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 2017-05-29, 04:48 #1 devarajkandadai     May 2004 4748 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)).
 2017-05-29, 07:36 #2 axn     Jun 2003 26·34 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.
2017-05-29, 08:02   #3
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

2×37×149 Posts

Quote:
 Originally Posted by axn x+k*f(x) == x (mod f(x)) ==> f(x+k*f(x)) == f(x) (mod f(x)). QED Looks like a trivial result.
Except at roots of f(x).

2017-05-30, 10:43   #4

May 2004

22×79 Posts

Quote:
 Originally Posted by axn x+k*f(x) == x (mod f(x)) ==> f(x+k*f(x)) == f(x) (mod f(x)). QED Looks like a trivial result.
Agreed the result is not earth-shaking. However it has a couple of interesting applications, one of which I would like to mention in this post:

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.

2017-05-30, 13:16   #5
Dr Sardonicus

Feb 2017
Nowhere

2×33×5×19 Posts

Quote:
 Originally Posted by devarajkandadai 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)).
Since A - B is an algebraic factor of A^n - B^n for every non-negative integer n, we have that if f(x) is a polynomial in K[x], where K is a field, then

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).

2017-06-01, 06:05   #6

May 2004

4748 Posts

Quote:
 Originally Posted by Dr Sardonicus Since A - B is an algebraic factor of A^n - B^n for every non-negative integer n, we have that if f(x) is a polynomial in K[x], where K is a field, then 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).
Merely saying " I am pretty sure Isaac....." will not do; can you quote any paper or book where in either Newton, Euler or any mathematician has mentioned this result?

2017-06-01, 08:55   #7
bhelmes

Mar 2016

32×41 Posts

Quote:
 Originally Posted by devarajkandadai 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)).
You are not the first mathematician,
i discovered it also, a little bit earlier than you, may be 10 years ago:
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

2017-06-01, 11:13   #8
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

2B1216 Posts

Quote:
 Originally Posted by devarajkandadai Merely saying " I am pretty sure Isaac....." will not do; can you quote any paper or book where in either Newton, Euler or any mathematician has mentioned this result?
The result is so trivial that any self-respecting mathematician would not even think of publishing it --- especially so because it is incorrect as you first stated it (see my subsequent correction).

2017-06-01, 13:15   #9
Dr Sardonicus

Feb 2017
Nowhere

2·33·5·19 Posts

Quote:
 Originally Posted by devarajkandadai Merely saying " I am pretty sure Isaac....." will not do; can you quote any paper or book where in either Newton, Euler or any mathematician has mentioned this result?
This will not do. You are making a claim of priority. It is incumbent on you to check the literature. I suggested Newton, whose work with polynomials is well known, both WRT derivatives of powers and "Newton's identities," but you refused to look. It is reasonable to conclude that won't look because you're afraid of what you might find.

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 2017-06-01 at 13:15

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