## Sunday, February 16, 2014

### Proof that there is only one prime number preceding a square number and any other power number (or "What is the use of factoring polynomials?")

Students are made to study how to factor polynomials but are never given an interesting task that involves using polynomial factorization. Here is a use that I find interesting.

Let's say we have the polynomial "x^2 - 1". This is a difference of two squares, which means that you can easily factorize it into "(x+1)(x-1)". But what does this factorization mean?

It means that any number which precedes a square number can always be factored into two numbers: one more than the number that was squared and one less than the number that was squared. For example, 15 precedes the square of 4, that is "4^2 - 1", so we can immediately conclude that "4+1" (5) and "4-1" (3) are factors of 15. Another example is 99 which precedes the square of 10, that is "10^2 - 1", so we can immediately conclude that "10+1" (11) and "10-1" (9) are factors of 99.

OK so maybe this is interesting, but can we use this for something more interesting?

Are there any numbers preceding a square number that are prime numbers? If there is such a number, than it's factors must be 1 and itself. So we are looking for a number of the form "x^2 - 1" such that it can only be factored into 1 and the number itself (which is prime). We said that any number of the form "x^2 - 1" can always be factored into two numbers, "x+1" and "x-1". So if this number, "(x+1)(x-1)", can only be factorized into 1 and itself, then surely one of these two factors must be 1. Which of these factors can be equal to 1? The "x+1" or the "x-1"? Since "x" is a positive number (the number that is squared), it can only be the "x-1" factor.

Will the "x-1" factor equalling 1 automatically make the other factor, "x+1", a prime number? No, because the "x+1" factor could itself be factorizable by other numbers. The fact that "x-1" equals 1 means that "(x+1)(x-1)" could be prime. But if "x-1" is not equal to 1 then, given that "x+1" cannot be equal to 1, "(x+1)(x-1)" is surely factorizable by "x-1", making it not a prime.

In order for "x-1" to equal 1, "x" must be 2. In this case, the number preceding a square number is "2^2 - 1" which is 3. Notice that the other factor, "x+1" would also equal 3 when "x" is 2 which makes sense since we said that one of the factors must be 1 and the other must be the number itself. If 3 a prime number? Yes. So when "x" is 2, "x-1" is 1 and "x+1" is a prime number. Therefore 3 is a prime number which precedes a square number.

Are there any other prime numbers which precede a square number? No, because if "x-1" is anything else but 1, then neither of the two factors will be 1, making the number factorizable into two numbers which are not 1 and itself, hence not being a prime number. This means that 2 is the only number "x" can be, which means that there are no other numbers which precede a square number that are prime.

It's interesting to also note that using the exact same method, one can proof that 7 is the only number preceding a cube number that is prime. Use this website to find the factors of numbers preceding other powers of "x".

It seems that there can only one number preceding an nth power of "x" that can be prime, for every power of "n". Can we prove this?

What will the factors of "x^n - 1" for any "n" be? Regardless of "n", will "x-1" always be a factor of "x^n - 1"? That is, will 1 always be a root of "x^n - 1"? Yes, because "1^n - 1" will always be 0, for all "n". So we can be sure that for "x^n - 1", regardless of "n", "x-1" will always be a factor. This means that "x^n - 1" will always be factored into "(x-1)(...)" where the "..." is some other polynomial factor. Since "x-1" is equal to 1 only when "x" equals 2, then "2^n - 1" is the only number preceding the power number "x^n" that can be a prime number. Of course it could be that the other number could be factored into other numbers hence making "(x-1)(...)" not a prime number when "x" is 2, but if "x" is not 2 then "(x-1)(...)" cannot be a prime number because "x-1" will be a factor.

So, regardless of what "n" is, there can only be one number which precedes "x^n" that can be prime, and that is "2^n - 1", since "x^n - 1" can always be factored into "(x-1)" and some other factor that cannot be equalled to 1. The other factor might not be prime, but if it is then it is the only one preceding "x^n" for all "x". Such prime numbers are called Mersenne primes.

Is this useful? This might be useful for testing whether a number is a prime number by checking if the number precedes a number raised to some power, x^n. If it does then it can only be a prime number if it is a number of the form of "2^n - 1", otherwise it will be always be factorizable by "x-1", regardless of what "x" or "n" are.