Here we go for this week!
Can a four dimensional topological sphere have two or more inequivalent smooth structures?

Kidding! I'm kidding! (Devil)

Here's the real one:

Solve the following system for (integers) x and y.
$\displaystyle \frac{1}{x} + \frac{1}{y} = -\frac{1}{2}$

$\displaystyle x^3 + y^3 = 7$

-Dan

Jun 8th 2014, 10:09 AM

JeffM

Re: Problem #9 - College Algebra

Quote:

Originally Posted by topsquark

Here we go for this week!
Can a four dimensional topological sphere have two or more inequivalent smooth structures?

Kidding! I'm kidding! (Devil)

Here's the real one:

Solve the following system for (integers) x and y.
$\displaystyle \frac{1}{x} + \frac{1}{y} = -\frac{1}{2}$

$\displaystyle x^3 + y^3 = 7$

-Dan

The first question would send me back to school for years. In fact, I never took college algebra either: I so hated my calculus course at Columbia that I took no other math courses there.

Letting s = x + y and p = xy we get the two equations

$\displaystyle \frac{s}{p}=-\frac{1}{2}$

$\displaystyle s\left(s^2-3p\right)=7$

which in turn lead to

$\displaystyle s^3+6s^2=7$

$\displaystyle s=1$

$\displaystyle p=-2$

$\displaystyle x+y=1$

$\displaystyle xy=-2$

Jun 15th 2014, 03:10 PM

topsquark

Re: Problem #9 - College Algebra

Thanks to all that helped out!

My solution is roughly the same as idea's and I have the unfortunate tendency to write out the first method I see, even if it isn't all that elegant. So here's my version in all it's multi-step glory!

Since we are looking for integral solutions, let's try the rational root theorem. After sorting the list a bit to get all factors of 56 we find that z = -2 is the only rational solution. (The other two are irrational.) Now there is a detail we need to (quickly) look at. There is no a priori restriction here that says z must be an integer, only x and y. However if we look at the other two solutions for the z equation we can easily see that z = -2 is the only way to go.

So $\displaystyle z = xy = -2$ gives $\displaystyle y = -\frac{2}{x}$