Prove every odd integer is the difference of two squares I know that I should use the definition of an odd integer ($2k+1$), but that's about it.
Thanks in advance!

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 $49=7^2$)
