# Number TheoryProving the primality of a quadratic over the natural numbers

#### mathworker

##### Well-known member
is there any way to prove or disprove the statement:
y=3x^2+3x+1 is prime for all x belongs to natural numbers...

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Re: help!!

is there any way to prove or disprove the statement:
y=3x^2+3x+1 is prime for all x belongs to natural numbers...
Yes, use the following theorem (Goldbach 1752):

If $f\in\mathbb{Z}[x]$ has the property that $f(n)$ is prime for all $n\ge 1$, then $f$ must be a constant.

#### chisigma

##### Well-known member
Re: help!!

is there any way to prove or disprove the statement:
y=3x^2+3x+1 is prime for all x belongs to natural numbers...
Is...

$$y = 3\ (x-x_{1})\ (x-x_{2})\ (1)$$

... where $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2} + x + \frac{1}{3} = 0$. The discriminat is <0, so that no real roots exists and that means that no real factor of y in (1) exists...

Kind regards

$\chi$ $\sigma$

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Re: help!!

Is...

$$y = 3\ (x-x_{1})\ (x-x_{2})\ (1)$$

... where $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2} + x + \frac{1}{3} = 0$. The discriminat is <0, so that no real roots exists and that means that no real factor of y in (1) exists...

Kind regards

$\chi$ $\sigma$
I don't undesrtand. We have $y(x)=3(x-a)(x-\bar{a})$ with $a=(-3+\sqrt{3}i)/6$, however $y(n)=3(n-a)(n-\bar{a})$ could be composite or prime.

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#### mathworker

##### Well-known member
Re: help!!

Is...

$$y = 3\ (x-x_{1})\ (x-x_{2})\ (1)$$

... where $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2} + x + \frac{1}{3} = 0$. The discriminant is <0, so that no real roots exists and that means that no real factor of y in (1) exists...

Kind regards

$\chi$ $\sigma$
but if we consider x^2+x+1=y whose discriminant too is less than zero..
for x=4 it gives 21 which is not prime... #### mathworker

##### Well-known member
Re: help!!

Yes, use the following theorem (Goldbach 1752):

If $f\in\mathbb{Z}[x]$ has the property that $f(n)$ is prime for all $n\ge 1$, then $f$ must be a constant.
but given function is not constant and i checked y =f(X) for some numbers and found it prime every time

#### PaulRS

##### Member
Re: help!!

but given function is not constant and i checked y =f(X) for some numbers and found it prime every time
$$\displaystyle 3\cdot 5^2 + 3\cdot 5 + 1 = 91 = 7\cdot 13$$ this is the smallest non-prime value.

Nonconstant polynomials can't be forever prime, indeed suppose $$\displaystyle f(x)=a_n\cdot x^n +\ldots + a_1\cdot x^1 + a_0$$

Then imagine it were always prime, and then $$\displaystyle f(x)-f(y)=a_n\cdot (x^n-y^n)+\ldots + a_1\cdot (x-y)$$ right?

Well, clearly $$\displaystyle x-y$$ divides $$\displaystyle x^n-y^n=(x-y)\cdot \left(x^{n-1}+x^{n-2}y+\ldots +x y^{n-2}+y^{n-1}\right)$$ for each $$\displaystyle n$$, and so $$\displaystyle (x-y)|(f(x)-f(y))$$

Thus if $$\displaystyle p$$ were a prime dividing $$\displaystyle f(r)$$ for some $$\displaystyle r\geq 1$$, we have $$\displaystyle \left( (k\cdot p + r) - r\right) | \left( f(k\cdot p + r) - f(r) \right)$$ for each $$\displaystyle k$$ and so, since $$\displaystyle p|f(r)$$, it follows that $$\displaystyle p | f(p\cdot k+r)$$ for each $$\displaystyle k\geq 1$$.
But $$\displaystyle f(p\cdot k+r)$$ for $$\displaystyle k\geq 1$$ can't all equal $$\displaystyle p$$, since $$\displaystyle f$$ would otherwise have to be constant. Hence some $$\displaystyle f(k\cdot p + r)$$ must be composite.

So with this construction in mind we could note that $$\displaystyle f(1)=7$$, and so $$\displaystyle f(7\cdot k + 1)$$ is nonprime for some $$\displaystyle k\geq 1$$, check for instance $$\displaystyle f(8)=217$$ which, as we saw, must be a multiple of $$\displaystyle 7$$.

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Re: help!!

i checked y =f(X) for some numbers and found it prime every time
For some, but not for all. #### mathworker

##### Well-known member
Re: help!!

For some, but not for all. thank you for providing the theorem and quick reply... 