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September 3rd, 2019, 10:44 AM   #1
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Math Focus: Elementary Math
Equation

Solve for integers : $\displaystyle x(-1)^{x+y}=y(-1)^{xy} $.
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September 3rd, 2019, 11:14 AM   #2
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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?
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September 3rd, 2019, 02:45 PM   #3
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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.
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