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The pairs $6h-1$ and $6h+1$ are not twin primes, that is at least one of them is factorizable/decomposable, where $h$ can be found from any of the following equations:

$$ \begin{align} h = 6t_1t_2 +7t_1+5t_2+6\;,\\ h = 6t_1t_2 +5t_1+5t_2+4\;,\\ h = 6t_1t_2 +7t_1+7t_2+8\;.\\ \end{align} $$

Here $h$ is the third integer coordinate of the points of the paraboloids obtained by integer values of $t_1$ and $t_2$, where $t_1$ and $t_2$ belongs to $N\cup\{0\}$.

Now, can you explain how to support the statement cited above, if I am correct. Or else, let me know where I am wrong.

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What is N?$\text{}$ – Fredrik Meyer Oct 31 '12 at 7:55
@ N is set of Natural numbers. N = {1, 2, 3, 4, ...} – vmrfdu123456 Oct 31 '12 at 8:28
@N union 0 is = {1,2,3,...} U {0} – vmrfdu123456 Oct 31 '12 at 8:28

I'll use $x,y$ for your $t_1,t_2$, and $h_1,h_2,h_3$ for your three formulas for $h$.

Now if $(6k-1,6k+1)$ is a pair of twin primes, then the product $36k^2-1$ should have only two prime factors. If the algebraic factorizations of these turn out to factor into three (or more) terms all clearly greater than 1, then you have what you want.

We have $36h_1^2-1=(5+6x)(7+6y)(36xy+42x+30y+37)$. So $h_1$ does what you want.

Also $36h_2^2-1=(5+6x)(5+6y)(36xy+30x+30y+23)$. So $h_2$ does what you want.

I'll leave you to do $h_3$...

EDIT: We can find other examples of such $h$ by the following method. Choose $a,b$ such that $ab=6k+1$ [$6k-1$ would also work]. Now compute $(6x+a)(6y+b)=6(6xy+bx+ay+k)+1.$ So if we define $h=6xy+bx+ay+k$ we will have $6h+1=(6x+a)(6y+b)$ and $6h-1=36xy+6bx+6ay+6k-1$. Thus with positive $x,y$ we can't have $6h+1$ and $6h-1$ both prime, since here we have in fact $6h+1=(6x+a)(6y+b)$ which is not prime.

Added note: In the above I assumed $x,y>0$. If one allows negative $x,y$ there may be a way to make one of the factors 1, while the other two are integer primes, by the latter meaning we allow $-2,-3,-5,...$ also as primes.

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! where you got (5+6x)(7+6y)(36xy+42x+30y+37)= 36h^2-1? – vmrfdu123456 Oct 31 '12 at 10:55
I put the thing into maple to do the factorization. – coffeemath Oct 31 '12 at 11:01
Actually we have $6h_1-1=(6x+5)(6y+7)$ explaining two factors of $36h_1^2-1$. And $6h_2+1=(6x+5)(6y+5)$, and for $h_3$ we have $6h_3+1=(6x+7)(6y+7)$. I think this "explains" the factorizations in a simpler way than using maple. – coffeemath Oct 31 '12 at 12:02
! Sorry for late response. Thank you for your quick reply and good explanation. Thanks a lot. – vmrfdu123456 Nov 1 '12 at 6:11

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