A two dimensional square box with a corner length L, the energy level of a particle with a mass #h^2/(8mL^2)(n_1^2+n_2^2)# . What is the quantum number of the first, second and third degenerate energy levels?

I don't know what you mean by corner length... I assume it is just an #L xx L# square, because that is what the given energy expression implies.

Degenerate energy levels are those for which the total energy is the same, i.e. if

#n = n_1 + n_2#
#n_1, n_2 = 1,2,3,4,5, . . . #

and

#E_(n_1n_2) = h^2/(8mL^2)(n_1^2 + n_2^2)#

then we want all combinations of increasing #n_1^2 + n_2^2# for which there exist more than one possibility for the same #E_(n_1n_2)#.

The quantum number we want is #n -= n_1+n_2#.

You should be working these out as trial and error, but here are the first 15 combinations, degenerate or not:

#ul(n_1" "" "n_2" "" "n_1^2+n_2^2" "" "E_(n_1n_2)" "" "" ""degeneracy"#
#1" "" "color(white)(/)1" "" "2" "" "" "" "color(white)(/)2h^2//8mL^2" "" "1#
#1" "" "color(white)(/)2" "" "5" "" "" "" "color(white)(/)5h^2//8mL^2" "" "color(blue)(bb2)#
#2" "" "color(white)(/)1#
#2" "" "color(white)(/)2" "" "8" "" "" "" "color(white)(/)8h^2//8mL^2" "" "1#
#1" "" "color(white)(/)3" "" "10" "" "" "color(white)(/.)10h^2//8mL^2color(white)(/)" "color(blue)(bb2)#
#3" "" "color(white)(/)1#
#2" "" "color(white)(/)3" "" "13" "" "" "color(white)(/.)13h^2//8mL^2color(white)(/)" "color(blue)(bb2)#
#3" "" "color(white)(/)2#
#1" "" "color(white)(/)4" "" "17" "" "" "color(white)(/.)17h^2//8mL^2color(white)(/)" "color(blue)(bb2)#
#4" "" "color(white)(/)1#
#3" "" "color(white)(/)3" "" "18" "" "" "color(white)(/.)18h^2//8mL^2color(white)(/)" "1#
#2" "" "color(white)(/)4" "" "20" "" "" "color(white)(/.)20h^2//8mL^2color(white)(/)" "color(blue)(bb2)#
#4" "" "color(white)(/)2#
#3" "" "color(white)(/)4" "" "25" "" "" "color(white)(/.)25h^2//8mL^2color(white)(/)" "color(blue)(bb2)#
#4" "" "color(white)(/)3#

Clearly, the first three degenerate levels, which are highlighted, have:

1. #n = n_1 + n_2 = 1 + 2 = bb3#
2. #n = n_1 + n_2 = 1 + 3 = bb4#
3. #n = n_1 + n_2 = 2 + 3 = bb5#

Note that even if #n# is the same for two combinations of #n_1# and #n_2#, they might NOT be degenerate levels. For example, #2+3 = 5#, and #1+4 = 5# as well, but #2^2 + 3^2 = 13# and #1^2 + 4^2 = 17#, so those levels are NOT the same energy.

Also, at most you can only have a 2-fold degenerate energy because there are only two permutations you can have of two numbers.