Prime-generating polynomials

[mfgoode]

Quote: ixfd64 I've read in a book that f(x) = x2 + x + 41 is a very good prime-generating polynomial. [/Quote]

QUOTE=mfgoode]From Memory I give a better one- n^2-79n+1601. Please check it out
Mally [/QUOTE]

Eulers formula is f(n) = n^2 +n +41 n=[0----39]
I have found a similar one; f(n) = n ^2 - n =41 n=[0----40]
The 3rd is as given above f(n) = n^2 -79n +1601 n=[0----79]

I want to stress that no rational algebraic formula can represent prime nos. only

Proof: Simple for all readers not knowing modular arithmetic.
If possible let the formula a+bx +cx^2 + dx^3 +---represent prime no.s only
Also Suppose that when x=m the value of the expression is p so that
p = a+bm +cm^2 +dm^3 -----
When x= m+ np the expression becomes a+b(m+np) +c(m+np)^2 +d(m+np)^3---
That is a+bm +cm^2 +dm^3 + a multiple of p or
p + a multiple of p
Thus the expression is divisible by p and is therefore not a prime. This is contrary to our original assumption.
Q.E.D. N.B. cm^2 is the same as c*m^2

Mally (and a smoke! )

[mfgoode]

Originally Posted by mfgoode

Quote: ixfd64 I've read in a book that f(x) = x2 + x + 41 is a very good prime-generating polynomial.

QUOTE=mfgoode]From Memory I give a better one- n^2-79n+1601. Please check it out
Mally [/QUOTE]

Eulers formula is f(n) = n^2 +n +41 n=[0----39]
I have found a similar one; f(n) = n ^2 - n =41 n=[0----40]
The 3rd is as given above f(n) = n^2 -79n +1601 n=[0----79]

I want to stress that no rational algebraic formula can represent prime nos. only

Proof: Simple for all readers not knowing modular arithmetic.
If possible let the formula a+bx +cx^2 + dx^3 +---represent prime no.s only
Also Suppose that when x=m the value of the expression is p so that
p = a+bm +cm^2 +dm^3 -----
When x= m+ np the expression becomes a+b(m+np) +c(m+np)^2 +d(m+np)^3---
That is a+bm +cm^2 +dm^3 + a multiple of p or
p + a multiple of p
Thus the expression is divisible by p and is therefore not a prime. This is contrary to our original assumption.
Q.E.D. N.B. cm^2 is the same as c*m^2

Mally (and a smoke! )[/QUOTE]


Here is one that defied mathem'cians from the 17th to the 19th century

((2)^N)/N = 2 . If so then N is prime.

Can anyone tell me when this formula breaks down

[mfgoode]

Originally Posted by mfgoode

Quote: ixfd64 I've read in a book that f(x) = x2 + x + 41 is a very good prime-generating polynomial.

QUOTE=mfgoode]From Memory I give a better one- n^2-79n+1601. Please check it out
Mally [/QUOTE]

Eulers formula is f(n) = n^2 +n +41 n=[0----39]
I have found a similar one; f(n) = n ^2 - n =41 n=[0----40]
The 3rd is as given above f(n) = n^2 -79n +1601 n=[0----79]

I want to stress that no rational algebraic formula can represent prime nos. only

Proof: Simple for all readers not knowing modular arithmetic.
If possible let the formula a+bx +cx^2 + dx^3 +---represent prime no.s only
Also Suppose that when x=m the value of the expression is p so that
p = a+bm +cm^2 +dm^3 -----
When x= m+ np the expression becomes a+b(m+np) +c(m+np)^2 +d(m+np)^3---
That is a+bm +cm^2 +dm^3 + a multiple of p or
p + a multiple of p
Thus the expression is divisible by p and is therefore not a prime. This is contrary to our original assumption.
Q.E.D. N.B. cm^2 is the same as c*m^2

Mally (and a smoke! )[/QUOTE]

Here is one that defied mathem'cians from the 17th to the 19th

(2^N)/N =2. If so then N is prime.
Can any one tell me when this formula breaks down? For what value of N?
Mally

[asdf]

Originally Posted by mfgoode

(2^N)/N =2. If so then N is prime.
Can any one tell me when this formula breaks down? For what value of N?

It only works for n=1,2 and 1 is not prime

[mfgoode]

Originally Posted by asdf

It only works for n=1,2 and 1 is not prime

Maybe you are confused by the sign ^. This means 'to the power of''
So in other words you raise 2 to the power of N and then divide by N. If the remainder is two then N is a prime.
e.g. Let N=5 which is a prime then 2^5 =32. Now divide by 5 and you get 2 as the remainder. Hence you conclude that 5 is prime. And so on and so forth
for no.s like 7,11, etc.
Hope this clarifies this problem
Mally

[garo]

The / sign is used to denote division. You should have used the "modulus" sign % or "mod" to denote the operation.

In any case, your question is a simple rehash of Fermat's little theorem which stated that for p to be prime the following condition is necesssary:

a^(p-1) = 1 mod p

Now multiply both sides by a and substitute a=2 and we get your statement back. And the answer to your question is 341 = 11.31

[mfgoode]

Originally Posted by garo

The / sign is used to denote division. You should have used the "modulus" sign % or "mod" to denote the operation.

In any case, your question is a simple rehash of Fermat's little theorem which stated that for p to be prime the following condition is necesssary:

a^(p-1) = 1 mod p

Now multiply both sides by a and substitute a=2 and we get your statement back. And the answer to your question is 341 = 11.31

Excellent Garo! I stand corrected
Thanks for showing my formula as Fermats little theorem in disguise
Mally