20140807, 19:17  #12  
Nov 2003
1D24_{16} Posts 
Quote:
Schinzel's (and hence BatemanHorn) conjecture. 

20140807, 19:26  #13  
Nov 2003
2^{2}×5×373 Posts 
Quote:
quickly realize that his expression h(at^2 + bt + c) is a quartic polynomial and that one can find values of (a,b,c) such that this polynomial is reducible End of discussion. Charles is correct. There are algebraic factorizations. Enough of this! Why on Earth anyone should imagine that n^2+n+41 is somehow "special" in this regard is beyond me. THIS IS ALL HIGH SCHOOL LEVEL MATH. 

20140807, 22:56  #14  
Aug 2006
3^{2}×5×7×19 Posts 
Quote:
Table of special primes which lets you trace through the various names given to these sorts of conjectures. I prefer to call the last conjecture BatemanHornStemmler; there was a paper with the first two authors and a more substantial (IMO) paper with all three, both published in the same year, and it seems fair to not omit credit where due. 

20140808, 05:13  #15  
Romulan Interpreter
Jun 2011
Thailand
2·4,679 Posts 
Quote:
Quote:
And yes, I wanted to post directly in the "useless posts" thread, but it seems that thread is now locked... (Why?). Please a moderator moves the post there. 

20140810, 04:43  #16 
"Matthew Anderson"
Dec 2010
Oregon, USA
2^{3}·89 Posts 
Thank you for all your constructive feedback. I will try to answer.
Let h(n) = n^2 + n + 41. Now if 41 divides n, then 41 divides h(n). Furthur, if n is congruent to 0 or 40 mod 41 then 41 divides h(n). I know this because of an excel calculation. 40^2 + 40 + 41 mod 41 is congruent to 0. also 0^2 + 0 + 41 mod 41 is congruent to 0. This means that there are certain cases that cause h(n) to be composite. So far, if n is congruent to 0 or 40 then 41 divides n. The question of CRGreathouse can be answered with the concepts of substitution and algebraic factorization. let n(z) = 1*z^2 + 0 + 40. Then the composition of functions goes like this  h(n(z)) = (z^2 + 40)^2 + (z^2 + 40) + 41. This expands to a messy 4th order polynomial, and I use the Maple commands factor(subs(n=z^2 + 40, h)) Now, h(n(z)) = (z^2 + z + 41)*(z^2  z + 41). The point of this is because of this algebraic factorization, I now have two quadratic polynomials. I evaluate both at only integers. That is to say z can be in the integers only. Both quadratics are always positive and greater than one. Composite numbers are the product of two integers, both greater than one. Now I have found an infinite set of values (z^2 + 40, for all z) that make h(n) composite. To elaborate, h(n) is only evaluated at integer values of n. When n is restricted, that is to say when n = z^2 + 40 for some integer z, then h(n) is composite. Regards, Matt C. Anderson 8/9/2014 
20140810, 11:33  #17 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 

20140811, 09:35  #18 
"Matthew Anderson"
Dec 2010
Oregon, USA
2^{3}·89 Posts 
Hi Science_man_88,
Thank you for your kind comment. To answer your question, letting z = 1 does not destroy the polynomial. h(n) and n(z) are important to me because when there is the composition of functions, the resulting 4th order polynomial can be factored using Maple, a computer algebra system that I use. Further, since both factors are integers greater than one, the polynomial, when evaluated for a given z, gives a composite number. Cheers, Matt 
20140811, 11:52  #19  
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
Quote:


20210322, 15:04  #20 
"Matthew Anderson"
Dec 2010
Oregon, USA
2^{3}·89 Posts 

20210408, 04:23  #21 
"Matthew Anderson"
Dec 2010
Oregon, USA
1310_{8} Posts 
Although insect populations change, life finds a way.
