SOLUTION: If one is the remainder when A to the second power is divided by 4, what would the remainder have to be if (A+5) to the second power is divided by 4?

Lemma 1:
The square of every odd positive integer leaves remainder 1 when
divided by 4:

Proof: Every odd integer can be expressed as 2n-1 for some positive integer n.

(2n-1)2= 4n2-4n+1=4(n2-n)+1 

That is 1 more than a multiple of 4 so it leaves remainder 1 when 
divided by 4.

Lemma 2:
The square of every even positive integer leaves remainder 0 when
divided by 4:

Proof: Every even integer can be expressed as 2n for some positive integer n.

(2n)2= 22n2 = 4n2  

That is a multiple of 4 so it leaves remainder 0 when 
divided by 4.

Therefore by those two lemmas, A can be any odd number (and cannot be an even
number).

Therefore (A+5)2 = [(2n-1)+5]2 = (2n+4)2 = [2(n+2)]2.

That's the square of an even integer and by lemma 2 above it has remainder 0
when divided by 4.

Answer: the remainder is 0

Edwin