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Two integers $m$ and $n$ are chosen at random with replacement from the natural numbers $1,2,.....9$. The probability that $m^2-n^2$ is divisible by $4$ is

  1. $\dfrac{41}{81}$
  2. $\dfrac{37}{81}$
  3. $\dfrac{2}{3}$
  4. $\dfrac{4}{9}$
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m2-n2=(m+n) (m-n)
so need to check cases for (m+n) div by 4 + (m-n) div by 4

consider for (m+n) below are 20 such cases                                         Now check for (m-n) div by 4 one such case is when m=n
(1,3) (1,7)                                                                                                                m=n we have 9 cases
(2,2)(2,6)                                                                                                                (9,5)(9,1) (8,4) (7,3) (6,2) (5,1)
(3,1)(3,5)(3,9)                                                                                                        (5,9) (1,9) (4,8) (3,7) (2,6) (1,5)
(4,4)(4,8)                                                                                                                Above are 12 cases
(5,3)(5,7)
(6,2)(6,6)
(7,1)(7,5)(7,9)
(8,4)(8,8)
(9,3)(9,7)

so Total (20+12+9)/81= 41/81

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