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%I need to add the picture of squares to this. 2N+1 on each side. 
 
 \noindent
  February 10, 2023\\
   Math Class

\begin{center}
{\large {\bf   Odd Squares Minus One Are Divisible by Eight}}
\end{center}

\noindent
 Theorem:  Suppose a square's side is an odd number. Then the area of the square minus one is divisible by eight.  
 
 \noindent 
  Proof: Let the side of the square have length X. Since X is odd, we can write it as 
 X = 2N+1 for some number N. What we must show is that $(2N+1)^2 -1 $ is divisible by 8. 
 \begin{eqnarray}
  (2N+1)^2 -1 &  = & (2N+1)(2N+1) -1\\
   && \nonumber\\
  & = &(2N+1) (2N) + (2N+1) (1) -1 \nonumber\\
   && \nonumber\\
  & = &2N \cdot 2N + 2N\cdot 1+   2N\cdot 1 + 1\cdot 1  -1\\
   && \nonumber\\
  & = & 4N^2 + 2N + 2N + 1 -1 \\
   && \nonumber\\
  & = &  4N^2 + 4N\\
   && \nonumber\\
  & = &4 (N^2+N)\\
   && \nonumber\\
  & = & 4 N (N+1)  
  \end{eqnarray}

 Clearly this is divisible by 4. But we can show that $N(N+1)$ is even, so  $N(N+1)$ is divisible by 2 and the whole thing is divisible by $4 \cdot 2 = 8$. The reason $N(N+1)$ is even is that either $N$ is even or $(N+1)$ is even, so $N(N+1)$ is an even number multiplies by an odd number. But anything multiplied by an even number is even. So $N(N+1)$ is even, so it is divisible by 2 and $4N(N+1)$ is divisible by 8.  QED. 


 
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