*The Lady’s Diary* was a popular magazine published in England from 1704 to 1841. It contained mathematical puzzles such as the following, published in 1798.

What two numbers are those whose product, difference of their squares, and the ratio or quotient of their cubes, are all equal to each other?

From Benjamin Wardhaugh’s new book A Wealth of Numbers: An Anthology of 500 Years of Popular Mathematics Writing.

See also my brief review of How to Read Historical Mathematics by the same author.

It’s a no brainer:

FindInstance[x y == x^2 – y^2 && x^2 – y^2 == x^3/y^3, {x, y}]

That’s exactly the solution one of the readers wrote in. ðŸ™‚

Curios, I got both x and y in minus from Mathematica 8 using the no brainer above.

Did I miss anything?

Some properties of the golden ratio may be quite helpful to find those numbers…

Still a divine, or should I say Golden, problem

Surprising result! At least to me.

Jorz,

Use FindInstance[x y == x^2 – y^2 && x^2 – y^2 == x^3/y^3, {x, y}, 4] to get all the solutions:

x -> -2.61803, y -> -1.61803

x -> 0.381966, y -> -0.618034

x -> -0.381966, y -> 0.618034

x -> 2.61803, y -> 1.61803

Fie to you, Mr Cook. An interesting and somewhat golden problem.

It’s also easy to use Algebra to eventually use the general quadratic equation to solve a quadratic in Y^2 (or X^2) to get:

Y= +or- SQRT((3 +or- SQRT(5))/2) which are all four solutions given by Tim H

Adding this to the math club list of challenges. THANK YOU!

This is a great little problem – thanks for posting this John.

I’ve provided a complete algebraic and graphical solution on my own blog if you’re interested. The solution is elegant! (danpearcymaths.wordpress.com)

Au-some

@Dan,

Graph looks like two Star Trek original series comm badges overlayed.

@Jeff,

I actually posted that up then left the apartment only to realise that wasn’t the full solution. There are four solutions to each of four possible equations. It looks like there are 16 different solutions but there are actually 4 depending on how you construct the equation. Each set of solutions are the vertices of a rhombus with centre (0,0).

You dont need graph or software to solve this one. Just use plain algebra.

xy = xÂ³/yÂ³

y^4 = xÂ²

x = yÂ² ———- (1)

xy = xÂ² – yÂ²

Substituting value of x from (1),

yÂ² * y = (yÂ²)Â² – yÂ²

y = yÂ² – 1

yÂ² -y – 1 = 0

Solve this quadratic equation to find y. Put value of y in (1) to find x.

What about infinity and 1?

How would one solve this w/o using algebraic symbol manipulation? Were women (or anyone, really) schooled in symbol manipulation in 18th century Britain? Seems like a much harder problem to work through verbally…

Peter: The book lists a couple solutions submitted by readers. Both used algebraic symbol manipulation and look little different from what someone would write now.