# A Problem of Yore

I was browsing old math journals a few days ago and got caught up in looking at puzzles/challenging problems sent in by professors to The Mathematical Gazette (the original publication of the Mathematical Association).

I got engaged in battle with a problem from it’s first year of publication (No. 1, Vol. 7, April 1896 – if you have JSTOR access, see the original problem here). The journal stated that it was a question “from recent Entrance Scholarship papers at Oxford and Cambridge.”

I think it’s a darn good problem, so take a stab at it. I’ll type my solution to it below the fold, but maybe you’ll get a better one? (I went down two wrong roads before I came up with this one…)

SOLUTION

Let $u=x+\frac{2}{x}$. Keep in mind that’s what we’re trying to find. If we calculate $u^3$ we get something nice: $u^3=x^3+\frac{8}{x^3}+6u$

Rearranging, we get $u^3-6u=x^3+\frac{8}{x^3}$.

But recall we are given $x^3=234\sqrt{2}+148\sqrt{5}$. Simplifying the right hand side of the equation above, we find it equals the very pretty $296\sqrt{5}$. (Do you see how? Remember to rationalize the denominator of $\frac{8}{x^3}$.)

Oh! So now we know $u^3-6u=296\sqrt{5}$. Clearly we know $u$ must be of the form $u=a+b\sqrt{5}$ to yield this answer. So we calculate: $(a+b\sqrt{5})^3-6(a+b\sqrt{5})=296\sqrt{5}$

Expanding out, we get $(a^3+15ab^2-6a)+(5b^3+3a^2b-6b)\sqrt{5}=296\sqrt{5}$

Equating the left hand side and right hand side, we see that $a^3+15ab^2-6a=0$ [Eqn 1] and $5b^3+3a^2b-6b=296$ [Eqn 2].

Let’s factor out an $a$ from Eqn 1: $a(a^2+15b^2-6)=0$.

Since I’m lazy, let’s first investigate the easy case, if $a=0$. (If it doesn’t help us, we’ll then consider the other possibility.) If $a=0$, and plugging that into Eqn 2, we get $5b^3-6b-296=0$.

This actually can be factored! (If you can’t see that, graph it and find that a nice root is 4.) It factors into: $(b-4)(5b^2+20b+74)=0$.

The only real root is $b=4$ (The quadratic yields imaginary roots.) So in fact we don’t need to deal with the other gross zero of Eqn 1 because we have a workable solution. (Phew. Factoring that cubic was hard enough.) We have now found that $a=0$ and $b=4$ and we’re done. We have solved for $u$ which was our goal: $u=a+b\sqrt{5}=4\sqrt{5}$

An interesting side note is that no one solved and sent in their solution for this problem in time. I searched through a number of subsequent issues to see if there was a published solution, to see if it differed from my own, but I found the journal’s policy: 1. Kate says:

I got up to u^3 – 6u = 296rad5, but then I got stuck. I didn’t see the insight to substitute a + brad5. I have this inability when a solution is available to not peek. Even though it takes all the fun out of it. Nice problem, thanks for sharing.

2. samjshah says:

Haha, I too am afflicted with this peeping tom disorder. Maybe we should call it “Peeping Pythagoras Disorder” or something cheesy.

However I am still pleased with myself for not looking at the solution to the following problem (from a recent IMO) while I let my subconscious brain mull over it:

Show $\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2} \geq 1$ if you know $xyz=1$ and all three variables are nonzero.

But I’m so frustrated that I might let the disorder take over!

3. polymath says:

I might be missing something here, but…If I solve the 2nd equation as a simple quadratic, and then cube my solution, and come up with x^3 equals 234*sqr(2)+148*sqr(5), doesn’t that complete the proof? It appears that this only solves the converse of the problem, however, I will have demonstrated that the cube root of 234*sqr(2)+148*sqr(5) indeed satisfies the required 2nd equation.

That seems too easy, but…???

4. samjshah says:

No no, you’re totally right! I I’m pretty sure that’s what I first did. But I think I wanted the challenge of going in the forward direction. Because if the problem asked you to find $x+2/x$ in simplest form, you wouldn’t be able to use your method.

But I definitely don’t see anything suspect with your method.