I just came across this in an article in The American Mathematical Monthly Vol. 1, No. 6, Jun., 1894. The article is taken from a paper by J. W. NICHOLSON., President and Professor of Mathematics at Louisiana State University.
To keep it simple I will present a very small example of the professor's theorem.
Pick a prime number, p (I'll use five because it keeps things short and easy) .
Now take the sum of the squares of every integer smaller than p and "voila", it is divisible by p
42 + 32 + 22 + 12 = 30, which is divisible by 5.
and it doesn't have to be a square, the same series using cubes gives:
43 + 33 + 23 + 13 = 100, which is also divisible by 5.
In general, the first baby rule says for prime p, (p-1)n + (p-2) n + ... + 1n will be divisible by p as for any power n smaller than p.
Go ahead, try a few of your own.
Now to kick it up a little... let's add in any constant, c, to the mix.
It is also true, that for prime p and n smaller than p:
(c+p-1)n+(c+p-2)n ...... + (c+0)n will also be divisible by p.
If I keep p = 5 and use c=2 the series would be :
(2+4)2+(2+3)2++(2+2)2+(2+1)2 + (2+0)2 =90 which is still divisible by 5.
Ok, and one to grow on: you can use a constant multiplier in front of the p-x terms... for example using a multiplier of three in each case we have
(2+3x4)2+(2+3x3)2+(2+3x2)2+(2+3x1)2 + (2+0)2 =410 which is still divisible by 5.
According to the Professor, this works only if, we use a prime p, As pointed out in the comments, there are some primes for which this is not true. I originally misstated what he wrote.
No justification was given, and I can provide none, so if you know how to prove this, I will welcome your proof and post it here.
A brief but rather opaque start of a proof:
ReplyDeleteThe group of units mod p is cyclic.
So we are summing 1 through p-1, which for p odd, means we have 0 mod p.
Now, any odd power won't cause a problem because we can still pair x with -x to get a sum of 0.
I think we can deal with all the powers by viewing all the numbers as powers of some generator mod p, but I'm too lazy to work out the details.
Anyway, the point of the group being cyclic mod p is that your addition of a constant and multiplication by a constant leaves your numbers the same mod p, so the second half of things is pretty much doing nothing.
As for the "if and only if", I'm not sure what exactly the statement is that's supposed to be reversible. For instance the sum of the squares of 1 through 24 is divisible by 25 even though 25 is not prime.
Has there been a proof done yet? I'm not really sure how to go about it either. But this definitely intrigues me. And I agree with Joshua here, how the "if and only if" wouldn't make sense since the sum of squares from 1 to 24 is 4900 (which, like he said, divisible by 25).
ReplyDeleteThe existence of a generator mod p is hard to prove but once you have it, you get that this sum is divisible by p pretty quickly. If you want me to write some of the details of that, let me know, though it probably won't do any good unless you know a bit of number theory. I haven't found a more elementary proof that doesn't depend on knowing that fact -- specifically I don't know any proof accessible to a high school student.
ReplyDeleteJoshua,
ReplyDeleteIf you want to write up a proof I will insert it into the original post as addendum.. much better than in comments. Even if the proof is not easily understandable by a HS student, exposing them to some more mature work is nice. I read lots of proofs that still leave me baffled, but enjoy extracting the tidbits of progress I can digest.
Thanks