3
$\begingroup$

I'm inspired in a recent post of this MSE. We denote the Euler's totient function in this post as $$\varphi(n)=n\prod_{p\mid n}\left(1-\frac{1}{p}\right).$$

Suppose we have three positive integers $a,b,c\geq 1$ that form a Markov triple, that is that satisfy $$a^2+b^2+c^2=3abc.\tag{1}$$

Additionally, suppose that when we apply the Euler's totient function to each term in LHS our Markov triple $(a,b,c)$ satisfies $$\varphi(a)^2+\varphi(b)^2+\varphi(c)^2=3\varphi(abc).\tag{2}$$

Question. Can you find all Markov triples satisfying $$\varphi(a)^2+\varphi(b)^2+\varphi(c)^2=3\varphi(abc)?$$ I can find four solutions, justify if these are all solutions or provide us a counterexample. Many thanks.

Thus I am asking about the characterization of the solutions $(a,b,c)$ with $a,b,c\geq 1$ instegers satisfying $(1)$ and $(2)$.

Computational fact. When $1\leq a,b,c\leq 500$ the only Markov triples satisfying $(2)$ (thus the solutions of our problem) are $(a,b,c)=(1,1,1),(2,1,1,),(1,2,1)$ and $(1,1,2)$.

$\endgroup$
  • 2
    $\begingroup$ If $\varphi(a)^2+\varphi(b)^2+\varphi(c)^3=\varphi(abc)$ holds then $(\varphi(a),\varphi(b),\varphi(c))$ forms a Markov triple. I don't know my observation is helpful. $\endgroup$ – Hanul Jeon Mar 22 '18 at 10:57
  • $\begingroup$ Yes you are right @HanulJeon , many thanks for your contribution. I suspect that the proof that there aren't more solutions that I'v eshowed should be easy. $\endgroup$ – user243301 Mar 22 '18 at 11:09
1
$\begingroup$

First, as mentioned in a comment by Hanul Jeon, if $(a,b,c)$ is Markov triple such that $\varphi(a)^2 + \varphi(b)^2 + \varphi(c)^2 = 3\varphi(abc)$, then $(\varphi(a), \varphi(b), \varphi(c))$ is another Markov triple. This is true because $a$, $b$, and $c$ are always pairwise relatively prime, so we can rewrite $\varphi(abc)$ as $\varphi(a)\varphi(b)\varphi(c)$. (I'll postpone the proof of this property of Markov triples, and another property we'll need, until the end.)

We can show that there are no such Markov triples, aside from the four already mentioned, by proving a stronger claim. If we define a Markov number to be a number that appears in any Markov triple, then for any Markov number $m$ except $1$ and $2$, $\varphi(m)$ is not Markov.

This follows from a factorization property of Markov numbers. If $m$ is a Markov number, then it is divisible by $2$ at most once, and all other prime divisors $p$ of $m$ satisfy $p \equiv 1 \pmod 4$. So when $m>2$ it must have some odd prime divisor $p$, and $\varphi(p) = p-1$ is divisible by $4$. Therefore $\varphi(m)$ is divisible by $4$, which is true of no Markov number.

So the only possible Markov triples that survive $\varphi$ are the ones composed only of $1$ and $2$.


As promised, here are proofs of the properties of Markov numbers and Markov triples we'll use. These are from the book Markov's Theorem and 100 Years of the Uniqueness Conjecture wherever I couldn't figure them out myself, and I'll cite the specific propositions used.

Corollary 3.4. Any two Markov numbers in a Markov triple are relatively prime.

Suppose not; let $(a,b,c)$ be the smallest counterexample (say, smallest by maximum), and assume $a>b>c$. Then it can be checked that $(b, c, 3bc-a)$ is also a Markov triple: $a$ and $3bc-a$ are the two roots of $x^2 - 3bcx + b^2 + c^2 = 0$. Also, setting $x = b$ in this quadratic yields a negative number, so $b$ is between $a$ and $3bc-a$, which means in particular $3bc-a < a$. So all numbers in $(b, c, 3bc-a)$ are smaller than $a$.

But now, if $d$ were a common divisor of two of $(a,b,c)$ then it would be a common divisor of two of $(b,c,3bc-a)$, so we get a smaller counterexample, contradiction.

Proposition 3.13 (paraphrased). If $m$ is a Markov number, then any prime divisor $p$ satisfies $p \equiv 1 \pmod 4$; if $m$ is even, then $m \equiv 2 \pmod 4$.

Let $(m,b,c)$ be a Markov triple. Then $b^2 + c^2 = 3mbc-m^2 \equiv 0 \pmod m$, so for any odd prime divisor $p$ of $m$, $b^2 \equiv -c^2 \pmod p$. Since $b$ and $c$ are neither divisible by $p$, they are invertible modulo $p$, so $-1$ is a quadratic residue modulo $p$. This means $p \equiv 1 \pmod 4$.

If $m$ is even, then $b$ and $c$ are odd. So $m^2 + b^2 + c^2 \equiv 2 \pmod 4$, which means $3mbc \equiv 2 \pmod 4$; this means $m$ is not divisible by $4$.

$\endgroup$
  • $\begingroup$ Many thanks, I am going to read and study your nice answer. $\endgroup$ – user243301 Mar 29 '18 at 6:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy