Easiest of All...

Let ' s assume there exists a non - constant polynomial F ( x ) such that eventually it takes all the

values of prime numbers .

Suppose for a particular value of x , say x = 1 , F ( x ) = P , where P is a prime number .

So we get , F ( 1 ) ≡ 0 ( mod P )

But for all integers Y , F ( 1 + PY ) = 0 ( mod P )

Hence , F ( 1 + PY ) cannot be a prime number as it is divisible by P .

So the only option left , is that F ( 1 ) = F ( 1 + PY ) .

But that also is not permissible as F ( x ) is not a constant function by our assumption .

So we conclude that there exists no such non - constant polynomial which can generate all the

prime numbers .

To prove that , F ( 1 + PY ) = 0 ( mod P ) ----- >

One can assume F ( x ) = a₀ x ⁿ + a₁ x ^{n - 1} + .............

So F ( 1 ) = a₀ + a₁ + a₂ + .............. = P

And F ( 1 + PY ) = a₀ ( 1 + YP ) ⁿ + a₁ ( 1 + YP ) ^{n - 1} + .........

= { a₀ + a₁ + ................ } + YP ( ...............) - - - - - - -> { by binomial expansion }
= P + YP ( ................... )

Clearly , F ( 1 + PY ) is divisble by P also .

That's for polynomials in a single variable. I remember reading about a polynomial in some 10 variables that generates primes. Lets see if I can locate that source

I located it at last! (whew!).

http://en.wikipedia.org/wiki/Formula_for_primes

He gives an example with 26 variables where +ve values taken by the polynomial are all the primes.

Note that he quotes a theorem by a Russian mathematician named Matiyasevich which already proves that such a polynomial will exist. So, a 10 variable one is also possible (thank god for that, i was wondering whether i was talking through my hat when I said that). That theorem is a landmark one as it decided a long standing problem named Hilbert's 10th problem

WOW! Big stuff up here....[152]