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I'm working on a fairly simple Physics problem (particle in an infinite potential cube), and I'm asked:

"Are any of the energy eigenvalues degenerate? If so, what is the degeneracy?".

The expression for the energy (ignoring a constant multiplier) comes to the form

$E = a^2 + b^2 + c^2$

Where a,b,c are all integers.

Degeneracy occurs when some arrangement of these values gives the same energy value as some different arrangement. (So any state without a = b = c will have at least triple degeneracy via permutations).

Is there some closed-form solution or sequence for how many ways an integer can be written as a sum of three integer squares?

Martin Sleziak
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Liam Bonds
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    I don't think there is a nice formula (in general, an even number of squares is easier). Also look at this: https://en.wikipedia.org/wiki/Legendre%27s_three-square_theorem – Kenta S Aug 09 '21 at 01:57
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    see https://mathoverflow.net/questions/3596/is-there-a-simple-way-to-compute-the-number-of-ways-to-write-a-positive-integer/22655#22655 – Will Jagy Aug 09 '21 at 02:12
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    See also https://math.stackexchange.com/questions/2068160/determining-the-number-of-ways-a-number-can-be-written-as-sum-of-three-squares – Gerry Myerson Aug 09 '21 at 02:59
  • Care to engage with the comments, Liam? – Gerry Myerson Aug 11 '21 at 13:36
  • Sorry not used to Stack-Exchange etiquette! Thanks for the link to the previously asked question. I'm curious as to the difficulty of the problem, is it too big to tackle, or is it just not particularly useful/desirable? – Liam Bonds Aug 12 '21 at 09:03
  • I'm planning to look through the index of integer sequences to see how far its already been worked out – Liam Bonds Aug 12 '21 at 09:09
  • Number of representations as a sum of three squares is harder than two squares or four squares. I'm not sure if it can be done without delving into modular forms, fairly advanced stuff. But it is considered quite mainstream Number Theory. It's tabulated out to $10,000$ at https://oeis.org/A005875 along with lots of links to the literature. – Gerry Myerson Aug 13 '21 at 10:26
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    There's a formula at oeis: For sums of 3 squares r_3(n): write (uniquely) -n=D(2^vf)^2, with D<0 fundamental discriminant, f odd, v>=-1. Then r_3(n) = 12L((D/.),0)(1-(D/2)) Sum_{d | f} mu(d)(D/d)sigma(f/d).

    Here mu is the Moebius function, (D/2) and (D/d) are Kronecker-Legendre symbols, sigma is the sum of divisors function, L((D/.),0)=h(D)/(w(D)/2) is the value at 0 of the L function of the quadratic character (D/.), equal to the class number h(D) divided by 2 or 3 in the special cases D=-4 and -3. - Henri Cohen (Henri.Cohen(AT)math.u-bordeaux1.fr), May 12 2010

     – Gerry Myerson Aug 13 '21 at 10:33

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