Just as a remark, proving that given a prime n, n^2 = 1(mod 24)[equivalent to n^2-1 is multiple of 24] is pretty easy:
i) prove that n^2 = 1(mod 3). Enumerating, n = {-1,1} (mod 3) (mod 3) [since n is prime, n != 0 (mod 3)]. n^2 = 1 (mod 3)
n^2 - 1 = 0 (mod 3). Exists m such that n^2-1 = 3m.
ii) prove that n^2 = 1(mod 8). Enumerating, n = {1,3,-3,-1} (mod 8) [ n != pair (mod 8) since then, it would be divisible by 2]. n^2 = {1,9,9,1} (mod 8) => n^2 = {1,8+1}(mod 8) => n^2 = 1 (mod 8). Exists k such that n^2 - 1 = k8.
iii) There exists 2 integers m,k such that k8 = m3. m must have an 8 factor and k a 3 factor. Then, there exists j such that j = m/8 = k/3 = (n^2-1)/24. qed.
ii) prove that n^2 = 1(mod 8). Enumerating, n = {1,3,-3,-1} (mod 8) [ n != pair (mod 8) since then, it would be divisible by 2]. n^2 = {1,9,9,1} (mod 8) => n^2 = {1,8+1}(mod 8) => n^2 = 1 (mod 8). Exists k such that n^2 - 1 = k8.
iii) There exists 2 integers m,k such that k8 = m3. m must have an 8 factor and k a 3 factor. Then, there exists j such that j = m/8 = k/3 = (n^2-1)/24. qed.