Proving the primality of a quadratic over the natural numbers

[mathworker]

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]

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]

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]

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.

 

[mathworker]

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]

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]

Re: help!

but given function is not constant and i checked y =f(X) for some numbers and found it prime every time

$$3\cdot 5^2 + 3\cdot 5 + 1 = 91 = 7\cdot 13$$ this is the smallest non-prime value.

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

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

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

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

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

 

[Fernando Revilla]

Re: help!

i checked y =f(X) for some numbers and found it prime every time

For some, but not for all. :)

 

[mathworker]

Re: help!

For some, but not for all. :)

thank you for providing the theorem and quick reply...:D