Question 1208751
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Show that x^2 + 3 is prime.
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<pre>
The precise meaning of this question is dark and unclear.


Therefore, I will re-formulate the request, to make it sensible.


    Show that polynomial  x^2 + 3  is a prime polynomial 
    in the ring of polynomials with real coefficients.


The solution is simple and beautiful, both at the same time.


Had the polynomial be non-prime (= composite) in this ring, 
it would be a product of two polynomials of degree 1, i.e. linear binomials

    x^2 + 3 = (ax+b)*(cx+d).


In this case, the polynomial would have real roots  {{{-b/a}}}  and  {{{-d/c}}}.


But the polynomial  x^2 + 3  has always positive values for all real values of x,
since x^2 is always non-negative, while the addend "3" is a positive constant.


This contradiction proves that the given polynomial is a prime polynomial 
in the ring of polynomials with real coefficients.
</pre>

Solved.


The same reasoning works for polynomials with rational coefficients or/and for
polynomials with integer coefficients.


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I don't know what is your level in Math - therefore, I fully admit that you 
may understand nothing from my explanation.

Why I may think so - because the problem in your post is worded mathematically incorrectly.


On the other hand, since you brought this problem, I may assume that you have all 
the necessary pre-requisites to understand my solution.


The rest depends on you - not on me.


With the full context this problem is for the first year undergraduate university 
Math student, who recently started learning Abstract Algebra.


Such a student should understand my solution adequately.