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Old 2014-08-07, 19:17   #12
R.D. Silverman
 
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Quote:
Originally Posted by CRGreathouse View Post
Matt, I don't see what you're getting at. Your essential claim is that is n^2 + n + 41 is composite, then ((some expression)) holds. But the expression does not involve n. Is that intentional, or was one of the variables in the expression supposed to be n?
Note that the Bunyanowski conjecture is merely one sub-case of
Schinzel's (and hence Bateman-Horn) conjecture.
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Old 2014-08-07, 19:26   #13
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Quote:
Originally Posted by CRGreathouse View Post
Maybe you meant for N to be n? In that case it's sensible to ask about, but I think there are probably algebraic factorizations for all n, even if n^2 + n + 41 is prime.
If Math would ever bother to learn some high school algebra, he would
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.
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Old 2014-08-07, 22:56   #14
CRGreathouse
 
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Quote:
Originally Posted by R.D. Silverman View Post
Note that the Bunyanowski conjecture is merely one sub-case of
Schinzel's (and hence Bateman-Horn) conjecture.
Yes. I keep a handy chart at the top of my page
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 Bateman-Horn-Stemmler; 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.
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Old 2014-08-08, 05:13   #15
LaurV
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Quote:
Originally Posted by R.D. Silverman View Post
If Math would ever bother to learn some high school algebra
Quote:
Originally Posted by R.D. Silverman
Hey Moron. Learn the difference between an error in grammar/spelling and a TYPO.
No, it wasn't a typo, you just shamelessly tried to steal my joke! (five posts before, where, in case some people here have their joke detector turned off, I wanted to say that Matt and Math are two different things... It seems no one got it...)

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.
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Old 2014-08-10, 04:43   #16
MattcAnderson
 
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"Matthew Anderson"
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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
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Old 2014-08-10, 11:33   #17
science_man_88
 
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Quote:
Originally Posted by MattcAnderson View Post
that is to say when n = z^2 + 40 for some integer z, then h(n) is composite.
I hope you don't mean you think z=1 doesn't destroy this polynomial.
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Old 2014-08-11, 09:35   #18
MattcAnderson
 
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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
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Old 2014-08-11, 11:52   #19
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Quote:
Originally Posted by MattcAnderson View Post
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
I realized this after posting, I don't know what I was thinking. I should hope I know about maple the OEIS has a line specifically for it ( not that I use it but I should have seen it enough by now).
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Old 2021-03-22, 15:04   #20
MattcAnderson
 
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Insects

https://en.wikipedia.org/wiki/Insect



I have dog
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Old 2021-04-08, 04:23   #21
MattcAnderson
 
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Although insect populations change, life finds a way.
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