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Old 2018-12-05, 02:59   #1
sweety439
 
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Nov 2016

42668 Posts
Default Schinzel's hypothesis H

For any polynomial f(x) in one variable with positive degree and integer coefficients:

f(x) = a_nx^n+a_{n-1)x^{n-1}+a_{n-2)x^{n-2}+...+a_2x^2+a_1x+a_0

with n>=1

Like The Sierpinski/Riesel-like problem, if the values of the polynomial f(x) have a single trivial factor for all x-values, then we take out this factor (e.g. x^2+x+2 become {1/2*x^2+1/2*x+1}, since all values of this polynomial are divisible by 2), besides, if f(x) is not irreducible over the integers, then we must factorize f(x) (e.g. x^2-1 become {x-1,x+1}, since x^2-1=(x-1)*(x+1)).

Here are some examples:

Code:
f(x)      the polynomials
x^4       {x,x,x,x}
x^2+1     {x^2+1}
x^2-1     {x-1,x+1}
x^3+1     {x+1,x^2-x+1}
x^4+1     {x^4+1}
x^4-1     {x-1,x+1,x^2+1}
x^2+x     {1/2*x,x+1} for even x, {x,1/2*x+1/2} for odd x
x^2+x+2   {1/2*x^2+1/2*x+1}
x^2-x-1   {x^2-x-1}
x^2-x-6   {x-3,1/2*x+1} for even x, {1/2*x-3/2,x+2} for odd x
x^3+2*x+9 {1/3*x^3+2/3*x+3}
x^5+x+1   {x^2+x+1,x^3-x^2+1}
This conjecture is for any f(x), there are infinitely many x such that all the polynomials take on prime values simultaneously.

Note: We also consider the negative primes (e.g. -2, -3, -5, -7, -11, ...) as primes, since they are primes in the ring Z.

Last fiddled with by sweety439 on 2018-12-05 at 03:33
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Old 2018-12-05, 03:07   #2
sweety439
 
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We take the corresponding from positive integers to integer-values polynomials:

if N=2^{a_0}*3^{a_1}*5^{a_2}*7^{a_3}*11^{a_4}*...

then we take b_k=A001057(a_k) for all k

and set the corresponding polynomial of N is

b_0+b_1x+b_2x^2+b_3x^3+b_4x^4+...

Last fiddled with by sweety439 on 2018-12-05 at 03:07
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Old 2018-12-05, 03:26   #3
sweety439
 
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Quote:
Originally Posted by sweety439 View Post
For any polynomial f(x) in one variable with positive degree and integer coefficients:

f(x) = a_nx^n+a_{n-1)x^{n-1}+a_{n-2)x^{n-2}+...+a_2x^2+a_1x+a_0

with n>=1

Like The Sierpinski/Riesel-like problem, if the values of the polynomial f(x) have a single trivial factor for all x-values, then we take out this factor, besides, if f(x) is not irreducible over the integers, then we must factorize f(x).

Here are some examples:

Code:
f(x)      the polynomials
x^4       {x,x,x,x}
x^2+1     {x^2+1}
x^2-1     {x-1,x+1}
x^3+1     {x+1,x^2-x+1}
x^4+1     {x^4+1}
x^4-1     {x-1,x+1,x^2+1}
x^2+x     {1/2*x,x+1} for even x, {x,1/2*x+1/2} for odd x
x^2+x+2   {1/2*x^2+1/2*x+1}
x^2-x-1   {x^2-x-1}
x^2-x-6   {x-3,1/2*x+1} for even x, {1/2*x-3/2,x+2} for odd x
x^3+2*x+9 {1/3*x^3+2/3*x+3}
x^5+x+1   {x^2+x+1,x^3-x^2+1}
This conjecture is for any f(x), there are infinitely many x such that all the polynomials take on prime values simultaneously.

Note: We also consider the negative primes (e.g. -2, -3, -5, -7, -11, ...) as primes, since they are primes in the ring Z.
For some conjectures:

twin prime conjecture: x^2-1
the fourth of Landau's problems: x^2+1
infinitude of prime quadruplets: x^4-10*x^2+9
infinitude of Sophie Germain primes: 2*x^2+x
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Old 2018-12-05, 03:27   #4
sweety439
 
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Quote:
Originally Posted by sweety439 View Post
For any polynomial f(x) in one variable with positive degree and integer coefficients:

f(x) = a_nx^n+a_{n-1)x^{n-1}+a_{n-2)x^{n-2}+...+a_2x^2+a_1x+a_0

with n>=1

Like The Sierpinski/Riesel-like problem, if the values of the polynomial f(x) have a single trivial factor for all x-values, then we take out this factor, besides, if f(x) is not irreducible over the integers, then we must factorize f(x).

Here are some examples:

Code:
f(x)      the polynomials
x^4       {x,x,x,x}
x^2+1     {x^2+1}
x^2-1     {x-1,x+1}
x^3+1     {x+1,x^2-x+1}
x^4+1     {x^4+1}
x^4-1     {x-1,x+1,x^2+1}
x^2+x     {1/2*x,x+1} for even x, {x,1/2*x+1/2} for odd x
x^2+x+2   {1/2*x^2+1/2*x+1}
x^2-x-1   {x^2-x-1}
x^2-x-6   {x-3,1/2*x+1} for even x, {1/2*x-3/2,x+2} for odd x
x^3+2*x+9 {1/3*x^3+2/3*x+3}
x^5+x+1   {x^2+x+1,x^3-x^2+1}
This conjecture is for any f(x), there are infinitely many x such that all the polynomials take on prime values simultaneously.

Note: We also consider the negative primes (e.g. -2, -3, -5, -7, -11, ...) as primes, since they are primes in the ring Z.
In fact, the only case of this conjecture that has been proved is when f(x) is polynomial of degree 1, no single case of this conjecture for f(x) with degree greater than 1 is proved, although numerical evidence in higher degree is consistent with the conjecture. (however, it is known that this conjecture is true for at least one of these 123 polynomials f(x): x^2+2*x, x^2+4*x, x^2+6*x, ..., x^2+246*x, although none of them have degree 1)

Last fiddled with by sweety439 on 2018-12-05 at 03:30
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Old 2018-12-05, 05:28   #5
Batalov
 
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I don't think you are quoting yourself elegantly enough.


When writing four posts in a row - it is imperative that the 2nd post quotes post #1 entirely, post #3 quotes both post #1 and post #2 in their entirety, and the 4th one - well, I hope you get the idea. Only then your messages will start getting to their audience.
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Old 2018-12-05, 15:25   #6
sweety439
 
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Quote:
Originally Posted by sweety439 View Post
For any polynomial f(x) in one variable with positive degree and integer coefficients:

f(x) = a_nx^n+a_{n-1)x^{n-1}+a_{n-2)x^{n-2}+...+a_2x^2+a_1x+a_0

with n>=1

Like The Sierpinski/Riesel-like problem, if the values of the polynomial f(x) have a single trivial factor for all x-values, then we take out this factor (e.g. x^2+x+2 become {1/2*x^2+1/2*x+1}, since all values of this polynomial are divisible by 2), besides, if f(x) is not irreducible over the integers, then we must factorize f(x) (e.g. x^2-1 become {x-1,x+1}, since x^2-1=(x-1)*(x+1)).

Here are some examples:

Code:
f(x)      the polynomials
x^4       {x,x,x,x}
x^2+1     {x^2+1}
x^2-1     {x-1,x+1}
x^3+1     {x+1,x^2-x+1}
x^4+1     {x^4+1}
x^4-1     {x-1,x+1,x^2+1}
x^2+x     {1/2*x,x+1} for even x, {x,1/2*x+1/2} for odd x
x^2+x+2   {1/2*x^2+1/2*x+1}
x^2-x-1   {x^2-x-1}
x^2-x-6   {x-3,1/2*x+1} for even x, {1/2*x-3/2,x+2} for odd x
x^3+2*x+9 {1/3*x^3+2/3*x+3}
x^5+x+1   {x^2+x+1,x^3-x^2+1}
This conjecture is for any f(x), there are infinitely many x such that all the polynomials take on prime values simultaneously.

Note: We also consider the negative primes (e.g. -2, -3, -5, -7, -11, ...) as primes, since they are primes in the ring Z.
Note that "x^2+x" is {1/2*x,x+1} for even x and {x,1/2*x+1/2} for odd x, however, if you only want even x, you can take {1/2*(2*x),(2*x)+1} = {x,2*x+1} ---> the polynomial "2*x^2+x"
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