341That'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
Can we find a non-const. Polynomial which is prime for all integers x?
The co-efficents of the Polynomial are all integers.
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7 Answers
1Let ' 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 .
11Yes, that's correct.
Alternately, we have for integers x' and k, P(x')=k.
Now substitute (x'+k) in place of x' and see the fun.
341
62prophet sir, is that really true!
I thought that there exists none! I would be really interested in reading this one..
One argument, that i could give is that when all the 10 variables take the same value, it reduces to a polynomial of one variable.. which cannot have all values as primes..
Hence a polynomial of more than one varaiable too cannot!
341I 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
341Ya, quite some level of abstraction. I was simply following links and I read this Matiyasevich theorem and in that link this 10th degree polynomial was mentioned. My joy and relief knew no bounds :D