Question

Prove the the equation n^2+n+41 is prime when the function has the value f(200)? Here is what I have

(200)^2+200+41=>40000+200+41=>40241
I break this down to prove that it is the equation is in fact 40(100+5+1)+1 leaving the whole problem divisible by 40 and 1 is still left. Is this a valid proof to state that 40241 is in fact a prime number?

Answer 1

No, you have only proven that the number is odd, and not divisible by 2. To be prime, it has not to be divisible by anything other than 1 and itself.

Answer 2

That's what my original impression was thanks. I could not think of any other way to start it.

Answer 3

I won't lie, I'm not completely sure how to do it either.
But don't forget your stadard types of proof:

direct proof
by induction
by contrapositive
by contradiction

Answer 4

I know the end result is true just trying to figure it out without using the Sieve of Erosthenes.

Answer 5

Proof by exhaustion?
In fact, this is the only thing I can think of. If I follow the Sieve of Erastothenes the next prime in fact is its square. You get 2,3,7,11,13... the first number to be crossed off after crossing off all that is divided by 2,3,and 5. Then the first number to be excluded from the list is 49 when dealing with primes. So, the question is do I need all prime squares until 200 or 40241?

Answer 6

49 when moving on to 7 as the next prime.

Answer 7

So the next number to be crossed off would be 121 when moving to 11.

Answer 8

Here is what I am thinking now without dividing all the primes up to 200 (45 division problems). If you follow the Sieve of Erastonthenes and observe the trends there are no 3 consecutive integers that are prime. So if I calculate n=40241, and find that (n-1), (n-2), (n+1), and (n+2) are in fact divisible by a prime number then in fact I have proven that 40241 is in fact prime. Do this sound valid?

Answer 9

I'm not completely sure what you mean, so I can't say, but it doesn't seem right.