First problem explanation!

So this is my first problem explanation. I wrote it mostly to help myself, because I need experience writing proofs and stuff like that. (When I find some more time, I'll make the expressions easier to read using latex or something)

Find all prime numbers p such that 32p + 1 is the cube of an integer.
Source: AwesomeMath 2008 Admission Test A (the deadline for passing it in is way overdue, so I figure it's ok to post a solution)

Things you should know beforehand:
Factorizations
Parity operations (Even * even = ?, Odd + even = ? Stuff like that)
Properties of prime numbers and prime factorizations

Rewrite this as 32p + 1 = n^3, where n is an integer. Subtract 1 from each side, and we get
32p = n^3 - 1. Using difference of cubes, we can factor the right side as (n-1)(n^2 + n + 1). Now we can work exclusively with the right side for a bit. Realize that this side (n-1)(n^2 + n + 1) must be even for it to be divisible by 32. So let's consider two cases: n is even, or, n is odd.

n is even: (For notation purposes, E is an even number and O is an odd number)
(E - 1)(E^2 + E + 1)
(O)(E + E + 1)
(O)(E + 1)
(O)(O)
O
Therefore, n cannot be even.

n is odd:
(O - 1)(O^2 + O + 1)
(E)(O + O + 1)
(E)(E + 1)
(E)(O)
E
therefore, since n cannot be even, and it is possible for n to be odd, n must be odd.

Let's also keep in mind from our work that (n-1) is even and (n^2 + n + 1) is odd.
We want our final product, (n-1)(n^2 + n + 1), to be divisible by 32. Thinking in term of prime factors, 32 = 2^5. So our product must have at least 2^5, or 5 two's, in its prime factorization. Also keep in mind that since (n^2 + n + 1) is odd, it will have no 2's in its prime factorization, so all of them must come from (n-1). Also, if n-1 has any other prime factors besides 2^5, then when we solve for p by dividing (n-1)(n^2 + n + 1) by 32, p will not be prime.

From this, we can conclude that (n-1) MUST equal 32, because our product must be divisible by 32, and (n^2 + n + 1) MUST be prime, because when we solve for it, it is equal to our p.

Letting n - 1 = 32, n = 33. Plugging n = 33 into n^2 + n + 1, we get 1123. We can easily check 1123 against a list of prime numbers, to discover that it is indeed prime. Therefore, the only prime p which satisfies the original argument is 1123.

Of course, we should always check our solution to make sure it works:

32p + 1 = n^3
32(1123) + 1 = n^3
35936 + 1 = n^3
35937 = n^3
n = cuberoot(35937) = 33