## Tuesday, 27 April 2021

### Two problems on cubic functions

Problem 1. Find the number of integer solutions to the equation
$m^3+n^3+30mn = 10^3$ with $mn \geq 0$.

I was given the above question today which is a pity because I've been drafting questions for some other contests. Questions like the above is definitely something I would consider.

At first glance recall the identity $(m+n)^3 = m^3 + n^3 + 3mn(m+n)$, so if $m+n = 10$ then the equation becomes $(m+n)^3 = 10^3$, and that gives 11 solutions $(10,0),...,(1,9), (0,10)$. This is however not all the solutions -- and that's why the question is not an easy one.

Now consider a more general factorization where $m+n = k$. Then the equation becomes
$k^3 + (30-3k)mn = 10^3$
$10^3 - k^3 = (10-k)(10^2+10k+k^2) = 3(10-k)mn$.

If $k\neq 10$ then we expand $k$ to obtain
$10^2 + 10(m+n) + (m+n)^2 = 3mn$
$m^2-mn+n^2-10m-10n+10^2 = 0$
$(m+10)^2+(n+10)^2-(m+10)(n+10) = 0$.

Since $LHS \geq 0$ we have the only extra solution $m = n = -10$, and there are 12 solutions in total.

*

When we see cubic equations in contests we know it's going to be nasty. Here is another one, from an university entrance exam in the late seventies or early eighties.

Problem 2. Sketch the graph of $x^3+y^3-3xy = 0$.

I recommend you to plot the graph using Wolframalpha because otherwise who in the world could sketch this by mathematically analyzing the function $x^3+y^3-3xy$?

Notice that the curve has an asymptote $y=-1-x$ (notice that $x^3+y^3-3xy = -1$ when $x+y=-1$, like what we did in problem 1, but mathematically showing the asymptote falls to the category of algebraic geometry, where you calculate the tangent at infinity. See this post for example). So we start with a change in coordinates $(x,y) = ((X+Y)/2, (X-Y)/2)$ (by the way why a multiple of 2 not $\sqrt{2}$ which gives an orthogonal transformation? Well 2 is easier to calculate with but more importantly, it cancels the coefficient of $X^3$) which rotates the asymptote into a vertical line. The new transformed equation is then $X^3-3X^2+3XY^2+3Y^2 = X^2(X-3)+3Y^2(1+X)=0$, so the graph becomes the plot of
$Y = \pm \sqrt{ \frac{-X^2(X-3)}{3(1+X)}}$ ,     (1)
which has domain (-1,3] and a vertical asymptote at $X = -1$ as expected. The roots of the function are of course $X = 0,3$, and the extrema (of the positive bit) is $\sqrt{2\sqrt{3}-3}$ at $X = \sqrt{3}$ (high school calculus by looking at the function $\frac{-X^2(X-3)}{3(1+X)}$). Finally, since the curve is smooth everywhere, you know that the cross at the origin are actually two smooth and symmetric curves and that should be enough to sketch the graph. (Even better, using implicit differentiation and (1) you know that
$(\frac{dY}{dX})^2 = \frac{X^2(2-X)^2}{4Y^2(1+X)^2} = \frac{3(2-X)^2(1+X)}{4(1+X)^2(3-X)} = 1$ at the origin. Therefore the two tangents at the origin are of slope $\pm 1$.)

This curve is pretty famous in fact -- its name is the Folium of Descartes. This is just another example that people always get their ideas for exam questions from the old stuffs regardless of the time.

This is the beauty of the old entrance exams: you get questions too analytic and too open-ended for high school or college Olympiad but too hard under modern high school or college curriculum. These are questions that we do not meet everyday, but I guarantee that a large portion of them are fun to play with.