Primey Polynomial

Let us assume that $f(p)=q^r$ , where $p,q$ are distinct (assume) primes and r is a natural number.

Then $t*q^{r+1}$ divides $f(p+t*q^{r+1})-f(p)$ , where $t$ is a non-negative integer.

(Since we know that $a-b$ divides $f(a-b)$ for a polynomial with integer coefficients where $a$ and $b$ are integers, not both zero.)

So $q^{r+1}$ and $q^r$ both divide $f(p+t*q^{r+1})-f(p)$ ,

Now we note that $p$ and $q$ being distinct primes, have their g.c.d as 1, so they are mutually coprime, so $p$ and $q^{r+1}$ are also coprime. In fact, this enables us to say that the arithmetic progression with first term $p$ and the common difference $q^{r+1}$ has infinitely many primes in it by Dirichlet's Theorem .

Now, so for infinitely many $t$ we will have $p+t*q^{r+1}$ =prime, since $p+t*q^{r+1}$ just represents the $(t-1)^{th}$ term of the above mentioned arithmetic progression.

So $f(p+t*q^{r+1})$ will also be the power of some prime, let it be $m^n$ , where $m$ is a prime.

Now we can see that $q^r$ divides $f(p+t*q^{r+1})-f(p)$ and $q^r$ divides $f(p)$ also as $f(p)=q^r$ , so $q^r$ also divides $f(p+t*q^{r+1})$ , that is $q^r$ also divides $m^k$

but since $m$ and $q$ are both primes, we have $m=q$ , and $q^r$ divides $q^k$ , so that we have $r<=k$ .

But again, we know that $q^{r+1}$ also divides the difference $f(p+t*q^{r+1})-f(p)$ , that is $q^k-q^r$ .

Now if $q^{r+1}$ divides $q^k$ then it also divides $q^r$ . But that is not possible.

So $q^{r+1}$ does not divide $q^k$ . So we have $q^r$ divides $q^k$ but not $q^{r+1}$ .

So we have $q^r=q^k$ .

Thus we have that any prime in the sequence of the arithmetic progression generating the prime numbers will be $q^r=c$ only.

So the polynomial $g(x)=f(x)-c$ has infinitely many roots, which means that the polynomial g(x) is constant, and identically equal to 0. But that means that we have $f(x)=c$ where c is a constant. But that is certainly not true, because the degree of the polynomial is given to be 5.

So we have a contradiction!!

The contradiction arises due to our initial assumption that p and q are distinct primes.

So that must be wrong.

So we have $f(p)=p^r$ .

Now we endeavour to prove that $f(x)=x^5$ actually.

We consider the sequence $f(p_{i})/{p_i}^5$ and try to compute the limit of the sequence as i tends to infinity, where $p_i$ refers to all the primes.

We know that the limit of this sequence is given by the leading coefficient of the polynomial f(x), say $a_0$ and it cannot be zero.

So we can say that after a certain natural number b, for all c>b, $f(p_c)=p^5$ , because if it is not so, then $f(p_c)=p^u$ where u>5 would lead to the limit being infinity, and if u<5, then the limit would be 0.

Thus we have that the polynomial $g(x)=f(x)-x^5$ has infinitely many roots, namely the primes ${p_c, p_{c+1},.... }$ and so on, which is an infinite set, because there are infinitely many primes and the above set only excludes finite number of them from $p_1$ to $p_{c-1}$ .

But this means that this is a constant and since there are roots of this polynomial as well, the constant is zero.

So we have $f(x)=x^5$ .

The interesting thing about this polynomial is that we know it's behaviour only for the prime numbers, but we can generalize it over the whole domain of real numbers.