Question 898721
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?
<pre>
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)<sup>2</sup>= 4n<sup>2</sup>-4n+1=4(n<sup>2</sup>-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)<sup>2</sup>= 2<sup>2</sup>n<sup>2</sup> = 4n<sup>2</sup>  

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)<sup>2</sup> = [(2n-1)+5]<sup>2</sup> = (2n+4)<sup>2</sup> = [2(n+2)]<sup>2</sup>.

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</pre>