What is the probability that the difference of squares of two positive integers up to $30$ is divisible by $3$ or $7$?

Question

If we choose any two numbers $a$ and $b$ from the integers $1$ through $30$, what is the probability of $a^2-b^2$ of being divisible by $3$ or $7$?

Answer 1

As said by peter , you can simplify (a^2-b^2) =(a-b)(a+b) ..

Then you can write the sample set for determining if either (a-b) is divisible by 3 or 7 or (a+b) is divisible by 3 or 7 ..

once you get the sample sets , you can easily calculate prob : no_of_samples/(30*30)

Though this is a hard method ,you would get your answers this way .

Answer 2

$$(a^2 - b^2) = (a+b)(a-b)$$ Since both $3$ and $7$ are prime, at least one of $(a+b),|a-b|$ must be $3$ or $7$. Note that (Correct me if I'm wrong) you will never get both $3$ AND $7$ unless $(a,b) \in \{(2,5),(5,2)\}$. Now It's a matter of counting the ways you can get $3$ or $7$ as a sum from the set, and as a difference from the set, excluding the special case once..

Answer 3

Note that $a^2-b^2$ is divisible by $c$ if and only if $$a\equiv b\equiv0\pmod{c}\qquad\text{ or }\qquad a\equiv\pm b\not\equiv0\pmod{c}.$$ Taking $c=3$ we find that the sum of the probabilities of these cases is $$\left(\frac{10}{30}\right)^2+\left(\frac{20}{30}\right)^2=\frac{500}{900}.$$ For $c=7$ there are $9$ integers congruent to $\pm1\pmod{7}$ and $9$ integers congruent to $\pm2\pmod{7}$, but only $8$ integers congruent to $\pm3\pmod{7}$, so the sum becomes $$\left(\frac{4}{30}\right)^2+2\cdot\left(\frac{9}{30}\right)^2+\left(\frac{8}{30}\right)^2=\frac{242}{900}.$$ Similarly the probability that $a^2-b^2$ is divisible by $21$ equals $$\left(\frac{1}{30}\right)^2+9\cdot\left(\frac{3}{30}\right)^2+\left(\frac{2}{30}\right)^2=\frac{85}{900}.$$ So the probability that $a^2-b^2$ is divisible by either $3$ or $7$ equals $$\frac{500}{900}+\frac{242}{900}-\frac{85}{900}=\frac{662}{900}=\frac{331}{450}.$$

Answer 4

There are 900 combinations, which you can calculate in an Excel table and then use the MOD(number,divisor) function to get the Modulo for 3 and 7. Then you can use "COUNTIF" function to count the relevant 0-Modulo pairs. This is of course not a Mathematical proof but can be used to verify.