k)^2=2^2k^2=4k^2[/itex] and not [itex]2k^2[/itex]. Also, make sure to add parentheses where necessary, because [itex]2k^2[/itex] is different to [itex](2k)^2[/itex] since exponents have higher precedence than multiplying, so [itex]2k^2=2(k^2)[/itex]. Finally this means that you should be able to factor out a 4 in [itex](2k)^2+16[/itex] to get [itex]4(k^2+4)[/itex].

    But alas this isn't going to help you solve your problem because what you've asked can be mistaken for a harder problem. What you're looking for is to find if the polynomial is irreducible as has been already mentioned.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: [solved] Is x^2+16 prime?
  1. Solve for x: 2x-x^2=3 (Replies: 10)

Loading...