Formulas for PrimesDate: 09/09/2002 at 23:44:49 From: Junny Tim Subject: Prime numbers 1. Ancient mathematicians thought that the formula n^2 + n + 41 always produced a prime number for any whole number n. How do I show that there is an infinite number of whole numbers from which the above formula does NOT produce a prime number? 2. How do I explain why the formula n^2 + 8n + 15 NEVER produces a prime number (assuming that n is a whole number)? Date: 09/10/2002 at 15:06:44 From: Doctor Paul Subject: Re: Prime numbers This quadratic is obviously not prime for n = 41 since 41^2 + 41 + 41 = 41 * (41 + 1 + 1) I think you'll find that a similar method will work for any multiple of 41. And certainly there is an infinite number of multiples of 41. > 2. How do I explain why the formula n^2 + 8n + 15 NEVER produces > a prime number (assuming that n is a whole number ) n^2 + 8*n + 15 = (n + 5) * (n + 3) is always a nontrivial factorization unless n = -4 or n = -2 (in either of these cases, we get one of the factors equal to the integer one, and thus we have the trivial factorization x = x * 1 that cannot be used to tell us whether or not the number x is prime). But a whole number cannot be negative, so we don't have to deal with this problem after all. I hope this helps. Please write back if you'd like to talk about this some more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ Date: 09/10/2002 at 20:26:28 From: Junny Tim Subject: Prime numbers I still have trouble understanding the problem. I'm more of a visual learner. Is it possible for you to rephrase your answers in another way for me to understand? Thank you for your help and cooperation. I appreciate it. I found this site very helpful. Date: 09/11/2002 at 12:27:10 From: Doctor Roy Subject: Re: Prime numbers Hi, I'll be giving a slightly different answer from Dr. Paul. For the first problem, we have: n^2 + n + 41 If we use n = 41, then n^2 is divisible by 41. n is divisible by 41 (since n *is* 41) and 41 is obviously divisible by 41. Since each part of the expression (all of n^2, n, and 41) are divisible by 41, then the entire sum must be divisible by 41. Notice that we can use any multiple of 41. If we use n = 2*41, or 3*41, etc., then n^2 and n are still divisible by 41, and the entire expression is divisible by 41. As for the second problem: n^2 + 8n + 15 There are two ways to see why the expression is never prime. The first is what Dr. Paul explained. n^2 + 8n + 15 = (n+5)*(n+3) This means that (n^2 + 8n + 15) is the product of two different numbers, (n+5) and (n+3). This means that is n is positive, and we have two different numbers that divide the expression. For example, if n = 7, then: n^2 + 8n + 15 = 7^2 + 8*7 + 15 = 49 + 56 + 15 = 120 (n+5) = (7 + 5) = 12 (n+3) = (7 + 3) = 10 This means that 120 = 12*10, so we have a factorization. Since any number that can be expressed as the product of two positive integers (without either being 1) is composite, n^2 + 8n + 15 can never be prime. There is one exception, of course. If either (n+5) or (n+3) = 1, then the resulting expression could be prime. However, because n must be negative for this to happen, we don't have to worry about it. The other way to see why (n^2 + 8n + 15) can never be prime is that it is always divisible by 3. Notice that 15 is divisible by 3, and you can prove that (n^2 + 8n) is always divisible by 3 for any positive n. - Doctor Roy, The Math Forum http://mathforum.org/dr.math/ |
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