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 September 3rd, 2019, 10:44 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 820 Thanks: 113 Math Focus: Elementary Math Equation Solve for integers : $\displaystyle x(-1)^{x+y}=y(-1)^{xy}$.
 September 3rd, 2019, 11:14 AM #2 Senior Member   Joined: Jun 2019 From: USA Posts: 376 Thanks: 202 x = 0 AND y = 0 or |x| = |y| AND (x = y NXOR ((x+y) mod 2 = xy mod 2)) Breaking it down: Case 1: x, y positive x and y are both odd or both even, and xy is even => x and y are both even Case 2: x, y negative Same as above, x and y are both even Case 3: xy negative a) x+y is odd, xy is even, OR b) x+y is even, xy is odd a) => x+y is odd -> one is odd, one even -> xy is even (violates |x|=|y|!) b) => xy is odd -> both odd -> x+y is even $\displaystyle (x,y) \in \left\{ (2a,2a), (2a+1,2a+1), (2a+1,-2a-1): a \in \mathbb(Z) \right\}$ So, (0,0), (1,1), (2,2), (3,3), (4,4), …, (-1,-1), (-2,-2), (-3,-3), (-4,-4), …, (1,-1), (3,-3), (5,-5), …, (-1,1), (-3,3), (-5,5), … Right? Thanks from idontknow
 September 3rd, 2019, 02:45 PM #3 Senior Member   Joined: Jun 2019 From: USA Posts: 376 Thanks: 202 Wait, I messed up in the final answer. It should be $\displaystyle (x,y) \in \left\{ (2a,2a), (2a+1,-2a-1): a \in \mathbb(Z) \right\}$ So, (0,0), (2,2), (4,4), …, (-2,-2), (-4,-4), …, (1,-1), (3,-3), (5,-5), …, (-1,1), (-3,3), (-5,5), … Somebody double-check me there. Thanks from idontknow

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