A promise

During my recent trip back home, I had a chance to talk with friends and family over dinner about the future of Greece. I heard about tales of political corruption and wasted opportunities within Greece, but I noticed also a growing resentment towards the rest of the world, who was mocking us, at best, and was out to get us, at worst. After I had listened for a while to the older generation, I asked them this: “So, what do you plan to do about all this?” They looked at me and an old, wise-looking man said: “There is nothing to be done. If you try to change anything, they will find you and silence you. I tried to change things once…” At this point, the younger generation, some still in high-school, others in college, nodded approvingly. But one young man, a young composer planning to study at the famed Berklee College of Music, was looking at me intently. What was he thinking? Why wasn’t he nodding with the old man? I don’t know. But, it was then that I turned to the old man and said: “There are wise men who see the world as it is. And there are fools who see it as it can be.”

For the young composer, who still dreams that his hands are capable of changing the world, I dedicate this post. To those who are struggling to see what lies ahead, I leave them with a problem to solve:

The redeemer: Find all pairs (a,b) of integers $latex age 1, bge 1$ that satisfy the equation $latex a^{(b^2)}=b^a$.

If you can solve this problem, you can change the world. That’s a promise.

PS: Good luck.

2017-01-13T10:06:03+00:00 September 13th, 2012|Real science, The expert's corner|6 Comments


  1. Dexter kim September 13, 2012 at 6:04 pm - Reply

    Currently I’ve found two sets. I’m not sure whether there would be more though.

    • Spyridon Michalakis September 13, 2012 at 6:28 pm - Reply

      Already on your way to changing the world. Now, can you show those are the only pairs (along with (1,1))?

      • Dexter kim September 13, 2012 at 6:38 pm - Reply

        I put a=b^c and then got the equality
        Well, it seems that when I use log on both sides, I get
        (log c)/(c-2) = log b
        which decreases as c increases. There should be some c that log b becomes smaller than log 2, then we have the upper limit of c. Well, up to here.

        • spiros September 14, 2012 at 12:58 pm - Reply

          Why would a have the form b^c? Can you prove that? What if they only have some common divisors?

  2. Steve Flammia September 13, 2012 at 10:17 pm - Reply

    Norbert and Aram and I discussed this briefly over lunch. We didn’t solve it completely, but I’ll offer a hint which we think might go in the direction of a complete solution. We proved that if b is a prime number then the only solutions are the ones already listed above by Dexter Kim. (We can show more than this, but I don’t want to give away too much. 🙂

  3. hmmmjenia September 14, 2012 at 12:04 pm - Reply

    During my recent trip back home, Russia was not in a smaller mess. What a coincidence – my first attempt to write something for this blog, done this morning, also turned to be travel notes. However, mine doesn’t have a clear message (“Change the world!”), so I’ve decided to keep it within my blog. Or do we want postmodernism* on Quantum Frontiers?

    *here stands for South Park.

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