20170529, 04:48  #1 
May 2004
2^{2}·79 Posts 
A question of history
I may be wrong perhaps I am the first mathematician to discover the following property of polynomials:
Let f(x) be a polynomial in x ( x belongs to Z, can be a Gaussian integer, or be a square matrix in which the elements are rational integers or Gaussian integers). Then f(x + k*f(x)) = = 0 mod(f(x)). 
20170529, 07:36  #2 
Jun 2003
13FF_{16} Posts 
x+k*f(x) == x (mod f(x))
==> f(x+k*f(x)) == f(x) (mod f(x)). QED Looks like a trivial result. 
20170529, 08:02  #3 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
3·3,613 Posts 

20170530, 10:43  #4  
May 2004
2^{2}·79 Posts 
Quote:
Indirect primality testing. Let f(x) be a quadratic polynomial in x ( x belongs to Z). For example let f(x) be x^2 + x +1. All x, other than those generated by 1 + 3k, 2 + 7k, 3 + 13k, 4 + 3k and 4 + 7k, 5 + 31k.....are such that f(x) is prime which need not be tested for primality. Unfortunately this is true only upto quadratic level. 

20170530, 13:16  #5  
Feb 2017
Nowhere
3×7×229 Posts 
Quote:
A  B is an algebraic factor of f(A)  f(B). I imagine this has been known for centuries; I'm pretty sure Isaac Newton knew it, certainly for the cases where K is the rational or real numbers. Of course, the result continues to hold in cases where K is not a field, but I'm not sure offhand just how far you can push it. If K is a commutative ring (with 1) I don't see any reason it wouldn't work. In particular, substituting x + k*f(x) for A and x for B, k*f(x) is an algebraic factor of f(x + k*f(x))  f(x). 

20170601, 06:05  #6  
May 2004
2^{2}·79 Posts 
Quote:


20170601, 08:55  #7  
Mar 2016
353 Posts 
Quote:
i discovered it also, a little bit earlier than you, may be 10 years ago: see http://devalco.de/quadr_Sieb_x%5E2+1.php#1 , and i do not claim to be the first. But it is indeed a good basic idea for prime generators, if you add the proof : f(x  k*f(x)) = = 0 mod(f(x)) (or k element Z) you have a good criteria for prime generators. Have a look at http://devalco.de/#106 and you will discover a little bit more of prime numbers or prime generators in quadratic progression. Nice Greetings from the primes Bernhard 

20170601, 11:13  #8 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2A57_{16} Posts 
The result is so trivial that any selfrespecting mathematician would not even think of publishing it  especially so because it is incorrect as you first stated it (see my subsequent correction).

20170601, 13:15  #9  
Feb 2017
Nowhere
3×7×229 Posts 
Quote:
I do know that in high school algebra, one of the exercises for learning mathematical induction was to prove that, for any positive integer n, a  b divides a^n  b^n. And while I will not claim that back then we were doing our homework with a stylus on damp clay, I will say that it was quite a number of years ago. So a result of which (a corrected form of) the one you claim is a trivial corollary, was relegated to the exercises in high school algebra long since. No mathematician worthy of the name would presume to claim it as an original result. The result I mention is also often used to prove the formula for summing a geometric series. That's been known for a while, too. Last fiddled with by Dr Sardonicus on 20170601 at 13:15 

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