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