Inspired by this question, which noted that for all natural numbers $a>2$, $(2^{15}-2^3)|(a^{15}-a^3)$.
My question deals with generalizing this:
Let let $a,b$ be integers such that $a>b\geq1$. For what values of $a,b$ is it true that for all coprime $x>y$ such that $x\nmid(y^a-y^b)$ and $y\nmid(x^a-x^b)$ can $\gcd(x^a-x^b,y^a-y^b)$ be written in the form $2^{a+c}-2^{b+c}$ for some integer $c$?
To analyze this, I declared $d=a-b$, and considered each value of $d$ separately. Restated in terms of $d$, the problem becomes:
A) For what values of $d$ does $(2^d-1)|(x^b(x^d-1))$ for all odd primes $x$
B) Are there any values of odd primes $x,y$, given that $(2^d-1)|(x^b(x^d-1))$ and $(2^d-1)|(y^b(y^d-1))$, and $\gcd(\frac{x^b(x^d-1)}{2^d-1},\frac{y^b(y^d-1)}{2^d-1})$ is divisible by an odd prime $z$ and both $x\nmid(y^d-1)$ and $y\nmid(x^d-1)$
First, for $d=1$, then any pair where $y-1$ is not a power of 2 and $x=2y-1$ provides a counterexample
When $d>1$ is odd, it appears (checking up to $d=99$) that $(2^d-1)\nmid(3^b(3^d-1))$. Assuming that the pattern holds, any solutionsmust have $d$ even.
Next, I considered the cases where $d=2(3n+1)$ for $n=0,1,2,...$. In each of these cases, if $x,y$ are twin primes of the form $18k\pm1$, then $2^d-1$ is not divisible by $9$, but both $x^d-1$ and $y^d-1$ are. As such, as long as $x\nmid(y^d-1)$ and $y\nmid(x^d-1)$, this provides a counterexample. For $y=17,x=19$, $x\nmid(y^d-1)$, and $y|(x^d-1)$ iff $n\not\equiv1\mod4$. For $y=71,x=73$, $x\nmid(y^d-1)$, and $y|(x^d-1)$ iff $n\not\equiv23\mod35$. Combined, we have counterexamples for $n\not\equiv93\mod140$. Going to one more such pair of twin primes, $y=107,x=109$, we can rule out $n\not\equiv35\mod53$, which combines to rule out for $n\not\equiv2473\mod7420$. It appears likely (although not proven), that there are no solutions when $d$ is of this form. Similar methodology can be used to show that for $d=2(3n+2)$, there are no possible solutions unless $(n\not\equiv2\mod4)\cap(n\not\equiv11\mod35)\cap(n\not\equiv17\mod53)$ (i.e. $n\not\equiv4946\mod7420$).
Lastly, for $d=6n$: For $d=6$ and $d=12$, $y=11,x=29$ gives a an odd prime as a common divisor that does not divide $2^d-1$. For $d>12$, it appears (I checked for up to $n=20$) that $(2^d-1)\nmid11^d(11^d-1)$.
My questions:
- Did I miss anything obvious?
- Can anyone provide a rigorous proof for the lack of divisibility that I checked numerically up to a point: that for odd $d>1$ , $(2^d-1)\nmid3^b(3^d-1)$ and that for multiples of 6 $d>12$, $(2^d-1)\nmid11^b(11^d-1)$
- For the even $d$ that are not multiple of 6, can anyone provide a proof that does not depend on the existence of infinitely many twin primes that are of the form $18k\pm1$, by finding some other pairs of primes that provide counterexamples
