The "Square-Cube" Age Problem |

The article that you are about to read in this page was written by the staff of the MATHCOUNTS Foundation and published in their Winter 1995 edition of their newsletter MATHCOUNTS NEWS. It is is reprinted here by permission.

Last year my age was a perfect square. Next year it will be a perfect cube. How old am I?

This particular MATHCOUNTS problem isn't a difficult one, until another dimension is added, as **Terrel Trotter, Jr.** of **Escuela Americana** in San Salvador, El Salvador did. He asked his students, "if we drop the idea of age and use larger numbers, can we find other examples of squares and cubes that have a difference of two?" At first glance, this seems like an easy answer, but it's anything but.

The problem appeared a few months ago in
Trotter's monthly classroom tabloid, *Trotter Math News*. Accompanying the problem was a reminder: "problem solving is not merely computing the sum
of two fractions or the product of two decimals, but rather doing whatever is necessary to answer a big question, using whatever method that may help, guessing, looking for patterns, using calculators --- the whole works."

That's great advice, but even with the use of calculators and spreadsheet software, the students came up empty-handed. When they reported their findings, Trotter confessed, "I don't know if there is indeed an answer!"

Not ready to accept defeat, Trotter went to the MATHCOUNTS head office in search of some answers.

MATHCOUNTS Curriculum Coordinator **Scott Stull** recalled, "when the problem was first written, we didn't have a strong argument that the answer was unique for all integers. However we did convince ourselves that the answer was unique for all *reasonable* integral ages for a human being."

To answer Trotter's question, Stull enlisted the help of **Richard Case, P.E.**, director of strategic development for IBM Corporation and MATHCOUNTS national judge, and also **Harold Reiter, Ph.D.**, professor of mathematics at The University of North Carolina at
Charlotte and MATHCOUNTS question writer.

The two initially agreed that another solution seemed possible but how they went about finding the alternate solution was quite different.

Case developed an algorithm to test the first *n* integers. He reduced the problem to finding solutions of the form {*x*, *y*} for the equation *y*^{3} - 2 = *x*^{2}. His algorithm involved taking an integer, cubing it, subtracting two, taking the square root, and evaluating whether or not the result was also an integer. He was able to confirm that the solution {5, 3} is unique for the first 10,201 positive integers (that's as high as his computer could go).

Reiter's approach was slightly different. He
developed his argument through number theory. Referencing an article by **John Stillwell** titled, *What Are Algebraic Integers and What Are They For?*, Reiter found an argument by **Euler** verifying that the original solution is the only solution:

A rigorous proof of this argument relies on
abstract algebra, and establishing the existence of prime factorization for the ring of algebraic integers *Z*[-2].

However, Stull warned, "this cannot be generalized
to say that there is only one solution to any problem of the form *y*^{3} - *j* = *x*^{2}." Take this problem for instance:

Two years ago the age of a certain tortise and that tortise's child were both perfect squares. In two years both of the ages will be perfect cubes. How old is each tortise?

Now that the first problem has been tamed, what about this one? Is there a solution? If there is, is it unique? A reward is offered to the first team (of four students) who sends in a solution and an argument verifying or disproving the uniqueness of the solution. To collect the bounty, send your solution to MATHCOUNTS, 1420 King St., Alexandria, VA 23314-2794.

*Good luck!*

P.S. (March 1999) Don't rush to send in your solution; the problem has already been solved and the reward claimed. But it's still a nice problem about the mommy and baby tortise, don't you think?

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