Find the probability that $m^2-n^2$ is divisible by $4$ Two integers are $m$ , $n$ chosen at random with replacement from the set of integers $1,2,3..9$ . Find the probability that $m^2-n^2$ is divisible by $4$ .
my solution : Two integers can be chosen (with replacement ) in $9\times 9=81$ ways .
$m^2-n^2 $ is divisible iff either both $m,n$ are  odd or both are even . 
There are $4\times 4=16 $ ways way of obtaining both even integers and $5\times 5=25 $ ways of obtaining both odd integer. 
So total number of favorable outcome $=16+25=41$ .
Hence required probability = $\frac{41}{81}$ .
Now i would like to slightly change the question :
Two integers are $m$ , $n$ chosen at random without replacement from the set of integers $1,2,3..9$ . Find the probability that $m^2-n^2$ is divisible by $4$
my solution when order consideration is relevant  : 
Total number of obtaining two integers $= 2\times 9\times 8=2\times 72$
Numbers ways of obtaining two even numbers $=2\times 4\times 3 $ and number of ways obtaining odd numbers $=2\times 5\times 4$ .
So total number of favorable cases= $2(4\times 3+5 \times 4)=2\times 32 $ .
Hence required probability $=\frac{2 \times 32 }{2 \times 72 }=\frac{32}{72}$
my solution when order is considered irrelevant 
Total numbers of cases $=9\times 8=72$
Numbers of ways obtaining both even and both odd integers is $4\times3=12$ and $5\times4=20$ respectably .
So number of favorable cases$=12+20=32$ .
Hence probability =$\frac{32}{72}$ 
Is my understanding in above problems correct ? 
Please point out if i calculated anything wrong?
Thank you 
 A: Everything is fine, the only thing wrong is :
When order is considered, the total ways will be $9 \times 8$ rather than  $2 \times 9 \times 8$. 
You have counted repeatedly, consider selecting $m$ and $n$ from set {$1,2$}.
Total ordered pairs are : (1,2), (2,1).
Id est $2 \times 1$, not $2 \times 2 \times 1$.
Although this doesn't alter your answer, because $2$ is cancelled from both the numerator and denominator.
Similarly, when order isn't considered, the total ways would be $\displaystyle \binom{9}{2}=\frac{9 \times 8}{2}$ rather than $9 \times 8$.
A: We will use the following three things:

1)
$(x^2-y^2) = (x-y)(x+y)$

2) Odd + Odd = Even

Odd - Odd = Even

Even + Even = Even

Even - Even = Even

Odd + Even = Even + Odd = Odd

Odd - Even = Even - Odd = Odd

3)
Even $\cdot$ Even = Multiple of $4$

Odd $\cdot$ Odd = Odd $\not = $ Multiple of $4$
Hence if $x,y $ both Odd or both  Even we have $(x^2-y^2) = (x-y)(x+y) =$ Even $\cdot$ Even = Multiple of $4$

And if one is Even and the other is Odd we have $(x^2-y^2) = (x-y)(x+y) =$ Odd $\cdot$ Odd $\not =$ Multiple of $4$
Hence for $(x^2-y^2) $ to  be divisible by 4 we need them to be "same" , with respect to divisibility by two. Think of this as there being $9$ cards. $4$ Red and $5$ Black. What is the probability that if, with replacement, we pick the same colour twice? Clearly this is $\frac{5}{9}\cdot \frac{5}{9} + \frac{4}{9}\cdot\frac{4}{9} = \frac{41}{81} \approx 0.5 $
