Question

Prove the following statement.

The square of any odd integer has the form 8m+1 for some integer m.

Answer 1

Answer:

You can prove this statement as follows:

Step-by-step explanation:

An odd integer is a number of the form $2k+1$ where $k\in \mathbb{Z}$. Consider the following cases.

Case 1. If $k$ is even we have: $(2k+1)^{2}=(2(2s)+1)^{2}=(4s+1)^{2}=16s^2+8s+1=8(2s^2+s)+1$.

If we denote by $m=2s^2+2$ we have that $(2k+1)^{2}=8m+1$.

Case 2. if $k$ is odd we have: $(2k+1)^{2}=(2(2s+1)+1)^{2}=(4s+3)^{2}=16s^2+24s+9=16s^{2}+24s+8+1=8(2s^{2}+3s+1)+1$.

If we denote by $m=2s^{2}+24s+1$ we have that $(2k+1)^{2}=8m+1$

This result says that the remainder when we divide the square of any odd integer by 8 is 1.