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 $49=7^2$)
You could add. 

For mod ((2x+1),4)= 1, 2x+1 has at least one more solution than (k^2+1)-k^2=2x+1 and k^2 will be odd and (k-n)^2 will be even. Opposite if  Mod((2x+1),4)= 3. n is always odd. 

k^2-(k+n)^2= 2x+1, if 2x+1 is a composite number, if not, 2x+1 only has the solution (k^2+1)-k^2=2x+1 and 2x+1 is a prime number. 

Sorry Mathjax is not working on my computer.