September 3rd, 2019, 09:44 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91  Equation
Solve for integers : $\displaystyle x(1)^{x+y}=y(1)^{xy} $.

September 3rd, 2019, 10:14 AM  #2 
Senior Member Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 
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 $\displaystyle (x,y) \in \left\{ (2a,2a), (2a+1,2a+1), (2a+1,2a1): a \in \mathbb(Z) \right\}$ 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? 
September 3rd, 2019, 01:45 PM  #3 
Senior Member Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 
Wait, I messed up in the final answer. It should be $\displaystyle (x,y) \in \left\{ (2a,2a), (2a+1,2a1): a \in \mathbb(Z) \right\}$ So, (0,0), (2,2), (4,4), …, (2,2), (4,4), …, (1,1), (3,3), (5,5), …, (1,1), (3,3), (5,5), … Somebody doublecheck me there. 

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