Let x, y be integers. Prove that if 3 does not divide x and 3 does not divide y, then 3 divides (x^2 - y^2). Note: If a and b are integers, then a divides b, written as such, if a != 0 and there exists an integer k such that b = ak. Note that though one uses the term \"divides\" the definition refers only to the operation of multiplication of integers. One also uses the terminology that \"b is a multiple of a\" if there exists an integer k such that b = ak. Note: This problem requires a verbose, proper, well-written, in-english solution/proof. This does not mean quick-and-dirty logic with a few token words tossed in. Solution (x^2 - y^2)=(x+y)(x-y) As 3 doesnt divide x and y .x,y is of the form 3k+1 or 3k+2 Case 1:x=3k+1,y=3k\'+1 This implies x+y is of the form 3k+1 , x-y is of the form 3k Case 2:x=3k+1,y=3k\'+2 This implies x+y is of the form 3k , x-y is of the form 3k+2 Hence 3 divides (x^2 - y^2) as it always divides x+y or x-y.

1. Let x, y be integers. Prove that if 3 does not divide x and 3 does not divide y, then 3 divides (x^2
- y^2).
Note: If a and b are integers, then a divides b, written as such, if a != 0 and there exists an
integer k such that b = ak. Note that though one uses the term "divides" the definition refers
only to the operation of multiplication of integers. One also uses the terminology that "b is a
multiple of a" if there exists an integer k such that b = ak.
Note: This problem requires a verbose, proper, well-written, in-english solution/proof. This does
not mean quick-and-dirty logic with a few token words tossed in.
Solution
(x^2 - y^2)=(x+y)(x-y) As 3 doesnt divide x and y .x,y is of the form 3k+1 or 3k+2
Case 1:x=3k+1,y=3k'+1 This implies x+y is of the form 3k+1 , x-y is of the form 3k Case
2:x=3k+1,y=3k'+2 This implies x+y is of the form 3k , x-y is of the form 3k+2 Hence 3 divides
(x^2 - y^2) as it always divides x+y or x-y