Finding Remainder of P Divided by 8
We are given that \(P = N^2\), where \(N\) is an odd integer. We need to find the remainder when \(P\) is divided by 8.
Representing an Odd Integer
An odd integer \(N\) can be expressed in the general form \(N = 2k + 1\), where \(k\) is any integer.
Calculating P = N^2
Substitute the expression for \(N\) into the equation for \(P\):
\(P = (2k + 1)^2\)
Expand the square:
\(P = (2k)^2 + 2(2k)(1) + 1^2\)
\(P = 4k^2 + 4k + 1\)
Simplifying the Expression for P
Factor out \(4k\) from the first two terms:
\(P = 4k(k + 1) + 1\)
Consider the product \(k(k + 1)\). Since \(k\) and \(k + 1\) are consecutive integers, one of them must be even. Therefore, their product \(k(k + 1)\) is always an even number.
Let \(k(k + 1) = 2m\), where \(m\) is an integer.
Substitute \(2m\) back into the expression for \(P\):
\(P = 4(2m) + 1\)
\(P = 8m + 1\)
Determining the Remainder
The expression \(P = 8m + 1\) shows that \(P\) is one more than a multiple of 8.
Therefore, the remainder when \(P\) is divided by 8 is 1.
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