see below 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?
A^2 if divided by 4 the remaining is 1
Then A^2 = 4n+1 .... now A = sqrt (4n+1)
Since 4n+1 is odd number, then sqrt (4n+1) also odd number , then A is odd number
Now (A+5)^2 = A^2 + 10A + 25
if we divided it by 4
then (A+5)^2/4 = A^2/4 + 10A/4 + 25/4
= A^2/4 + 2(5A/2) + 25/4
now we need to obtain the remaining:
1) now we know that the remaining of A^2/4 is 1
2) the remaining of 5A/2 is 1 since 5A is odd number because A is odd and 5 is odd (odd*odd) = odd
then the remaining of 10A/4 is 2 (1) = 2
3) the remaining of 25/4 is 1
Now add all remainders , you will get 4 remaining, but since we are dividing by 4 , then the remaining is 0
then the remainings of dividing (A+5)^2 by 4 equals 0
If A^2 = 4q+1, then to find the remainder if we divide A+5 by 4.
A^2= 4q+1 , holds, when q=2, 5, 12, 20, etc. for which A = 3, 5, 7, 9, 11,
So, A+5 gives a remainder: 3+5= 8 remainder is 0.
A=5, remainder whenyou divide 5+5 by 4 is 2,
A=7, remainder when you divide 7+5 by 4 is 0,
A=9, remainder when we divide 9+5 by 4 is 2.
A=11, remainder when we divide by11+5 by 4 =0
So all the odd number square is giving a remainder 1 when we divide by 4.
So every umber 4n-1 (n is an ineger) , 4n-1+5 = 4n-4 = 4(n-1) is divisible by 4 and the remainder is zero.
For every number 4n+1 ( where n is an integer), 4n+1+5 = 4n+6 = 4n+6 = 4(n+1)+2 , if we divide by 4, we get a remainder 2.