Question

In quantum mechanics and Pythagorean triples.

Answer 1

In quantum mechanics the squared length of the angular momentum vector is NOT \(L^2\) but instead it turns out to be \(L(L+1)\). We can use the 'rising factorial notation' to write this as:

\[L^{\bar 2} = L(L+1)\]

Just to give another example of rising factorials,

\[x^{\bar 5} = x(x+1)(x+2)(x+3)(x+4)\]

So now I got to wondering. Are there any integers that satisfy this equation:

\[a^{\bar 2} + b^{\bar 2} = c^{\bar 2}\]

and is there any interesting relationship to the integer solutions of this?

\[a^2+b^2=c^2\]

Answer 2

a(a+1) + b(b+1) = c(c+1)

a^2 + b^2 = c^2 + c - a - b


Pythagorean triples that satisfy c = a + b should work. Let me see if there are any such pythagorean triples

Answer 3

This is exactly what I had in mind when I came up with this question that made me decide to ask it actually, it turns out that there are NO pythagorean triples that satisfy this! >:))

Answer 4

clearly c^2 is not a^2 + b^2
So scratch that

Answer 5

I mean
c = a + b implies c^2 = a^2 + b^2 has only trivial solutions

Answer 6

Yeah, my reasoning was the other way around but same result as you. If it's a pythagorean triple then that means it obeys the triangle inequality \(a+b > c\) so \(a+b \ne c\).

I'm sorta having fun looking into what the solutions of this look like just finished a program to brute force some of them, the first "quantum triple" if I can call them that goofy name is:

\[2^{\bar 2}+2^{\bar 2}=3^{\bar 2}\]

Answer 7

\[ 2^{\bar 2} + 2^{\bar 2} =3^{\bar 2} \]\[ 3^{\bar 2} + 5^{\bar 2} =6^{\bar 2} \]\[ 4^{\bar 2} + 9^{\bar 2} =10^{\bar 2} \]\[ 5^{\bar 2} + 6^{\bar 2} =8^{\bar 2} \]\[ 5^{\bar 2} + 14^{\bar 2} =15^{\bar 2} \]\[ 6^{\bar 2} + 9^{\bar 2} =11^{\bar 2} \]\[ 6^{\bar 2} + 20^{\bar 2} =21^{\bar 2} \]\[ 7^{\bar 2} + 27^{\bar 2} =28^{\bar 2} \]\[ 8^{\bar 2} + 10^{\bar 2} =13^{\bar 2} \]\[ 9^{\bar 2} + 13^{\bar 2} =16^{\bar 2} \]\[ 9^{\bar 2} + 21^{\bar 2} =23^{\bar 2} \]\[ 10^{\bar 2} + 26^{\bar 2} =28^{\bar 2} \]\[ 11^{\bar 2} + 14^{\bar 2} =18^{\bar 2} \]\[ 11^{\bar 2} + 20^{\bar 2} =23^{\bar 2} \]\[ 12^{\bar 2} + 17^{\bar 2} =21^{\bar 2} \]\[ 12^{\bar 2} + 24^{\bar 2} =27^{\bar 2} \]\[ 14^{\bar 2} + 14^{\bar 2} =20^{\bar 2} \]\[ 14^{\bar 2} + 18^{\bar 2} =23^{\bar 2} \]\[ 15^{\bar 2} + 21^{\bar 2} =26^{\bar 2} \]\[ 17^{\bar 2} + 22^{\bar 2} =28^{\bar 2} \]\[ 18^{\bar 2} + 25^{\bar 2} =31^{\bar 2} \]\[ 20^{\bar 2} + 26^{\bar 2} =33^{\bar 2} \]\[ 21^{\bar 2} + 29^{\bar 2} =36^{\bar 2} \]\[ 23^{\bar 2} + 30^{\bar 2} =38^{\bar 2} \]\[ 25^{\bar 2} + 27^{\bar 2} =37^{\bar 2} \]

Answer 8

If I didn't mess up my program, that should be the first 30 of them, I guess my next thing I'm interested in is seeing if there are infinitely many or if there are "primitive" ones like there are for Pythagorean triples :P

Answer 9

https://repl.it/C8E2

That's my code if you're interested in playing around.

Answer 10

Interesting, turns out that this is the same thing as asking,

What triangular numbers when summed together make another triangular number?

since all of the terms here are two consecutive numbers, one of them must be even so they are all divisible by 2.
\[\frac{a(a+1)}{2}+\frac{b(b+1)}{2}=\frac{c(c+1)}{2}\]

Answer 11

Triangular numbers... this firm looks nice

Answer 12

Triangular numbers... this form looks nice

Answer 13

Yeah interesting interplay here cause the sum of two consecutive triangular numbers is a perfect square, so kinda funky stuff going on right now.

Answer 14

https://drive.google.com/file/d/0B7ss_Y8M_VEgRHBSZmlDWnZvQmc/view?usp=drivesdk

Answer 15

Ahhh Gauss strikes again. So it seems that he's already proven that every number is the sum of 3 triangular numbers, including 0.

That means every triangular number is the sum of 3 triangular numbers. Clearly this is true when 2 of the numbers are 0, and it also may be true when none of them are 0, but I am interested specifically in pushing this around so that it's true only when one of them is 0! Cool and good to know!

I think those two identities are going to interest me quite a bit :D