# [solved] Is x^2+16 prime?

1. Aug 3, 2015

### JR Sauerland

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. Aug 3, 2015

### dirk_mec1

Well if x = 2k, with k some natural number then we get (2k)^2 +16. Is that prime?

3. Aug 3, 2015

### JR Sauerland

I honestly have no idea 2k2+16... It seems like we can factor out a 2. 2(k2+8)?

4. Aug 3, 2015

### Ray Vickson

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.

5. Aug 10, 2015

### thelema418

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.

6. Aug 11, 2015

### Mentallic

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

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

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.