Let $n_1, n_2$ integers. Show that if $$n_1|2^{n_2}-1 \, \, n_2|2^{n_1}-1$$ then $n_1=n_2=1$.
I want a little hint for this problem. I tried to use the proof of
There is no $n>1$ such that $n$ divides $2^n-1$
but I got nothing.
Asked
Active
Viewed 90 times
1
1 Answers
1
Assume that both $n_1,n_2$ are odd and $>1$. Let $p_1$ be the smallest prime divisor of $n_1$, $p_2$ the smallest prime divisor of $n_2$. Then $p_1$ divides both $2^{n_2}-1$ and $2^{p_1-1}-1$, hence $p_1$ divides $2^{\gcd(n_2,p_1-1)}-1$. Similarly, $p_2$ divides $2^{\gcd(n_1,p_2-1)}-1$. If $p_2\leq p_1$, $p_2\mid 2^{\gcd(n_1,p_2-1)}-1$ implies $p_2=1$, contradiction. Similarly, if $p_1\leq p_2$, $p_1\mid 2^{\gcd(n_2,p_1-1)}-1$ implies $p_1=1$, contradiction.
This is exactly the proof you probably mentioned, with a minor twist.
Jack D'Aurizio
- 353,855
-
I get too close. Thanks – EQJ Oct 21 '16 at 00:16
-
You are missing something, why does $p_1$ divide $gcd(n_2,p_1-1)$? – EQJ Oct 21 '16 at 00:50
-
@YTS: oh, sorry. I wrote $\gcd(\ldots,\ldots)$ in place of $2^{\gcd(\ldots,\ldots)}-1$. Now fixed. – Jack D'Aurizio Oct 21 '16 at 01:17
-
Why if $p_2|2^{gcd(n_1,p_2-1)}-1$ and $p_2\leq p_1$ implies $p_2$. – EQJ Oct 21 '16 at 01:26
-
If $p_2\leq p_1$, $p_2-1\leq p_1-1$ splits as a product of primes $< p_1$. Since the smallest prime divisor of $n_1$ is $p_1$, $\gcd(n_1,p_2-1)=1$. – Jack D'Aurizio Oct 21 '16 at 01:28