First we calculate the square of the numbers from 1 to 10 modules 6

$$ \begin{aligned} & 1^2 \equiv 1(\bmod 6) \\ & 2^2 \equiv 4(\bmod 6) \\ & 3^2 \equiv 9 \equiv 3(\bmod 6) \\ & 4^2 \equiv 16 \equiv 4(\bmod 6) \\ & 5^2 \equiv 25 \equiv 1(\bmod 6) \\ & 6^2 \equiv 36 \equiv 0(\bmod 6) \\ & 7^2 \equiv 49 \equiv 1(\bmod 6) \\ & 8^2 \equiv 64 \equiv 4(\bmod 6) \\ & 9^2 \equiv 81 \equiv 3(\bmod 6) \\ & 10^2 \equiv 100 \equiv 4(\bmod 6) \end{aligned} $$

Next we determine possible values of $x^2+y^2$ modulo 6 which are $0,1,3$ and 4 consider,

$$ \begin{aligned} & 0-0 \equiv 0(\bmod 6) \\ & 1-1 \equiv 0(\bmod 6) \\ & 3-3=0(\bmod 6) \\ & 4-4 \equiv 0(\bmod 6) \\ & 1-4 \equiv-3 \equiv 3(\bmod 6) \\ & 4-1 \equiv 3(\bmod 6) \\ & 1-3 \equiv-2 \equiv 4(\bmod 6) \\ & 3-1 \equiv 2(\bmod 6) \\ & 4-3 \equiv 1(\bmod 6) \\ & 3-4 \equiv-1 \equiv 5(\bmod 6) \end{aligned} $$

$$ \begin{aligned} & 0-1 \equiv-1 \equiv 5(\bmod 6) \\ & 0-3 \equiv-3 \equiv 3(\bmod 6) \\ & 0-4 \equiv-4 \equiv 2(\bmod 6) \\ & 1-0 \equiv 1(\bmod 6) \\ & 4-2 \equiv 2(\bmod 6) \\ & 2-4 \equiv-2 \equiv 4(\bmod 6) \end{aligned} $$

From the above only cases where $\left|x^2-y^2\right|$ is divisible by 6 occur, when $x^2 \equiv y^2$ (mod 6), these pairs are

$$ \begin{aligned} & x^2 \equiv y^2 \equiv 0(\bmod 6) \\ & x^2 \equiv y^2 \equiv 1(\bmod 6) \\ & x^2 \equiv y^2 \equiv 3(\bmod 6) \\ & x^2 \equiv y^2 \equiv 4(\bmod 6) \end{aligned} $$

Now, counting the number of possible pairs $(x, y)$ that satisfy each condition.

$0(\bmod 6)$ is satisfied by 6 .

$1(\bmod 6)$ is satisfied by $1,5,7$.

$3(\bmod 6)$ is satisfied by 3,9 .

$4(\bmod 6)$ is satisfied by $2,4,8,10$

The total number of favorable outcomes

0 has 1 pair $(6,6)$

1 has $3 \times 3=9$ pairs $(\{1,5,7\} \times\{1,5,7\})$

3 has $2 \times 2=4$ pairs $(\{3,9\} \times\{3,9\})$

4 has $4 \times 4=16$ pairs

$$ (\{2,4,8,10\} \times\{2,4,8,10\}) $$

Total number of favorable outcomes

$$ =1+9+4+16=30 $$

Total possible outcomes $=10 \times 10=100$

Thus, the probability that $\left|x^2-y^2\right|$ is divisible by 6 is

$$ \begin{aligned} & =\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }} \\ & =\frac{30}{100}=\frac{3}{10} \end{aligned} $$

Probability

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