**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)

Graduates: Miguel Hurtado (ECE)

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)