Solution of Math Problem No.3

Problem: Show that a real number q is rational if and only if there are three distinct
integers, , such that forms a geometric progression.

Solution (by Richard B. Eden, Math . Graduate student, Purdue)

Suppose q is a rational number . If q = 0, we can choose . So now
suppose , not necessarily in lowest terms , where r, s ∈ Z, s ≠ 0 and r ≠ 0. We can
also as sume since we can multiply r and s by the same constant.

Let . These three integers are distinct since r ≠ 0 and
s ≠ -1,-2. In this case,

really do form a geometric sequence.

Now suppose form a geometric sequence with distinct
integers. This means , which implies

If , so , we can write and for
some d ∈ Z. We then have, from the above equation ,

so d = 0 and . However, are all distinct. Therefore,
and

which is a rational number.

Also solved by :

Undergraduates: Noah Blach (Fr. Math), Nathan Claus (Fr. Math)

Others: Brian Bradie (Christopher Newport U. VA), Hoan Duong (San Antonio College),
Elie Ghosn (Montreal, Quebec), Brian Huang (Jr. Saint Joseph's HS, IN), Gerard D.
Ko & Swami Iyer (U. Massachusetts, Boston), Steven Landy (IUPUI Physics), Sorin
Rubinstein (TAU faculty, Israel), Steve Spindler (Chicago), Kevin Ventullo (IIT, Chicago)

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