SOLUTION: Show that x^2 + 4 is prime.

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Question 1207338: Show that x^2 + 4 is prime.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52799)   (Show Source): You can put this solution on YOUR website!
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Show that x^2 + 4 is prime.
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If the given polynomial  be a composite polynomial, it would be a product of linear polynomials

     = (ax+b)*(cx+d).


In this case, it would have two roots,    and  .


But from the other side, this polynomial    is always positive over real numbers and has no real roots.


This contradiction PROVES that the given polynomial is a PRIME polynomial.


We proved that this polynomial is prime not only over the ring of integer numbers, 
but even over the much wider field of real numbers (!).

Solved.



Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Let's look at an example or two before diving into this current question.


Example 1: Find the factorization of x^2+5x+6 over the real numbers.
Solution: (x+3)(x+2)
We can use trial-and-error to find the factors. We're looking for two numbers that add to 5 and multiply to 6.
This is because (x+p)(x+q) = x^2+(p+q)x+p*q has the x coefficient p+q while the constant term is p*q.
Since x^2+5x+6 factors to (x+3)(x+2), it is not prime over the real numbers.
A prime polynomial only has factors of 1 and itself.


Example 2: is x^2+7x+10 prime over the set of real numbers? If no, then what is the factorization?
Solution: No it is not prime. It factors to (x+5)(x+2)
Again we could use trial-and-error to find the factors.
Note how 2+5 = 7 and 2*5 = 10.
Another approach that is more methodical without guess-and-check is to use the quadratic formula.
Use that formula to solve x^2+7x+10 = 0 for x. I'll skip steps and leave them for the student to do.
The two solutions are x = -5 and x = -2.
x = -5 leads to x+5 = 0 which gives the factor x+5
x = -2 leads to x+2 = 0 which gives the factor x+2


Example 3: Is x^2-7 prime over the rational numbers? If no, then what is the factorization?
Solution: Yes it is prime over the rational numbers.
Solve x^2-7 = 0 to get x = sqrt(7) and x = -sqrt(7)
We arrive at factors ( x-sqrt(7) ) and ( x+sqrt(7) ), but sqrt(7) isn't in the set of rational numbers.
Therefore we conclude that x^2-7 is prime over the rational numbers. However, it is not prime over the real numbers.

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Now onto your current question.
We want to show that x^2+4 is prime.
This portion "is prime" is a bit vague since we haven't defined what set of numbers we want to work with.
I'll assume your textbook is implying "is prime over the real numbers".

Solving x^2+4 = 0 leads to x = 2i and x = -2i where i = sqrt(-1) is an imaginary number.
This means x-2i and x+2i are the factors.
But these factors are not purely real numbers.

If your teacher asked you to factor x^2+4 over the complex numbers, then x^2+4 = (x-2i)(x+2i), which shows x^2+4 is not prime over the complex numbers.

But it is prime over the real numbers since the roots to x^2+4 = 0 are not in the set of real numbers.
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