The difference of two consecutive perfect squares is always odd I am working on another homework assignment about proofs. The question is:
Prove or find counterexample: the difference of two consecutive perfect squares is odd?
There is no counterexample correct? I am thinking this is always true. If I were to do 7^2-6^2 the answer is odd. I am unsure of how to start the proof though. I am new to proofs and not sure what to really do

Well of course it is.
Consider $n^2$ and the next greater perfect square would be $(n+1)^2$, which factors to $n^2+2n+1$. Now what would the difference between them be?
$n^2+2n+1-n^2=2n+1$.  
All odd integers conform to 2n+1 where n is an integer.
There is also this to consider.
$1=1^2$
$1+3=2^2$
$1+3+5=3^2$
$1+3+5+7=4^2$
$1+3+5+7+9=5^2$
$1+3+5+7+9+11=6^2$
$1+3+5+7+9+11+13=7^2$
...and so on
