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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 Tags equation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post max233 Algebra 2 March 31st, 2017 10:40 PM DarkX132 Algebra 3 September 26th, 2014 11:15 PM Sonprelis Calculus 6 August 6th, 2014 11:07 AM PhizKid Differential Equations 0 February 24th, 2013 11:30 AM mich89 Algebra 3 January 9th, 2013 02:22 PM

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