# An analytical solution to a Jewish problem

Maybe you have heard about the article of Khovanova and Radul, Jewish Problems, in which they collect tricky mathemical problems that were alledgedly designed to prevent Jews (to which they were specifically given) from passing the oral entrance exams of Moscow State University . With such a story and such a naïvely equivocal title, you can count on all the comments-section jewishologists of the world to promote your article 🙂

What I found interesting is this article is that, of the 21 problems given, I could answer most of the analytical ones, while I was completely unequipped to solve the geometrical problems. Has geometry always been that hard ? My impression is that, at least here in France, its teaching is slowly being replaced by linear algebra . Maybe it is simply less needed today. Take civil engineering : a beam used to be some kind of parralelepipede, now it is a grid of finite elements !

As an illustration, let us have a look at the fifth problem: solve the equation

$\sin^7(x) + \dfrac{1}{\sin^3x} = \cos^7(x) + \dfrac{1}{\cos^3x}.$

What kind of math problem do you see here ? (let’s just hope it has nothing to do with geometry !) The authors of the article give a solution based on some transformation of the equations and a few trigonometricks (at some point they use the variable t=sin(x)cos(x) and retransform it into sin(2x)/2). What I first saw when I read the question (and I bet that’s what you saw too if you have my kind of training) was an analytical problem involving the function

$f(t) = t^7 + \dfrac{1}{t^3}$

for which it is asked to find all the (u,v) which are the sine and cosine of a same angle and verify f(u)=f(v) ! An obvious situation where this will work is when u=v, which leads us to the solutions $t = \pi/4$ and $t = -3\pi/4$, which are the only angles whose sine is equal to the cosine. Now, are there any other solutions ? In the end what is asked is whether the function f is bijective, or not, or just enough ! Let us compute its derivative:

$f'(t) = 7t^6 - 3 \dfrac{1}{t^4} = \dfrac{7t^{10}-3}{t^4}$

This is only positive outside the interval $[-\sqrt[10]{\frac{7}{3}} , +\sqrt[10]{\frac{7}{3}}]$. We can now sketch the function:

Since u and v are sine and cosine, they will belong to the interval [-1,1]. In red I have represented two non-bijectivity zones of the function on this interval: if u and v are two different numbers of [-1,1] verifying f(u)=f(v), then the two of them must belong to one of these red zones. Now, notice that
$f(\sqrt{\frac{1}{2}}) = (\sqrt{\frac{1}{2}})^7 + \sqrt{2}^3 > \sqrt{2}^3 = 2\sqrt{2} > 2$
This shows that $\sqrt{\frac{1}{2}}$ is placed left to the red zone in the right (see figure). A consequence is that, if u and v belong to the right red zone, then

$u^2 + v^2 > \sqrt{\frac{1}{2}}^2 + \sqrt{\frac{1}{2}}^2 = 1$

so u and v cannot be the sine and the cosine of the same angle (or their squares would sum up to 1). By imparity of the function, the same can be said about the left red zone. As a conclusion, it is not possible that f(u)=f(v) if u and v are the sine and cosine of a same angle and have different values. So $t = \pi/4$ and $t = -3\pi/4$ are the only two solutions to the given equation. Did anyone see another solution ?