prime number and quadratic question

[apple2357]

I find the wording of the this question a bit complicated.
Is it just after a counter example? can i just say n=0? Or if c isn't 1 , then you can just use n=c ?
I am entirely sure i understand the question? Can anyone offer an interpretation?

 

[JeffM]

Zero is a non-negative integer. There are an infinite number of trivial counter-examples.

[math]a, \ b, p, \ q \in \mathbb Z^+, \ n = 0, \ p > 1 < q, \text { and } c = pq \implies \\ a(0)^2 + b(0) + c = pq \implies p(0) \text { is not prime.}[/math]
If the problem meant to say that n is a positive integer

[math]a = 1, \ b = 2, \ c = 7, \text { and } n = 2 \implies\\ p(n) = 1(2^2) + 2(2) + 7 = 4 + 4 + 7 = 15 = 3 * 5, \text { not prime.}[/math]

 

[Dr.Peterson]

I find the wording of the this question a bit complicated.
Is it just after a counter example? can i just say n=0? Or if c isn't 1 , then you can just use n=c ?
I am entirely sure i understand the question? Can anyone offer an interpretation?

There is some subtlety in the wording if you are not accustomed to logic:

​

You need to show that, no matter what values you pick for a, b, and c, there will be some non-negative integer n for which p(n) is not prime, so that it would be false to say it is always prime.

So you can't start by picking a value of n to make a counterexample; if you take n=0, the p(0) will be prime if c happens to be prime, so that isn't a counterexample for all a, b, c.

But you might take separate cases: if c is prime, then n=0 is counterexample; can you find a counterexample if c is composite? (I'm not saying that's the way to go, just that it is one possibility.)

 

[apple2357]

Reminded of x^2+x+41 which i think generates quite a few primes

 

[apple2357]

If c is non-zero then plug in c to get ac^2+bc+c which is not a prime. Otherwise c=0 and take n=4.

Will that answer it?

 

[Dr.Peterson]

Reminded of x^2+x+41 which i think generates quite a few primes

Yes, if you make a table of values, it takes a long time to find a non-prime.

x​	x^2+x+41​
0​	41​
1​	43​
2​	47​
3​	53​
4​	61​
5​	71​
6​	83​
7​	97​
8​	113​
9​	131​
10​	151​
...​	...​

And the reason it eventually makes a non-prime may be a good example of how to solve this problem! Do you recall seeing, or can you determine, a value of x that doesn't generate a prime?

EDIT:

If c is non-zero then plug in c to get ac^2+bc+c which is not a prime. Otherwise c=0 and take n=4.

Will that answer it?

Yes, you got the idea! f(c) = ac^2 + bc + c = c(ac + b + 1), which is not prime (because ac+b+1 > 1!).

Once you see it for this example, it's easy to find a counterexample for any given a, b, c.