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…)


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:


Rearranging, we get


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:


Expanding out, we get


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


This actually can be factored! (If you can’t see that, graph it and find that a nice root is 4.) It factors into:


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:


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. 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. 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. 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. 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.

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