TITLE: Finding triplets $(a,b,c)$ such that $\sqrt{abc}\in\mathbb N$ divides $(a-1)(b-1)(c-1)$
QUESTION [57 upvotes]: When I was playing with numbers, I found that there are many triplets of three positive integers $(a,b,c)$ such that 

$\color{red}{2\le} a\le b\le c$
$\sqrt{abc}\in\mathbb N$
$\sqrt{abc}$ divides $(a-1)(b-1)(c-1)$

Examples : The followings are positive integers.
$$\frac{(2-1)(8-1)(49-1)}{\sqrt{2\cdot 8\cdot 49}},\ \frac{(6-1)(24-1)(529-1)}{\sqrt{6\cdot 24\cdot 529}},\frac{(7-1)(63-1)(3844-1)}{\sqrt{7\cdot 63\cdot 3844}}$$
Then, I began to try to find every such triplet. Then, I found 
$$(a,b,c)=(k,km^2,(km^2-1)^2)$$
where $k,m$ are positive integers such that $k\ge 2$ and $km^2\ge 3$, so I knew that there are infinitely many such triplets. However, I can neither find the other triplets nor prove that there are no other triplets. So, here is my question. 

Question : How can we find every such triplet $(a,b,c)$?

Added : There are other triplets : $(a,b,c)=(k,k,(k-1)^4)\ (k\ge 3)$ by a user user84413, $(6,24,25),(15,15,16)$ by a user Théophile. Also, from the first example by Théophile, I got $(2k,8k,(2k-1)^2)\ (k\ge 3)$.
Added : $(a,b,c)=(k^2,(k+1)^2,(k+2)^2)\ (k\ge 2)$ found by a user coffeemath. From this example, I got $(k^2,(k+1)^2,(k-1)^2(k+2)^2)\ (k\ge 2)$.
Added : I got $(a,b,c)=(2(2k-1),32(2k-1),(4k-3)^2)\ (k\ge 5)$.
Added : I got $(a,b,c)=(k,(k-1)^2,k(k-2)^2)\ (k\ge 4)$.
Added : A squarefree triplet $(6,10,15)$ and $(4,k^2,(k+1)^2)\ (k\ge 2)$ found by a user martin. 
Added : user52733 shows that $(6,10,15)$ is the only squarefree solution.

REPLY [7 votes]: Still in extended comment territory, but I wanted to confirm @martin's conjecture that $(6, 10, 15)$ is the only squarefree solution.
Notations: 

$\mathbb{N} = \lbrace 1, 2, 3, \dotsc \rbrace$.  
The statement "$d$ divides $n$" is abbreviated $d \mid n$ when approriate.

The first statement is a quick rewrite.
Lemma 1:  Suppose $a, b, c \in \mathbb{N} \setminus \lbrace{1\rbrace}$ satisfy
\begin{equation} \tag{*}
\sqrt{abc} \in \mathbb{N}.
\end{equation}
Then the following are equivalent.
\begin{align}
\sqrt{abc} & \mid (a-1)(b-1)(c-1) \tag{1a}\\
\sqrt{abc} & \mid ab + bc + ca - (a + b + c) + 1  \tag{1b}
\end{align}
Proof.  Suppose that $a, b, c \in \mathbb{N} \setminus \lbrace{1\rbrace}$ satisfies (*). Multiplying out,
$$(a-1)(b-1)(c-1) = abc - (ab + bc + ca) + (a + b + c) - 1,$$
and certainly, whenever $\sqrt{abc} \in \mathbb{N}$, $\sqrt{abc}$ divides $abc$.  Thus, if (1a) holds, then $\sqrt{abc}$ divides both $abc$ and $(a-1)(b-1)(c-1)$, and hence divides
$$abc - (a-1)(b-1)(c-1) = (ab + bc + ca) - (a + b + c) + 1.$$
So (1b) holds.  The reverse direction is similar.  $\square$

We now note some divisibility conditions.  We start with an observation.
Lemma 2: Let $a, b, c \in \mathbb{N} \setminus \lbrace{1\rbrace}$ satisfy (*), and let $p$ be a prime divisor of $abc$.  Then either $p$ divides at least two of $a$, $b$, and $c$, or $p^2$ divides one of $a, b, c$.
The proof is an undergraduate number theory exercise, following from, e.g., Lemmas 2.1 and 2.4 of Jones and Jones, Elementary Number Theory.
The natural question is, "Can $p$ divide all three of $a$, $b$, and $c$?"  The answer is no, and the argument holds for all divisors $d$ to an extent.
Lemma 3: Suppose $a, b, c \in \mathbb{N} \setminus \lbrace{1\rbrace}$ satisfy (*) and (1b).  Fix $d \in \mathbb{N}$.

If $d$ divides $a$ and $d$ divides $b$, then $d$ divides $c - 1$.  
If $d^2$ divides $a$, then $d$ divides $(b-1)(c-1)$.

Proof, part 1.  Suppose $a, b, c \in \mathbb{N} \setminus \lbrace{1\rbrace}$ satisfy (*) and (1b), and also suppose that for some $d \in \mathbb{N}$, $d$ divides both $a$ and $b$.  Then certainly $d^2$ divides $abc$, so $d$ divides $\sqrt{abc}$.  Since (1b) holds and divisibility is transitive, $d$ divides $(ab + bc + ac) - (a + b + c) + 1$.  Since $d$ divides $a$ and $b$, it clearly divides $ab$ and $bc$ and $ac$.  Hence, $d$ divides
$$N := ab + bc + ac - (a + b).$$
Yet we already said that $d$ divides
$$(ab + bc + ac) - (a + b + c) + 1 = N - c + 1.$$
Thus, $d$ divides
$$N - (N - c + 1) = c - 1.  \quad \square.$$
Proof, part 2 (Modular Arithmetic).   Suppose $a, b, c \in \mathbb{N} \setminus \lbrace{1\rbrace}$ satisfy (*) and (1b).  By Lemma 1, (1a) holds.  Suppose that for some $d \in \mathbb{N}$, $d^2$ divides $a$.  If $d = 1$, then by $b, c \in \mathbb{N} \setminus \lbrace{1\rbrace}$, $d$ divides $(b-1)(c-1)$, so assume $d \neq 1$.  Since $d^2 \mid a$, $d^2 \mid abc$ and $d \mid \sqrt{abc}$.  By (1a) and transitivity of division, $d$ divides $(a-1)(b-1)(c-1)$, or equivalently,
$$(a-1)(b-1)(c-1) \equiv 0 \pmod{d}.$$
Yet $d^2$, and hence $d$, divides $a$, so $(a-1) \equiv -1 \pmod{d}$.  Yet by $d \geq 2$, $-1$ is certainly a unit in $\mathbb{Z}/d\mathbb{Z}$, since $(-1)^2 = 1$.  Hence, [the equivalence class of] $(a-1)$ is a unit in $\mathbb{Z}/d\mathbb{Z}$, hence not a zero-divisor.  Thus, the only way for $(a-1)(b-1)(c-1) \equiv 0 \pmod{d}$ is if $(b-1)(c-1) \equiv 0 \pmod{d}$, i.e., $d$ divides $(b-1)(c-1)$.  $\square$
Lemma 3, part 1 (hereafter, Lemma 3.1) shows in particular that $\gcd(a, b)$ divides $(c-1)$ for any solution $(a, b, c)$ (and the corresponding statements by permuting the letters).  Lemma 3, part 2 is weaker, but is slightly stronger if $d$ is a prime:
Corollary 4: Suppose $a, b, c \in \mathbb{N} \setminus \lbrace{1\rbrace}$ satisfy (*) and (1b), and suppose that $p$ is some prime divisor of $abc$.  Then at least one of $a$, $b$, and $c$ is congruent to $0 \pmod{p}$, and at least one of $a$, $b$ and $c$ is congruent to $1 \pmod{p}$.  
Proof.  Suppose $a, b, c \in \mathbb{N} \setminus \lbrace{1\rbrace}$ satisfy (*) and (1b), and suppose that $p$ is some prime divisor of $abc$.  By Lemma 2, either $p$ divides 2 of $a$, $b$, and $c$, or $p^2$ divides one of $a$, $b$, and $c$.  In the former case, without loss of generality $p$ divides $a$ and $b$, and applying the $d = p$ case of Lemma 3.1, $p$ divides $c-1$, i.e., $c \equiv 1 \pmod{p}$.  In the latter case, without loss of generality, $p^2$ divides $a$, and applying the $d = p$ case of Lemma 3.2, $p$ divides $(b-1)(c-1)$.  Since $p$ is prime, $p \mid (b-1)$ or $p \mid (c-1)$, i.e., $b \equiv 1 \pmod{p}$ or $c \equiv 1 \pmod{p}$.  $\square$
For the next divisibility condition, we note that the easiest way to satisfy (*) [apart from $a$, $b$, $c$ each being square] is for $c = ab$.  We show that such a combination never gives a solution to (1b).  We need the following rule, using the order properties of $\mathbb{N}$.
Lemma 5: If $m, n \in \mathbb{N}$ and $m$ divides $n$, then $m \leq n$.
The proof is a standard undergraduate number theory fact (e.g., Jones and Jones, Elementary Number Theory, Exercise 1.3(d)).  
Lemma 6: Let $a, b \in \mathbb{N} \setminus \lbrace{1\rbrace}$, and let $c := ab$. Then $a, b, c$ do not satisfy (1b).
Proof.  Let $a, b \in \mathbb{N} \setminus \lbrace{1\rbrace}$, and let $c := ab$.  Then (*) holds, with $\sqrt{abc} = \sqrt{a^2 b^2} = ab$ (by $a, b > 0$).  Suppose, by way of contradiction, that $a, b, c$ do satisfy (1b).  Then $ab$ divides
\begin{align}
ab + ac + bc - (a + b + c) + 1 &= ab + a^2 b + ab^2 - (a + b + ab) + 1\\
& = ab(1 + a + b - 1) - (a + b) + 1\\
& = ab(a + b) - (a + b) + 1
\end{align}
Yet $ab$ clearly divides $ab(a + b)$.  Thus, $ab$ divides
$$ab(a + b) - [ab(a + b) - (a + b) + 1] = a + b - 1.$$
In particular, $a \mid a + b -1$, and $a \mid a$, so $a \mid b - 1$; similarly, $b \mid a - 1$.  Since $a$, $a-1$, $b$, and $b-1$ are positive integers, by Lemma 5, $a \leq b - 1$ and $b \leq a - 1$, so $a \leq a - 2$, a contradiction.  $\square$
Sadly, although our divisibility checks reduce our workload, they are not yet a characterization; as shown by a computer search, the cases $(3, 4, 27)$, $(7, 8, 14)$, and $(6, 15, 40)$ [among others] allow $\sqrt{abc} \in \mathbb{N}$ and satisfy all the divisibility checks above, but do not satisfy (1a).

Now we get to the main proof.  Our first argument is that the square freeness, combined with Lemmas 2 and 6, force a large amount of common factors.
Proposition 7: Let $a, b, c \in \mathbb{N} \setminus \lbrace{1\rbrace}$ satisfy (*) and (1b).  If in addition, $a$, $b$, $c$ are squarefree, then we may write
\begin{align}
a &= PQ\\
b & = PR\\
c & = QR
\end{align}
where $P, Q, R \in \mathbb{N} \setminus \lbrace{1\rbrace}$ are pairwise coprime.  
Proof.  Since $a$ is squarefree, for any prime $p$, $p^2 \nmid a$.  Thus, $a$ is a product of distinct prime factors.  Yet by Lemma 2, for any prime $p$ dividing $a$, $p$ must therefore divide one of $b$ or $c$; by Lemma 3.1, it divides exactly one of $b$ or $c$.   Thus, we may write $a = PQ$ where
$$P = \prod_{\substack{ p \text{ prime}\\ p \mid a \text{ and } p \mid b }} p; \quad Q = \prod_{\substack{ p \text{ prime}\\ p \mid a \text{ and } p \mid c }} p,$$
where the empty product is $1$.  Similarly, since $b$ and $c$ are also squarefree, then we may write $b = PR$ and $c = QR$, where
$$ R = \prod_{\substack{ p \text{ prime}\\ p \mid b \text{ and } p \mid c }} p.$$
Obviously, $P$, $Q$, and $R$, being factors of squarefree numbers, are squarefree.  Also, $P$ and $Q$ are coprime, else $a$ would not be squarefree; similarly $P$ and $R$ are coprime, and $Q$ and $R$ are coprime.
Moreover, $P \neq 1$; else, $a = Q$, $b = R$, and $c = QR = ab$, which by Lemma 6 cannot hold.  Similarly, $Q \neq 1$ and $R \neq 1$.  $\square$
Now, it suffices to show that by the power of Lemma 3.1, any solution with the above decomposition cannot hold.
Proposition 8: Fix $P, Q, R \in \mathbb{N} \setminus \lbrace{1\rbrace}$, and define
\begin{align}
a &:= PQ\\
b &:= PR\\
c & := QR
\end{align}
Then (*) holds.  If in addition, $P$, $Q$ and $R$ are pairwise coprime, and $a, b, c$ satisfy (1b), then up to permutations, $P = 2$, $Q = 3$, and $R = 5$; i.e., $(a, b, c)$ is $(6, 10, 15)$ up to permutations.
Proof.  Fix $P, Q, R \in \mathbb{N} \setminus \lbrace{1\rbrace}$, and define $a$, $b$, $c$ as above.  Then clearly $abc = (PQR)^2$, so $\sqrt{abc} = PQR$ is in $\mathbb{N}$. (*) holds.  
Suppose in addition that $P$, $Q$ and $R$ are pairwise coprime, and that (1b) holds; i.e., $PQR$ divides
\begin{align}
ab + ac + bc - (a + b + c) + 1 & = P^2QR + PQ^2R + PQR^2 - (PQ + PR + QR) + 1\\
& = PQR(P + Q + R) - (PQ + PR + QR) + 1
\end{align}
Yet clearly, $PQR$ divides $PQR(P + Q + R)$.  Hence, $PQR$ divides
$$PQR(P + Q + R) - \left[ PQR(P + Q + R) - (PQ + PR + QR) + 1 \right] = PQ + PR + QR - 1.$$
Since $P, Q, R \in \mathbb{N} \setminus \lbrace{1\rbrace}$, $PQ + PR + QR - 1 \geq 2$, so we may apply Lemma 5 and get that
\begin{equation} 
0 < PQR \leq PQ + PR + QR - 1 \tag{2}
\end{equation}
Claim 1.  At least one of $P$, $Q$, and $R$ is less than $3$.
Suppose, by way of contradiction, that $P \geq 3$, $Q \geq 3$, and $R \geq 3$.  Then 
\begin{align}
PQ + PR + QR - 1 & = \frac{PQR}{R} + \frac{PQR}{Q} + \frac{PQR}{P} - 1\\
& \leq \frac{PQR}{3} + \frac{PQR}{3} + \frac{PQR}{3} - 1\\
& = PQR - 1 < PQR
\end{align}
Yet this contradicts (2).
Since $P, Q, R$ are each at least $2$, at least one of them, say $P$, is equal to $2$.  Since $Q$ and $R$ are coprime to $P$, they must be at least $3$.
Claim 2.  One of $Q$ and $R$ is equal to $3$.  
Suppose by way of contradiction that $Q > 3$ and $R > 3$.  Then by $Q$ coprime to $P$, $Q \neq 4$, so $Q \geq 5$; similarly, $R \geq 5$.  Thus,
\begin{align}
PQ + PR + QR - 1 & = \frac{PQR}{R} + \frac{PQR}{Q} + \frac{PQR}{P} - 1\\
& \leq \frac{PQR}{5} + \frac{PQR}{5} + \frac{PQR}{2} - 1\\
& = \frac{9}{10} \, PQR - 1 < PQR
\end{align}
Again, we contradict (2).  
Thus, without loss of generality, $Q = 3$.  Now the second inequality in (2) reads
\begin{align}
6R &\leq 6 + 2R + 3R -1, &&\text{or}\\
6R & \leq 5R + 5, &&\text{or}\\
R & \leq 5
\end{align}
Since $R > 1$, and $R$ is coprime to $2$ and $3$, then $R = 5$.  $\square$
Combining Propositions 7 and 8, we have our result.