List 'em Out!

Notice $0^2 \equiv 0\pmod3$ and $1^2, 2^2 \equiv 1\pmod{3}$ . We must have $x^2 \equiv y^2 \pmod 3$ .

If $x^2 \equiv y^2 \equiv 0 \pmod3$ , there are $\dbinom{10}{2}$ ways to select $x,y$ from the $\dfrac{30}{3}$ multiples of three.

If $x^2 \equiv y^2 \equiv 1 \pmod{3}$ , there are $\dbinom{20}{2}$ ways to select $x,y$ from the $30-10 = 20$ non-multiples of three.

Our answer is $\dfrac{\dbinom{10}{2} + \dbinom{20}{2}}{\dbinom{30}{2}} = \dfrac{45 + 190}{435} = \dfrac{47}{87} \implies \boxed{134}$ .

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We know that $x^{2} \equiv 1,0 \mod{3}$ . Hence the possibilities for $x^{2} - y^{2}$ not to be divisible by $3$ are just $x \equiv 0 \mod{3} \wedge y \not\equiv 0 \mod{3}$ and the other way round, therefore $20*10 + 10*20$ ordinate couples. All of the ordinate couples are $30*29$ . So , $1- \frac{20*10 + 10*20}{30*29} = \frac{47}{87}$ gives us the probability , and $47 + 87 = 134$ the result.

We can see that all number squares are of the form 3k+1 or 3k for some integer k. So both nos x&y should either be both divisible my 3 in which case their squares will be of the form 3m &3n and thus their difference is divisible by 3 Or the 2 nos may both be not divisible by 3 in which case their squares are 3m+1 & 3n+1 form their difference again being divisible by 3 So x&y can be chosen in (10 choose 2)+ (20 choose 2) ways out of total possible 30 choose 2 ways thus answer is 47/87 i.e 47 + 87 = 134

For any $x$ , we have $x^2\equiv0\pmod3$ or $x^2\equiv1\pmod3$ . Hence $x^2-y^2$ is divisible by $3$ if and only if both $x,y$ are divisble by $3$ or both $x,y$ are not divisble by $3$ .

There are $10$ numbers less than $30$ divisible by $3$ , so there are $\binom{10}{2}$ options to choose two of those.

There are $20$ numbers less than $30$ not divisible by $3$ , so there are $\binom{20}{2}$ options to choose two of those.

There are total of $\binom{30}{2}$ options to choose a pair, so the probability we seek is: $\frac{\binom{10}{2}+\binom{20}{2}}{\binom{30}{2}}=\frac{235}{435}=\frac{47}{87}=\frac{a}{b}$ So $a+b=47+87=134$

Look for number: $x \equiv 0 \pmod{3}$ $x \equiv 1 \pmod{3}$ $x \equiv 2 \pmod{3}$

Case I: If we choice number remainder 2 as first number, there is two subcase, first subcase pair remainder of $(xy)$ are $(2,2$ . Second subcase pair remainder of $x,y)$ are $2,1$

probability $x^2-y^2 \equiv \pmod{3}$ is $\displaystyle \frac{10}{30} \times \frac{9}{29}+\frac{10}{30} \times \frac{10}{29}=\frac{19}{87}$

case II If we choice first number is $1 \equiv \pmod{3}$ It's same case I, because case I and case II symetric relation so probability is

(\displaystyle \frac{9}{29}+\frac{10}{30} \times \frac{10}{29}=\frac{19}{87})

Case III

If first number $x \equiv 0 \pmod{3}$

Probability for this case is $\displaystyle \frac{10}{30} \frac{9}{29}=\frac{9}{29}$

Now, Probability $x^2-y^2$ divisible by 3 is:

$P(A)=2 \times \frac{19}{87}+ \frac{9}{87}=\frac{47}{87}$

So, $a+b=47+87=\fbox{134}$ as desired.