Prove every odd integer is the difference of two squares

Question

I know that I should use the definition of an odd integer ($2k+1$), but that's about it.

Thanks in advance!

Answer 1

Step 1: pick an odd number (like $n=13$ here)

Step 2: bend it in "half" (any odd number $n$ can be written as $2k+1$, and $13=2\cdot 6 + 1$)

Step 3: fill in the blank space

Step 4: Count squares. (Here, the blue square has area $36=6^2$, while the whole square has area $49=7^2$)

Answer 2

Hint: Consider the difference of two consecutive squares. What is $(k+1)^2-k^2$?

Answer 3

HINT: $$\begin{align} &2k + 1 \\= & 1\cdot(2k + 1) \\ =& \left(k + 1 - k \right)\left(k + 1 + k\right) \\ = & \cdots\end{align}$$

Answer 4

Eric and orlandpm already showed how this works for consecutive squares, so this is just to show how you can arrive at that conclusion just using the equations.

So let the difference of two squares be $A^2-B^2$ and odd numbers be, as you mentioned, $2k+1$. This gives you $A^2-B^2=2k+1$.

Now you can add $B^2$ to both sides to get $A^2=B^2+2k+1$. Since $B$ and $k$ are both just constants, they could be equal, so assume $B=k$ to get $A^2=k^2+2k+1$. The second half of this equation is just $(k+1)^2$, so $A^2=(k+1)^2$, giving $A = ±(k+1)$, so for any odd number $2k+1$, $(k+1)^2-k^2=2k+1$.