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# A General Problem on functions

## Problem

Let $\mathbb{Z}$ denote the set of all integers (as usually it do ).

Consider a function $f : \mathbb{Z} \rightarrow \mathbb{Z}$ with the following properties:
$f (92+x) = f (92-x)$
$f (19 \times 92+x) = f (19 \times 92 -x)$
$f (1992+x)=f (1992-x)$ for all $x \in \mathbb{Z}$. Is it possible that all positive divisors of $92$ occur as values of $f$?
A happy note: $19 \times 92$ is actually 1748 and it is written to retain symmetry in problem.

## Solution

The answer is $No$.
Since $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfies:
$f (92+x) = f (92-x)$ . . . (1)
$f (1748+x) = f (1748-x)$ . . . (2)
$f (1992+x)=f (1992-x)$ . . . (3)
Then, we have
$f(488+x)=f(244+244+x) \\ =f(1992-1748+244+x) \\ =f(1992-(1748-244-x)) \\ =f(1992+1748-244-x), \ \textbf {by \ (3) } \\ =f(1748+(1992-244-x)) \\ =f(1748-1992+244+x), \ \textbf {by \ (2) } \\ = f(x)$ . . . (4)

Again we have
$f(40+x)=f(1992-1952+x) \\ \, =f(1992-4 \times 448 +x) \\ \, =f(1992+x), \ \textbf {by repeated application of (4)} \\ \, =f(1992-x), \ \textbf {by (2)} \\ \, =f(1992 -4 \times 448 -x), \ \textbf{by repeated app. of (4)} \\ \, =f(40-x)$ . . . (5)
So, $f(104+x)=f(52+52+x) \\ \, =f(92-40+52+x) \\ \, =f(92+40-52-x), \ \textbf {by (1)} \\ \, =f(40+92-52-x) \\ \, =f(40-92+52+x), \ \textbf {by (5)} \\ \, =f(x)$ . . . .(6)

It’s your test of breaking and arranging numbers.

Similarly we can prove that $f(8+x)=f(x)$ by this way:
Since, $8=3\times 488-14 \times 104$
Therefore; $f(8+x) = f(3\times 488-14 \times 104+x) \\ \, =f(-14 \times 104+x), \ \textbf{by repeated app. of (4)} \\ \, =f(x) \ \textbf{by repeated app. of (6)}$ .. …..(7)

This shows that $f$ is periodic by $period=8$ (the least we could get) and all possible values of $f$ are in the list ${f(0), f(1), f(2), ..., f(7)}$. Finally $f(4+x)=f(92-88+x)=f(92-8\times 11+x) =f(92+x)$ by periodicity
$=f(92-x)$ by (1).
$=f(92-8\times 11 +x) =f(4-x)$
In general, $f(4+x)=f(4-x)$
In particular, $f(7)=f(1) \ \, f(6)=f(2) \ \, f(5)=f(3)$.
Hence all possible values of function $f$ are ${f(0), f(1), f(2), f(3)}$ and $f(4)$. In particular, $f$ assumes no more than 5 function values. However, 92 has 6 positive divisors 1, 2, 4, 23, 46 and 92. Hence, the all positive divisors of 92 cannot occur as values of $f$. That’s why the answer is “No”.