[Guest]

Given a polynomial of the form,

F(x) = c₀ + c₁x¹ + c₂x² + .. + c_nxⁿ

such that F(N) is a positive integer for any positive integer N,

Then, F(x) can not generate only primes for x = 1, 2, 3..

This is a reasonably well known result.

Can you prove it?

[Guest]

Then, F(x) can not generate only primes for x = 1, 2, 3..

what?

[Guest]

what?

Then, F(x) can not generate only primes for x = 1, 2, 3....

Hope it is clear

---------------

"..Which part of this don't you understand?.."

- Death

[Guest]

When x = yc₀ (c₀ times any y) then the polynomial will have c₀ and 1 + c₁y + c₂c₀¹y² + .. + c_nc₀^n-1yⁿ as factors.

[superprismatic]

When x = yc₀ (c₀ times any y) then the polynomial will have c₀ and 1 + c₁y + c₂c₀¹y² + .. + c_nc₀^n-1yⁿ as factors.

Suppose c₀ = 1. Then what?

[Guest]

Is this right?

If such a polynomial existed, then F(1) would evaluate to some prime p.

Consider F(1 + kp) where k is a positive integer.

By the Fundamental Theorem of Algebra we can write F(x) as the product of linear terms ( I neglect quadratic terms because I'm confident that we are dealing with only real numbers).

F(x) = b0(x-b1)(x-b2)...

F(1) = b0(1-b1)(1-b2)...

F(1+kp) = b0(1-b1+kp)(1-b2+kp)...

Distributing the kp's out:

One step:

b0[ (1-b1)(1-b2+kp)(1-b3+kp)... + kp(1-b2+kp)(1-b3+kp)...]

Working on the left term once again:

b0[ (1-b1){ (1-b2)(1-b3+kp)(1-b4+kp)... + kp(1-b3+kp)(1-b4+kp)...} + kp(1-b2+kp)(1-b3+kp)...]

By examining this pattern, we can see that we can remove the kp's from the original product term with the consequence of adding an additional term that is a multiple of kp.

Assuming that the polynomial is of finite degree, we can continue to perform this algorithm on the expression and end up with

b0(1-b1)(1-b2)(1-b3)... + b0*kp*(...Some big expression...)

The left term here is simple F(1) which is also equal to p.

F(1+kp) therefore equals p + b0*kp*(...)

= p*(1 +k*b0*(...))

which is divisible by p.

These F(1+kp) will be composite numbers with p as a factor unless they are all equal to p, in which case you have simply a constant function F(x) = p;

This trivial case does not seem like a fair answer to the question.

Edited August 20, 2009 by mmiguel1

[Guest]

Is this right?

If such a polynomial existed, then F(1) would evaluate to some prime p.

Consider F(1 + kp) where k is a positive integer.

By the Fundamental Theorem of Algebra we can write F(x) as the product of linear terms ( I neglect quadratic terms because I'm confident that we are dealing with only real numbers).

F(x) = b0(x-b1)(x-b2)...

F(1) = b0(1-b1)(1-b2)...

F(1+kp) = b0(1-b1+kp)(1-b2+kp)...

Distributing the kp's out:

One step:

b0[ (1-b1)(1-b2+kp)(1-b3+kp)... + kp(1-b2+kp)(1-b3+kp)...]

Working on the left term once again:

b0[ (1-b1){ (1-b2)(1-b3+kp)(1-b4+kp)... + kp(1-b3+kp)(1-b4+kp)...} + kp(1-b2+kp)(1-b3+kp)...]

By examining this pattern, we can see that we can remove the kp's from the original product term with the consequence of adding an additional term that is a multiple of kp.

Assuming that the polynomial is of finite degree, we can continue to perform this algorithm on the expression and end up with

b0(1-b1)(1-b2)(1-b3)... + b0*kp*(...Some big expression...)

The left term here is simple F(1) which is also equal to p.

F(1+kp) therefore equals p + b0*kp*(...)

= p*(1 +k*b0*(...))

which is divisible by p.

These F(1+kp) will be composite numbers with p as a factor unless they are all equal to p, in which case you have simply a constant function F(x) = p;

This trivial case does not seem like a fair answer to the question.

Brilliant! I should have clarified that F(x) is not a constant function.

The solution that I know is similar in spirit, but a little simpler on analysis.

Use reminder theorem.

..is, F(x + q.F(x)) is divisible by F(x) for any q, by reminder theorem.

i.e, if F(x) = 0, F(x+q.F(x)) = F(x) = 0.

[Guest]

My comments do not need a spoiler.

1. I loved that accidental typo about the "Reminder" theorem, methinks. Unintentionaly brilliant that !!!

2. I remember that some medi-aeval monk thought that the quadratic n*n + n + 41 was prime for all positive integers. It actually seems to work for the first few integers, but not when n reaches 40.

[Guest]

My comments do not need a spoiler.

1. I loved that accidental typo about the "Reminder" theorem, methinks. Unintentionaly brilliant that !!!

Oops! Sorry, I stand corrected