Modular arithmetic proof (check please?)

A simple Q from courant's book:
Question as stated in the book:
We have seen that

x^{2}\equiv (p-x)^{2}(mod p)

. Show that these are the only congruences among the numbers

1^{2},2^{2},3^{2},...,(p-1)^{2}

.

The question one its own I found quite ambigious, as I think p must be a prime, which is implied by the previous pages but not explicitly stated. So, my (hopefully equivalant) interpretation is: considering p is a prime, show that no other number pairs from the set

{1^{2},2^{2},3^{2},...,(p-1)^{2}}

are congruent module p other than

x^{2}\equiv (p-x)^{2}(mod p)

.

Proof:

Assume

x^{2}\equiv (p-y)^{2}(mod p)

, for

1\leq x\leq p-1

, and

1\leq p-y\leq p-1\Rightarrow 1\leq y\leq p-1

, and

x+y\neq p

Then, we have

x^{2}\equiv (p-y)^{2}(mod p)\Rightarrow (x+p-y)(x-p+y)\equiv 0(mod p)\Rightarrow p|x-y

p|x+y

as p is a prime.

We know x-y is less that p , so p divides this only if x=y, gives the congruent pair stated in the question.
Also, x+y is positve, less than 2p and not equal to p, so p can not clearly divide this.

Hence no other congruent pair exist. Question was quite easy, so hopefully no mistakes.