TargetIIT

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 .

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

One can assume F ( x ) = a[ss]0[/ss] x [p]n[/p] + a[ss]1[/ss] x [p]n - 1[/p] + .............

So F ( 1 ) = a[ss]0[/ss] + a[ss]1[/ss] + a[ss]2[/ss] + .............. = P

And F ( 1 + PY ) = a[ss]0[/ss] ( 1 + YP ) [p]n[/p] + a[ss]1[/ss] ( 1 + YP ) [p]n - 1[/p] + .........

                        = { a[ss]0[/ss] + a[ss]1[/ss] + ................ } + YP ( ...............) - - - - - - -> { by binomial expansion }
                        = P + YP ( ................... )

Clearly , F ( 1 + PY ) is divisble by P also .[/hide]        

Yes, that's correct.

Alternately, we have for integers x' and k, [im]http://codecogs.izyba.com/gif.latex?P%28x%27%29%3Dk[/im].

Now substitute [im]http://codecogs.izyba.com/gif.latex?%28x%27+k%29[/im] in place of x' and see the fun.