Diophantine Equations (Problem 5)

Number Theory Level pending

x y − y x = 1 x^y - y^x = 1 The equation above has 2 integer solutions. One of them is ( x , y ) = ( 3 , 2 ) (x,y) = (3,2) . What is the other solution?

( 1 , 2 ) (1,2) ( 2 , 1 ) (2,1) ( 2 , 3 ) (2,3)

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1 solution

When x = 3 , y = 2 x = 3, y = 2 : 3 2 − 2 3 = 9 − 8 = 1 3^2 - 2^3 = 9 - 8 = 1

When x = 2 , y = 3 x = 2, y = 3 : 2 3 − 3 2 = 8 − 9 = − 1 2^3 - 3^2 = 8 - 9 = -1

When x = 1 , y = 2 x = 1, y = 2 : 1 2 − 2 1 = 1 − 2 = − 1 1^2 - 2^1 = 1 - 2 = -1

When x = 2 , y = 1 x = 2, y = 1 : 2 1 − 1 2 = 2 − 1 = 1 2^1 - 1^2 = 2 - 1 = 1

Therefore, the other solution is: x = 2 , y = 1 x = 2, y = 1

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