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Let $q(n)$ be a nonconstant polynomial with integer coefficients, and let $c=q(0)$ be the constant term of $q$. Show that if $q$ is nonconstant and $c \gt 1$, then there are infinitely many $q(n)$ $\in \mathbb N$ that are not primes.

Hint: You may assume the familiar fact that the magnitude of any non constant polynomial, $q(n)$, grows unboundedly as $n$ grows.

How to solve it using the given hint? As it is stated that polynomial has integer coefficients, so that means polynomial can be decreasing as well. But in that case we may not get mapping to infinitely many natural numbers.

coderoach
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3 Answers3

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First, please note that some authors allow for negative numbers to be prime. Maybe your problem allows for that.

Actually, the thing that matters the most in the end is the behavior of the leading coefficient of the polynomial, say, $a_nx^n$ ; as $\mathbb N \rightarrow \infty$ , the poly will go either to $(+/-) \infty $. Since your coefficients are natural numbers, your polynomial will go to $+\infty$ as you approach $+ \infty $. This means that, in the process, it will take-on infinitely-many values. Then, let $c:=q(0)$. Then $f(n.c)$, for n in $\mathbb N$ will be a multiple of c, i.e., $f(n.c)$ will never be prime .

FBD
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  • what if the domain is integers, would the answer be same. – coderoach Aug 12 '13 at 19:31
  • Yes, because the integers includes the naturals then. But if you allow your coefficients to be any integer, then the issue is a bit more complicated. – FBD Aug 12 '13 at 19:35
  • how'd u solve for $-{n^3} + 1$, i mean if we do not allow -ve primes. Is it possible. – coderoach Aug 12 '13 at 19:42
  • If you allow for negative numbers to be composite, then notice that -$n^3+1$ will always be composite for n>1, using the factorization: $(n^3-1)=(n-1)(n^2+n+1)$ , as the product of two factors, each of which is larger than one. This is a special case of the factorization: $a^n-b^n=(a-b)(a^{n-1}+a^{n-2} b+...+b^{n-1} )$ – FBD Aug 12 '13 at 19:47
  • If you do not allow for negative primes, than it is still possible, if you allow for negative integers as inputs, since if $n<0$, then $^-(-n)^3$ will be >0. If your inputs are only natural numbers, then the expression $-n^3-1$ will only produce negative numbers, so it is not then possible to generate (positive) primes at all. – FBD Aug 12 '13 at 19:57
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Suppose that $q(n)$ is prime for some $n$. Notice that $q(x)=a_mx^m+a_{m-1}x^{m-1}+\ldots+a_1x+c$. What can you say about $q(cn)$?

Your result will hold for any $c$. Split into cases of $c$ being an integer and noninteger. There is also the special case of $c=\pm 1$.

Alex R.
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  • Do you mean set $q(0) = c$, where $c$ is prime... – coderoach Aug 12 '13 at 19:21
  • I think you stated c>1 in your top paragraph. – FBD Aug 12 '13 at 19:32
  • The value of $c$ is immaterial to the result. For any $c$, $q(n)$ achieves infinitely many nonprime numbers (be they negative or positive). The proof is slightly trickier when $c=\pm 1$. You also need to consider non-integer $c$ which isn't too difficult. – Alex R. Aug 12 '13 at 19:36
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If there is no constant coefficient, then $n\mid q(n)$ for all $n$. If $q(n)$ is exactly $n$, then I claim that you are done (do you see why?). If not, then In general $q(n)$ will differ fom $n$ infinitely often, and you are done again.

If there is a constant coefficient $c$, then $q(nc)$ will be divisible by $c$ always.

In both cases, since polynomials take on arbitrarily large values, the above argument shows that $q(n)$ will take on infinitely many non-prime values.

davidlowryduda
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