Equations involving Euler's $\varphi$

Question

Let $x_1,x_2,\ldots, x_n, y_1,y_2,\ldots, y_n$ be some positive integers and let $p$ be a prime number. It is possible to show that if $\sum\limits_{i=1}^nx_i\varphi(p^i)=\sum\limits_{i=1}^ny_i\varphi(p^i)$ and $\sum\limits_{i=1}^nx_i=\sum\limits_{i=1}^ny_i$, then $x_i=y_i, \ \forall \ i=1,2,\ldots, n$ ? I tried to prove it by induction over $n$ (it worked for n=1, n=2 since we have two equations). However, I couldn't go further and use the inductive hypothesis. Any hint/idea is appreciated.

Answer 1

With $n \gt 2$ and a particular $p$, you have too many free variables for the two equalities to force pairwise equality

For a counterexample with $n=3$ and $p=2$ and $x_1 =5, x_2=6, x_3=1, y_1 =7, y_2=3, y_3=2$ (so all the $x_i$ and $y_i$ distinct - there are smaller examples), you have

• $\varphi(p^1)=\varphi(2)=1$
• $\varphi(p^2)=\varphi(4)=2$
• $\varphi(p^3)=\varphi(8)=4$

and thus the two equalities are

$$\sum\limits_{i=1}^nx_i\varphi(p^i)=5\times 1+ 6 \times 2+ 1 \times 4 = 21=7\times1+ 3\times 2+ 2\times 4 = \sum\limits_{i=1}^ny_i\varphi(p^i)$$ and $$\sum\limits_{i=1}^nx_i=5+6+1=12 =7+3+2=\sum\limits_{i=1}^ny_i$$