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Number Theory Proving the primality of a quadratic over the natural numbers

mathworker

Active member
May 31, 2013
118
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
Jan 29, 2012
661
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
Feb 13, 2012
1,704
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
Jan 29, 2012
661
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.
 
Last edited:

mathworker

Active member
May 31, 2013
118
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...:confused:
 

mathworker

Active member
May 31, 2013
118
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
Jan 26, 2012
37
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
Jan 29, 2012
661

mathworker

Active member
May 31, 2013
118