# Problem of the Week # 278 - Aug 29, 2017

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#### Ackbach

##### Indicium Physicus
Staff member
Here is this week's POTW:

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Find all pairs of real numbers $(x,y)$ satisfying the system of equations
\begin{align*}
\frac{1}{x} + \frac{1}{2y} &= (x^2+3y^2)(3x^2+y^2) \\
\frac{1}{x} - \frac{1}{2y} &= 2(y^4-x^4).
\end{align*}

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#### Ackbach

##### Indicium Physicus
Staff member
Re: Problem Of The Week # 278 - Aug 29, 2017

This was Problem B-2 in the 2001 William Lowell Putnam Mathematical Competition.

Congratulations to Kiwi for his correct solution, which follows. Also, an honorable mention to Opalg for a correct but indirect solution.

[EDIT] See below for a correction.

Taking the first equation and subtracting 1/3 of the second equation gives:
$\frac 1x + \frac 1{2y}-\frac 1{3x}+\frac 1 {6y}=3x^4+3y^4+10x^2y^2-\frac 23 y^4 + \frac 23 x^4$

$\therefore \frac 2{3x} + \frac 2{3y}-3x^4-3y^4-10x^2y^2=-\frac 23 y^4 + \frac 23 x^4$

Now notice that the LHS has the property that we can interchange x and y without affecting the result. Therefore, the RHS must have this property and:

$-\frac 23 x^4 + \frac 23 y^4=-\frac 23 y^4 + \frac 23 x^4$

$\therefore \frac 43 y^4= \frac 43 x^4$

$\therefore y^4= x^4$

So $x=\pm y$ and $x^2=y^2$

Making substitutions into the second of the given equations gives:

$\frac 1x \pm \frac 1{2x}=0$ so x has no solution.

In conclusion the given set of equations has no solution.

#### Ackbach

##### Indicium Physicus
Staff member
Re: Problem Of The Week # 278 - Aug 29, 2017

A user has pointed out that Kiwi 's solution is not actually correct. The step that I initially thought was the clever step (reasoning by reversing the $x$ and $y$ on the LHS and therefore also on the RHS) is incorrect. The correct solution, attributed to Kiran Kedlaya and his associates, follows:

By adding and subtracting the two given equations, we obtain the equivalent pair of equations
\begin{align*}
2/x &= x^4 + 10x^2y^2 + 5y^4 \\
1/y &= 5x^4 + 10x^2y^2 + y^4.
\end{align*}
Multiplying the former by $x$ and the latter by $y$, then adding and subtracting the two resulting equations, we obtain another pair of equations equivalent to the given ones,
$3 = (x+y)^5, \qquad 1 = (x-y)^5.$
It follows that $x = (3^{1/5}+1)/2$ and $y = (3^{1/5}-1)/2$ is the unique solution satisfying the given equations.

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