Pythagorean triplets which satisfy Euler's totient function

Question

Let $\varphi(x)$ be the Euler totient function and $a,b$ and $c$ be natural numbers.

Question 1: Are there infinitely many non-trivial solutions of
$$ \varphi(a)^2 = \varphi(b)^2 + \varphi(c)^2 $$

$$ \varphi(a^2) = \varphi(b^2) + \varphi(c^2) $$ A trivial solution is one which is obtained multiplying a smaller solution with a constant natural number.

The first few solutions are

(1004, 802, 604)
(1012, 782, 644)
(1050, 840, 630)
(1056, 816, 672)
(1084, 866, 652)
(1100, 850, 700)
(1136, 904, 688)
(1144, 884, 728)
(1188, 918, 756)
(1200, 960, 720)

Question 2: Is there a triplet with at least one of the three numbers $a,b$ and $c$ odd?

Related question: Pythagorean triples that “survive” Euler's totient function

Dunno if it's useful, but the second equation can be rewritten as $a\phi(a)=b\phi(b)+c\phi(c).$ — Hagen von Eitzen, Oct 06 '19 at 08:12
which is the same as certain rational coefficients on $a^2$,$b^2$ and $c^2$ — , Oct 06 '19 at 14:10

score 1 · Accepted Answer · answered May 29 '20 at 21:48

For Question 1, here is an infinite family of solutions for $n \in \mathbb{N}$. None these can be obtained by scaling another, so each of them can be reduced to a 'non-trivial' solution, as you call it.

\begin{align} a &= (n^2+1)^2(n^2-1)(2n) \\ b &= (n^2+1)(n^2-1)^2(2n) \\ c &= (n^2+1)(n^2-1)(2n)^2. \end{align}

This family was generated by making a family of scaling-different triples using the general formula for Pythagorean triples (which is $a = t(p^2+q^2)$, $b = t(p^2-q^2)$, and $c = t(2pq)$), and then scaling them up to ensure that $a$, $b$, and $c$ have the same prime factors, since then

\begin{equation} \frac{\varphi(a)}{a} = \frac{\varphi(b)}{b} = \frac{\varphi(c)}{c} = \frac{\varphi(a^2)}{a^2} = \frac{\varphi(b^2)}{b^2} = \frac{\varphi(c^2)}{c^2}. \end{equation}

Pythagorean triplets which satisfy Euler's totient function

1 Answers1