Finding triplets $(a,b,c)$ such that $\sqrt{abc}\in\mathbb N$ divides $(a-1)(b-1)(c-1)$

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.

• Why the downvotes? – Daniel W. Farlow Apr 25 '15 at 17:52
• I really don't understand the downvote. The matter seems interesting, it is not homework, the OP tells his thoughts... I have no idea of what would be the solution, but I'm pretty interested. – ajotatxe Apr 25 '15 at 17:55
• Sigh, would the downvoter care to comment? – Sufyan Naeem Apr 25 '15 at 18:18
• A couple of triples that don't fit the pattern: $(6,24,25)$ and $(15,15,16)$. – Théophile Apr 25 '15 at 19:24
• Another family is sequences of three consecutive squares i.e. $(a,b,c)=(k^2,(k+1)^2,(k+2)^2)$ with $k \ge 2$ for which the ratio becomes $(k-1)(k+1)(k+3).$ – coffeemath Apr 27 '15 at 2:48

Too long for a comment:

In addition to the rather lengthy

\begin{align} &(m^2,\\ &((-1)^{2 k} \left(2 (-1)^k k m+(-1)^{k+1} (m+2)+m-6\right)^2)/16,\\ &\left((-1)^k \left(2 (-1)^k k m+(-1)^{k+1} (m+2)+m-6\right)+1\right)^2/4)\\ \end{align}

we also have $(a,b,c):$

\begin{align} &\left(k^3+k^2+k+1,k^3+k^2+k+1,k^4\right)\\ &\left(k^4+k^2+1,k^4+k^2+1,k^6\right)\\ &\left(k m^2,k m^2 \left(k m^2-2\right)^2,\left(k m^2 \left(k m^2-3\right)+1\right)^2\right)\\ \end{align}

and for $f(n)=(n-1)^2$ we also have

\begin{align} &\left(k^2,f^{2 n-1} \left((k m+1)^2\right),f^{2 n} \left((k m+1)^2\right)\right)\\ \end{align}

where $f^n$ is $f$ iterated $n$ times for $n \geq 1.$

However, even for fixed $a,$ the above formulae don't catch all of the solutions (and they say nothing of non-square $a$ combinations), and yet for each $a$ there seem to be multiple (infinite?) solutions.

Examples: case $a=8:$

A straightforward brute-force search for $(8,b,c);\ (b,c)<1000$ gives triples

$(8,2,49),(8,8,49),(8,18,49),(8,18,289),(8,32,49),(8,32,961),(8,49,72),(8,49,288),(8,289,392),(8,392,529),$

where it is immediately apparent that the same numbers recur a number of times. Removing the $8$ and graphing shows the connectedness more clearly:

Searching for $c$ only, using the distinct elements from the initial search (eg $(8,49,c)$, etc.) up to $10^5$ reveals further connections:

$(8,49,c)$ for example turns up $6$ triplets: $(8,49,2),(8,49,8),(8,49,18),(8,49,32),(8,49,72),(8,49,288)$

It may be more pertinent to ask then, are there infinitely many triplets for fixed $a?$ Certainly where $a$ is square, this is the case, but it is less clear whether this is the case when it is not.

It may also be worthwhile pursuing the idea of primitive pairs $(a,b).$

• Very nice.Yes, just one or all the three a, b, c can be a square (never just two). And it is possible to find solution for the family of triplets (a,b,c) having some restriction at the start. But the question ask for EVERY triplet and I am convinced it is impossible because of the main equation with three unknown with severe condition. We can say the hardest problem in this is to find a solution, if it exists, with non of a, b, c square. Till now neither I did not see any example nor a proof of it doesn't existe. – Piquito Apr 30 '15 at 13:30
• @LuisGomezSanchez I must admit, I am leaning towards that view, since there seems to be evidence of both infinite progressions and arbitrarily long finite ones. This seems to be more of a combinatorial problem. – martin Apr 30 '15 at 13:47
• All the "Added" posted by Mathlove give infinitely many solutions because these Addeds are identities really. On the other hand, respect of square-free triplets, not matter of calculator in hand and patience but show a method of finding it. I have given a method of deduction for a family but it have been refused I think for bad written. (Sorry for bad english without traductor of Google now; I progress anyway on this) – Piquito Apr 30 '15 at 14:52
• You are an artist, Martin. – Piquito Apr 30 '15 at 16:26
• What I find fascinating is that when you plug in $f_k(n)$ into $a,\,b,\,c$ and do the division you get the integer $n ( k^2-1) ( k n+3)$ so it works for any choice of $k,n$ where $k \ge 2$. – Alexander Vlasev May 1 '15 at 19:36

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

$$\tag{*} \sqrt{abc} \in \mathbb{N}.$$

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. $\square$

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 $d \in \mathbb{N}$.

1. If $d$ divides $a$ and $d$ divides $b$, then $d$ divides $c - 1$.
2. 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 $$0 < PQR \leq PQ + PR + QR - 1 \tag{2}$$

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.

• I think I can get what you mean. Thank you for your answer. – mathlove Aug 28 '15 at 22:47

(Too long for a comment.)

The two solutions,

$$a,\,b,\,c = k^2,\;(k+1)^2,\;(k+2)^2$$

$$a,\,b,\,c = 2^2,\;k^2,\;(k+1)^2$$

by users coffeemath and martin, respectively, are special cases of the more general solution,

$$a,\,b,\,c = k^2,\;(km\pm1)^2,\;(km\pm2)^2$$

where coffeemath's had $m=1$, while martin's had $k=2,\, m = \frac{n}{2}$.

We study solutions, if any, of the shape $$(a,b,c) = (x^2 P + 1,~ y^2 P + 1,~ z^2)$$. We need $$z^2 \left(x^2 y^2 P^2 + (x^2 + y^2) P + 1 ) \right)$$ be a square. Namely, we need \begin{align} 2 x^2 y^2 P &= -(x^2 + y^2) + \sqrt{ (x^2 + y^2)^2 + 4(w^2 - 1)x^2 y^2} \\ \implies P &= \frac{ -(x^2 + y^2) + u }{ 2 x^2 y^2} \end{align} where we have $$(x^2 + y^2)^2 + (2wxy)^2 = (2xy)^2 + u^2$$

The equation $$X^2 + Y^2 = V^2 + W^2$$ like the Pythagorean one has also infinitely many solutions. For arbitrary integer $$M,N,r,s$$ we have

\begin{align} 2X &= r M + s N & 2Y &= r N – s M \\ 2V &= r N + s M & 2W &= r M – s N \end{align} making

\begin{align} 2(x^2 + y^2) &= r M + s N & 4wxy &= r N – s M \\ 4xy &= rN + sM & 2u &= rM – sN \end{align}

The main equation $$(z^2 - 1) x^2 y^2 P^2 = Zzw \quad (?)$$ with $$Z$$ integer becomes $$4(z^2 - 1)s^2 N^2 = Zz (r^2 N^2 - s^2 M^2)$$ and we have to choose conveniently, if possible, among the arbitrary parameters in order this equation be verified, in particular $$z$$ must divide $$4s^2 N^2$$ for instance. No time for me now. If someone wants to continue go on.

• Sorry, I don't see how this contributes to a solution. Also, $abc$ can certainly be square without any of $a$, $b$, $c$ being squares (2,3,6). Do you mean to say you can prove that the divisibility condition precludes this? – Erick Wong Apr 29 '15 at 7:01
• @Erick Wong The case of all, a, b, c, square-free should be excluded from the equation (I would be stupid to affirm such a thing regardless of the equation). – Piquito Apr 29 '15 at 14:05
• Why should $a,b,c$ square-free be excluded? And what does that very specific condition have to do with $a,b,c$ not being squares? – Erick Wong Apr 29 '15 at 15:03
• I was wrong, as I am accustomed to dealing with coprime numbers in Diophantine equations. However my viewpoint still stands, although I'm still looking for a solution as generally as possible.I'll edit my answer. Thanks. – Piquito Apr 29 '15 at 15:21