Primes

Factual Questions
#1

Is it possible to prove. (4n+1)^2+4n/2

Can never produce a prime number?

#2

Yes. It can be factorized (easily) with two terms so that it will always be the product of two numbers.

#3

You didn’t put in enough parentheses to allow us to understand what you’re talking about. Is that this?:

Multiply n by 4.
Add 1.
Square it.

Then multiply n by 4.
Divide it by 2.

Add those two results.

Or is it this?:

Multiply n by 4.
Add 1.
Square it.
Then multiply n by 4.
Add those two numbers.

Divide the result by 2.

(Yes, I know how to express this with parentheses. I’m just not sure mla3 does.)

#4

In my opinion (4n+1)^2+4n/2 is not ambiguous and is the same as

(4n/2) + (4n+1)^2

#5

Why would mla3 write (4n/2) instead of 2n?

#6

My guess is that it is homework and that is why I did not give the complete answer but your guess is as good as mine.

#7

So what is the factorization?

#8

Hmm, I guess you don’t want to give the factorization because that would be doing the homework for mla3.

#9

This isn’t homework don’t be silly. I get 90% for maths thank you. When you multiply it out you get. 16n^2 + 10n +1.

When n is a multiple of three you get a number which is equal to 6p+1. (All elements in these equations are whole numbers.) This of course does not not mean that the number has to be prime only that it may be prime. The trouble I am having is that when I run this on an excel spreadsheet. None of the numbers that are produced from the equation result in a prime number. I was wondering if there was anything that I had missed.

#10

When its factorized you get the following.
(8x+1)(2x+1)

#11

There’s your answer.

#12

The question wasn’t how you factorize. The question was why does

16n^2 +10n + 1

never produce a prime number when n is a a whole number?

When you subtract one from the statement you get a multiple of 6 only when n is a multiple of 3. When n is a multiple of 3 and larger than 3 you will get a multiple of four when you subtract one from the above statement. So I can think of no logical reason why the above statement will never result in a prime number.

#13

I think you’re having a bit of a cognitive disconnect here.

If you can factor a number, it’s obviously not prime. That’s basically the definition of prime, whose only factors are 1 and itself.

There is one exception here: if one of the factors is one. In this case, neither factor (8x+1) or (2x+1) equals 1 for a whole number value of x, so no problem.

And this is a very popular method for showing something isn’t prime - by providing a factorization. Providing a factorization is precisely equivalent to showing something isn’t prime.

ETA: The (2x+1) part is a particularly good one. This term means there is always an odd number that divides the number, as odd numbers can generally be expressed as 2n+1.

#14

The fact that you can factor it means it’s not prime. Whatever number your equation produces will factor to at least those two factors.

#15

Actually, to clarify what I meant by (2n+1), it’s pretty neat, since it means that the N-th number you generate with this polynomial will always be divisible by the (N+1)-th odd number.

So, the n = 1 number will be divisible by 3, the n= 2 number will be divisible by 5, the n=3 number will be divisible by 7, and so on.

#16

Thanks didn’t thin k of that

#17

Okay, I feel dumb for asking this since I should really know this, but how does this prove it’s not prime for all rationals?

It’s obvious it proves it for all integers. And I’m pretty sure that if x is irrational it will never be prime since none of the coefficients are irrational (and thus it probably will never be whole). However, I’m not clear how we can be sure that the product will never be prime for a rational-but-not-integral x. Can somebody explain it?

#18

Yeah, I worked something out on paper. Take this factorization:

(x + 1)(2x + 1)

Let x = 1/2

(1/2 + 1)(2 * 1/2 + 1)
(3/2)(1+1)
(3/2)(2)

= 3

Which is prime.

#19

Primes are whole numbers by definition, not rationals.

#20

If it can be factorized it cannot be prime, by definition of prime. Now, if you can find a number which can and cannot be factorized at the same time you really got something there. I would first try selling it to the US military because they could make heads explode with something like that :slight_smile: