This question is from "Chartrand, Polimeni and Zhang Mathematical Proofs": Chapter 8, Exercise 8.81.

Since this is an existence question, a simple example suffices to complete a proof. The book provides

Proof Let $a = 6/5$, $b = 10/3$ and $c = 15/2$. Then $ab = 4$, $ac = 9$, $bc = 25$ and $abc = 30$

I am interested however in specifying the general conditions for existence. While trying to solve the exercise I could not easily come up with an example, so I tried to find the general form for a, b and c so that an existence is possible.

My question is how would one go about this?

From the solution one can see for example the following relationships: \begin{align} a=6/5=p/q \end{align} \begin{align} b = 2q/(p/2) \end{align} \begin{align} c = 3q/(p/3) \end{align} etc.

Also note that $ab = 4$, $ac = 9$, $bc = 25$ are all perfect squares. Now I wonder if that is a precondition as well but not sure how to go about showing this.

  • $\begingroup$ Squares? No...If $(a,b,c)$ is a good triple, then so is $(na,b,c)$ for any $n\in \mathbb Z$ (well, at least any $n$ for which $na\not \in \mathbb Z$). So take your triple and multiply $a$ by $7$. $\endgroup$
    – lulu
    Aug 4, 2023 at 9:58
  • $\begingroup$ As an exercise, I suggest: prove that if $(a,b,c)$ is a good triple, then $a,b,c\in \mathbb Q$. That's true, but I don't think it's entirely obvious. $\endgroup$
    – lulu
    Aug 4, 2023 at 9:59
  • $\begingroup$ thanks. What exactly is a "good triple"? you mean one that is a solution to this exercise? $\endgroup$
    – yomath
    Aug 4, 2023 at 10:10
  • $\begingroup$ Yes, that's what I mean. $\endgroup$
    – lulu
    Aug 4, 2023 at 10:11
  • $\begingroup$ No problem. Should say: In the counterexample I gave, of course $\gcd(ab,ac)=7>1$. You could tighten your assumptions and require that $\gcd(ab,bc)=1, \gcd(ab,ac)=1, \gcd(bc,ac)=1$. Maybe in that case they are squares? Haven't thought about that. $\endgroup$
    – lulu
    Aug 4, 2023 at 10:15

4 Answers 4


Set $x:=ab$, $y:=bc$, $z:=ca$ and $p:=abc$. Note that $$ a=xz/p, \quad b=xy/p, \quad \text{ and }\quad c=yz/p. $$ Then it is enough to find four values of $x,y,z,p$ such that $a,b,c$ are not integers. The relation between those integers is that $ p^2=xyz. $ If you want that $a,b,c\notin \mathbb{Z}$, then the integer $p$ does not divide any of the products $xy$, $yz$, $zx$.

A class of examples: Pick three pairwise coprime integers $r,s,t \ge 2$ and set $$ x=r^2, y=s^2, z=t^2 \,\,\text{ and }\,\,p:=rst. $$ Then $$ a=rt/s, \quad b=rs/t, \quad \text{ and }\quad c=st/r $$ are not integers.

  • $\begingroup$ Thanks. I think several key steps I missed out on in my understanding were the relationship $p^2 = xyz$, that p should not divide any of the products xy.. etc. and especially also to select $r,s,t $ such that they are pairwise coprime. $\endgroup$
    – yomath
    Aug 4, 2023 at 11:06

We'll look at $a, b, c$ in $\mathbb{Q}$.

Let $p_i$ be the set of prime numbers that appear in $a, b, c$, with at least one strictly positive or strictly negative exponent among $a, b,$ or $c$.
Let $a_i, b_i, c_i$ be those exponents, i.e. $a = \prod_i p_i^{a_i}$, etc.

In order to meet the question, we must have:

  • some $a_i$, some $b_i$, some $c_i$ are strictly negative (otherwise $a, b$ or $c$ is an integer)
  • $\forall i, a_i + b_i \ge 0, b_i + c_i \ge 0, c_i + a_i \ge 0$
    Note that this entails $\forall i, a_i + b_i + c_i \ge 0$, i.e. $abc$ is an integer, too.

Given any $a_i$, it is always possible to find $b_i$ and $c_i$ that meet the conditions, which are rather large.

For a given $i$, we can represent $X_i = (a_i, b_i, c_i)$ in $\mathbb{R}^3$.
The conditions "$\forall i, a_i + b_i \ge 0, b_i + c_i \ge 0, c_i + a_i \ge 0$" are equivalent to: $X_i$ is in a cone, intersection of 3 semi-spaces.
The condition "some $a_i$, some $b_i$, some $c_i$ are strictly negative" means that $\exists i, X_i$ is not in the first orthant.

Note also that given a solution, it is always possible to find an infinity of other solutions by changing the $p_i$.

An example:
$(a_1, b_1, c_1) = (3, -2, 4)$
$(a_2, b_2, c_2) = (-1, 1, 2)$
$(a_3, b_3, c_3) = (1, 1, -1)$
with $p_1 = 2, p_2 = 3, p_3 = 5$
This gives: $a = \frac {40} 3, b = \frac {15} 4, c = \frac {144} 5$
$ab = 50, bc = 108, ca = 384, abc = 1440$


Another approach starts by observing that $a$, $b$, and $c$ must be rational but not integer, in which case we can express them them as $$a = \frac {a_n} {a_d}, b = \frac {b_n} {b_d}, c = \frac {c_n} {c_d}$$, with each numerator $x_n$ being a nonzero integer, each denominator $x_d$ being an integer greater than 1, and corresponding $x_n$, $x_d$ pairs being coprime.

We then observe that $ab = \frac {a_n}{a_d}\frac {b_n}{b_d} = m$ for some integer $m$, and therefore ${a_n}{b_n} = m{a_d}{b_d}$. But by construction, $a_n$ and $a_d$ have no common factors, so $a_d$ must divide $b_n$. Likewise, $b_d$ must divide $a_n$. By analogous arguments, it follows from $ac$ and $bc$ being integers that $a_d$ and $b_d$ each divide $c_n$, and $c_d$ divides both $a_n$ and $b_n$.

In that case, there must be nonzero integers $n_a$, $n_b$, and $n_c$ such that $$\DeclareMathOperator{\lcm}{lcm} a_n = n_a \lcm(b_d, c_d), b_n = n_b \lcm(a_d, c_d), c_n = n_c \lcm(a_d, b_d)$$ The ($x_n$, $x_d$) pairs are coprime by construction, so $a_d$ must be coprime with all factors of $a_n$, including $n_a$ and $\lcm(b_d, c_d)$ in particular, and therefore also with $b_d$ and $c_d$. Analogously, $b_d$ must be coprime with $n_b$ and $c_d$, and $c_d$ with $n_c$.

Since the denominators are all pairwise coprime, their pairwise least common multiples are just their products. Using that, together with the two previous sets of equations, we find that

Every solution must have the form $$a = \frac{n_a b_d c_d}{a_d}, b = \frac{n_b a_d c_d}{b_d}, c = \frac{n_c a_d b_d}{c_d}$$ , where

  • $a_d$, $b_d$, and $c_d$ are mutually coprime integers greater than 1, and
  • $n_a$, $n_b$, and $n_c$ are nonzero integers, coprime with $a_d$, $b_d$, and $c_d$, respectively.

Now consider the product $a b c$. Substituting from the previous formulas, we get $$abc = \frac{n_a b_d c_d}{a_d} \frac{n_b a_d c_d}{b_d} \frac{n_c a_d b_d}{c_d} = n_a n_b n_c a_d b_d c_d$$, which is an integer. Thus, the conditions above are not only necessary, but also sufficient for $a$, $b$, and $c$ satisfying the requirements.

Example: if we choose the first three primes for the denominators and 1 for $n_a$, $n_b$, and $n_c$ then that gives us $$ a = \frac{1 \times 3 \times 5}{2}, b = \frac{1 \times 2 \times 5}{3}, c = \frac{1 \times 2 \times 3}{5}$$, or $$ a = \frac{15}{2}, b = \frac{10}{3}, c = \frac{6}{5}$$ Looks familiar ....

  • $\begingroup$ Please use the editing sandbox if you need to make many edits over a short period. $\endgroup$ Aug 5, 2023 at 15:46

Let $p,q,r$ be three distinct positive integers such that $pqr$ is a perfect square $s^2,$ then the three distinct real numbers $$a:=\frac sr,\;b:=\frac sq,\;c:=\frac sp$$ satisfy $$ab=p,\;ac=q,\;bc=r,\;abc=s,$$ and none of them is an integer as long as $p,q,r$ don't divide $s.$


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