Prime divisors

[Copernicus]

I have a question about frequency of divisors. Lets say we have a number N, which the next lower number is prime, for example 48. One less than 48 is 47. 47 is a prime.

If I take 48 to a power. For example 48^2 is 2304. If I subtract 1 from this number it is 2303. One of the factors of 2303 is 47.

Why is it, that, no matter what power I use, for example 48^30=27367625686405765816997946520518481655117436 5773824 or another power and subtract 1, that one of the factors almost always, 47?

[grant hutchison]

Because the binomial expansion of (x+1)^n is 1+[a whole lot of stuff evenly divisible by x]

Grant Hutchison

[DonM435]

Thanks.
I was working along those lines:

n^2 - 1 = (n - 1) * (n + 1)
n^3 - 1 = (n - 1) * (n^2 + n*1 + 1^2)
n^4 - 1 = (n − 1) * (n^3 + n^2 + n + 1)
...

I had to look up the last two, but it seems there's always an (n - 1) factor in there. But that's as far as I got before headache set in!

[grant hutchison]

Specifically:


Which is equivalent to:


So every time you take an integer x, add 1, raise it to any integer power, and then subtract 1, you'll end up with a number evenly divisible by x.

Grant Hutchison

[DonM435]

That would seem to work for any integer, not merely (prime_number +1), right?

[Copernicus]

Thank all and Grant.

[DonM435]

Y'r welcome.

-- All

[grant hutchison]

That would seem to work for any integer, not merely (prime_number +1), right?

That's right.
If you multiply (1+x) by itself any number of times, there will be only one term in the expansion that isn't a multiple of x, and that'll be the term derived from multiplying all the ones together. Which will, of course, be 1.
That'll be more obvious if you're interested in prime factors, but it's true for any integer.

Grant Hutchison