Duh.

Well, it is arguably leaner.

p prime and >3.
p^2 - 1 = (p-1)*(p+1).

p-1, p, p+1 are 3 consequtive numbers so one of them is divisible by 3 and since p is prime , 3 divides p-1 or p+1 and definitely divides their product p^2-1.

p-1 and p+1 are even numbers and so is their product .

With p-1 = 2n and p+1 = 2(n+1) it gives

p^2 - 1 = 2n*2(n+1)= 4 n(n+1) = 24*(n(n+1)/6) =24k with k=n(n+1)/6

(i) Does any number in this sequence end in 2, 4, 7, or 9?

Consider the same sequence written in base-16 notation: 0, 1, 3, 6, A, F, 15, ...

Here's mine:

Suppose all numbers are expressed in base m.

Then the rightmost digit R(x) of any integer x is given by

R(x) == x (mod m).

(I write it this way because <chessgames.com> won't let me use the regular equivalence symbol.)

Define S(n) = 1 + 2 + ... + n = n(n+1)/2.

Let a = S(m-1) = m(m-1)/2 and a' = S(m) = m(m+1)/2.

Then a' - a = m, so R(a') = R(a).

Moreover, if

b = S(m-2) = a - (m-1)

and

b' = S(m+1) = a' + (m+1),

then b' == b (mod m), so

R(b) = R(b').

Similarly, if c = S(m-3) and c' = S(m+2), then R(c) = R(c'), and so on.

In other words, the sequence of the first m rightmost digits repeats in reverse order in the next m rightmost digits.

Hence the rightmost digit of S(2m-2) is the same as the rightmost digit of S(1) = 1, and the rightmost digit of S(2m-1) is 0.

Of course, this follows directly from the fact that S(2m-1) = m(2m-1) is a multiple of m.

Since S(2m) = m(2m+1) is also a multiple of m, its rightmost digit is 0 also. Furthermore, S(2m+1) = S(2m) + 2m+1, so R[S(2m+1)] = 1 =R[S(1)], and S(2m+2) = S(2m+1) + 2m+2, etc., so R[S(2m+2)] = 3 = R[S(2)]. The sequence of rightmost digits keeps repeating endlessly with periodicity 2m. If a digit does not appear in the rightmost place in the first m terms, it will never appear.

Now suppose that the base is a power of 2. Let m = 2^p.

Can two numbers k = S(n) and k' = S(n') have the same rightmost digit?

We can restrict consideration to cases 0 < n < n' < m because of the way the sequence repeats.

The condition that k and k' have the same rightmost digit is k' == k (mod m), or

k' - k = n'(n'+1)/2 - n(n+1)/2 = qm

for some number q. If we set n' = n+d, this becomes

(n+d)(n+d+1) - n(n+1) = 2qm.

This equation can be rewritten as

d(2n+d+1) = 2^(p+1) q.

The left hand side of the equation must be divisible by 2^(p+1) = 2m.

If d is even, then 2n+d+1 is odd, so d must be divisible by 2^(p+1). But d < m by definition, so this is impossible.

If d is odd, then 2n+d+1 = n+n'+1 must be divisible by 2^(p+1). But n+n'+1 < 2m, so this too is impossible. Hence R(k) and R(k') are different.

Therefore all m digits must occur in the rightmost position of the first m terms of the sequence.

Suppose you have a chessboard made of some flexible, stretchable material. If you fasten the left edge and the right edge of the board together so that the a-file and h-file are adjacent, a king, for example, could move directly between the two files. (You don't have to spoil a perfectly good vinyl chessboard for this; just imagine that they are adjacent, i.e., hold these relationships in your mind.) Now the chessboard has the topology of a cylinder. A second way to turn the chessboard into a cylinder is to identify the top and bottom edges, so that a king can move directly between the first and eighth ranks.

Now imagine *both* pairs of opposite sides joined, left to right and top to bottom. This turns the chessboard into a torus. It can be done either by identifying the left and right edges first and then the top and bottom, or by identifying the top and bottom edges first and then the left and right. I'll call these two kinds of torus “horizontal” and “vertical,” respectively. In the first kind the files become major circumferences and the ranks become minor circumferences. In the second kind it's the other way around. The two are distinct but completely equivalent topologically.

Now a king on a1 is able to move directly to a8, b8, h1, h2, and h8, as well as the usual moves to a2, b1, and b2. A knight on a1 can move directly to b7, c8, g2, h3, g8, and h7, as well as the usual moves to b3 and c2. But there is nothing special about a1. No square has any property that isn't shared by all the others. None is in a privileged position.

Depending on which way the board is curved in the first step, the playing surface can end up either on the inside or the outside of the torus. So altogether there are four ways to make the topology of the chessboard toroidal: “horizontal inside,” “horizontal outside,” “vertical inside,” and “vertical outside,” or HI, HO, VI, and VO, respectively.

Imagine that some kind of intelligent two-dimensional life form lives on the playing surface of the torus. It can distinguish left and right, forward and backward, but it has no way of perceiving up and down, the third dimension. It understands topology. It therefore knows that it is living on a toroidal surface, but it can no more visualize a three-dimensional torus than I can visualize a hypersphere. Assume that it has thoroughly explored its 2D world. It has traversed all the major and minor circumferences (ranks and files, and the diagonals as well). It has even labeled them so that it can tell where it is (but of course it doesn't use algebraic notation, because that would presuppose a way of telling which are ranks and which are files).

(a) Show, or at least convince yourself, that it can't determine by its own devices which kind of torus it is living on (HI, HO, VI, or VO).

(b) Suppose it somehow learns (let's say, from an alien life form) that it is living on the outside surface. How can it then determine whether it is an HO or VO torus?

There are two ways to make a circuit that comes back to the starting point, one that encircles the hole in a doughnut (a major circumference) and one that doesn't (a minor circumference). The torus has an inner surface and an outer surface.

(a) Does a 2D being have any way to tell which is the major circumference and which is the minor? Does it have any way to tell whether it is living on the inner surface or the outer?

(b) If it somehow learns which surface it is on, can it then tell which is the major circumference and which is the minor?

https://www.youtube.com/watch?v=N9q...

A couple of times in the video, Weird Al and some other dude I don't recognize are dancing in front of a white screen. The screen has a mathematical equation on it.

Q1) Fill in the blanks:

What is shown is the __________ equation for a __________ .

Q2) (A two-part question) Is there one or more errors in the equation? If so, what is/are the errors?

Since the kinetic energy term is written using the Laplacian, it's three-dimensional. (The angular dependence of the wave function hasn't been factored out.) The argument r of psi, the wave function, should therefore be bold-face to show this.

I typed a long reply. Then, before posting, I realized although your answer has multiple merits going for it, there are a couple of things that still leave me puzzled and unsure about <aw> the person.

Oh, and unless my eyes or memory deceived me, you missed one error.

I apologize for bothering you this day, especially since my selfish purpose was thwarted.

Before posting I did a little googling and discovered that there is widespread confusion about the definition of the center-of-mass coordinate system and the reduced mass mu. That discovery was reflected in my response.

The answers to part (a) (as rephrased) are No and No,

At the time I posted I thought that the answers to parts (b) and (c) are Yes and Yes.

Since then I have thought some more and I now think I was wrong. There is no way for 2D individuals living on a toroidal surface to discover which is the minor and which is the major circumference, even if they know whether they are on the inner or outer side, or to know which side they are on even if they know which circumference is which.

I am happy to agree that you were "spot-on" (as the British Soccer commentators say) in regards to the reduced mass. In fact, I trust you on that particular point more than I trust myself.

In any event, I wasn't much interested in that portion of your answer. And I did no math or googling whatsoever to confirm my impression that you're spot-on. Color me lazy or unconcerned. Guilty of both.

<Unfortunately, <chessgames.com> doesn't allow me to use the Greek alphabet or common mathematical symbols. This put a crimp in my style.>

You and me both.

<What's the one error?>

In the kinetic energy term of the Hamiltonian, Planck's constant should be the reduced Planck's constant, aka h-bar. I knew I could be wrong, since I am having to remember 18 years ago (the last time I solved any quantum mechanics problems). So I checked two trusted sources, and they validated my recollection.

https://www2.ph.ed.ac.uk/~ldeldebb/...

In the kinetic energy term of the Hamiltonian, Planck's constant should be the reduced Planck's constant, aka h-bar.>

You're right, of course.

Now find another one.

Your discussion about the radial Schrodinger equation may have suggested h to my unconscious, particularly after I started thinking about letters of the alphabet because of your hint.