Math Is Fun Forum

Here is it:
Let
F[x]=ax^2+bx+c and F[x] is not factorisible.
Proof that in the rank
F[1], F[2], ...
exist infinitely many prime numbers.

I'll try something...

Why are we disputing something fundamental?

Merry Christmas Ricky and Mathsy!

I have a problem:
What picture is better than e^Pii?
This is e^Pii with christmas hat on e!
But when I try to change my picture, I get the message that the picture has been refreshed and then I go back. But then I see my previous picture. Has anybody had the same problem as mine?

Yes. It must work.

2+4+8= (8-1)/(2-1) or 2+4+8=2(8-1)/(2-1), e.a
14=7 or 14=14?

I'm sure that

Why first x in the numerator shouldn't be there?

For now the injuiry only proofs that the mathematics theory is correct.
The males must be statisticaly equal to the females.

Here's my simplification

The question isn't so hard but there are a lot of things to consider.

Sorry, if I made a mistake, but I think:

But

, so

and

Thus we have:
S=7(1-1/2-1/4)-(1/4-1/5-1/25)=7/4 - 1/100 = 87/50 = 1,74

When I think for a while, I found an interesting fact:
The question is
Which remainders by K suppose that p is eventually prime?
So, if  tru
If a number is from the form

if GCD[K,G]>1 then L is truely non-prime.
So, if all prime, greater than K are from the form

then for all 1<i<K-1 GCD[i,K]>1
Let find the chain of K-s
The firs member is 4:
GCD[4,2]=2
The second is 6:
GCD[6,2]=2
GCD[6,3]=3
GCD[6,4]=2
Is there third?
I think there isn't.
Here's what have to be prooved:
There isn't number H greater than 6 for such every prime numbers, greater or equal to H-1 are from the form

I think Ricky is right. Here's a direct proof:
Let
u^2-v^2 = p
We'll find u and v by p:
u^2-v^2 = p
=> (u+v)(u-v) = p, but p is prime, so his divisors are only 1 and p, so:
u+v=p && u-v=1
=> u=v+1; 2v+1=p; v= (p-1)/2, but p is odd, so v is integer.
So:
v= (p-1)/2
u=1 + ((p-1)/2)= (p+1)/2.