# A William Lowell Putnam Problem and its Solution

Problem:

Curves A, B, C and D are defined in the plane as follows: $A = \{ (x, y): x^{2}-y^{2}=\frac{x}{x^{2}+y^{2}}\}$ $B = \{ (x, y): 2xy + \frac{y}{x^{2}+y^{2}}=3\}$ $C = \{ (x, y): x^{3}-3xy^{2}+3y=1\}$ $D = \{ (x, y): 3x^{2}y -3x - y^{3}=0\}$.

Prove that $A \bigcap B = C \bigcap D$

(1987 William Lowell Putnam Mathematics Competition)

Solution:

Let $z= x+iy$. The equations defining A and B are the real and imaginary parts of the equation $z^{2}=z^{-1}+3i$, and similarly the equations defining C and D are the real and imaginary parts of $z^{3}-3iz=1$. Hence, for all real x and y, we have $(x,y) \in A \bigcap B$ if and only if $z^{2}=z^{-1}+3i$. This is equivalent to $z^{3}-3iz=1$, that is, $(x, y) \in C \bigcap D$.

Thus, $A \bigcap B = C \bigcap D$.

Isn’t that an elegant solution? What do you think?

Nalin Pithwa

PS: Solution published in “Complex Numbers from A to …Z” by Titu Andreescu and Dorin Andrica

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