The  square root of a prime number is irrational

I searched the CTK archives and did not find these proofs.

Prove for any prime p, sqrt(p) is irrational.

Case 1 by contradiction.
Assume sqrt(p) = a/b for integers a,b gcd(a,b) = 1.
Then p = a^2/b^2 and b^2p = a^2. This implies p divides a.
Let a = pr for some integer r. Then pb^2 = p^2r^2 and
b^2 = pr^2. This implies p divides b contradicting gcd(a,b) = 1.
Therefore the assumption that sqrt(p) = a/b was false and sqrt(p)
is irrational.

Case 2. (Euclid assuming uniqueness of integer factorization into primes
except for the order of the factors)

Assume sqrt(p) = a/b. Then p = a^2/b^2 or pb^2 = a^2. Since p has
only 1 prime factor and every square number has an even number of
prime factors, we have a number with an odd number of prime factors
= a number with an even number of prime factors, a contradiction.
Therefore, the assumption that sqrt(p) = a/b is false and sqrt(p)
is irrational for all primes p.

A side note.

This argument is obvious if you assume uniqueness of factorization. I
tried it on a college exam having no idea about the Fundemental
Theorem. The graduate student grader gave me  0 points for the
problem. I protested to no avail. This was 1961.  On a 20 point
problem what should I have got? Since uniqueness of integer factorization
is used to establish irrationality of the sqrt(p) for prime p, I wonder if some how
we could use the proof in case 1 to prove unuqueness of factorization?

Looks like the case of converse not true, but, it is a thought.

Cino