Prove that if a is a positive integer of the form 4n+3, then at least one prime divisor of a is of the form 4n+3.

I'm suck D: The book gave a hint to prove the contrapositive, but I'm still not sure what to do. I thought of a proof and at first I thought it was right, but then on re-examination I see I made a mistake. I'll type it up because maybe there is some way to scavange my wrong proof and make it right.

Let S={a in N|a=4t+3 and a is not divisible by a prime of the form 4n+3}[/tex]

Suppose S =/= empty set, then by the Well Ordering Principle there exists a smallest element. Call it n. n can't be a prime because it would then be divisible by a prime of the form 4t+3. Hence n is composite and there exists integers x and y s.t. n=xy where 1<x<n and 1<y<n. Since x<n then x is not in S (this is where I made my mistake), then x is divisible by a prime of the form 4t+3 (I can't make this conclusion, right? Because x may not even be of the form 4t+3... I'll finish what I wrote out). Call this prime p. But since p|x and x|n, then p|n - a contradiction. Thus S= empty set.

Help would be appreciated. Thanks!