Showing $gcd(2^m-1,2^n+1)=1$
$begingroup$
A student of mine has been self-studying some elementary number theory. She came by my office today and asked if I had any hints on how to prove the statement
If $m$ is odd then $gcd(2^m-1,2^n+1)=1$.
It's been a while since I took number theory and I'm not sure what to do. She said she is learning about congruences, primitive roots, and power residues. She has not taken any group theory.
elementary-number-theory divisibility greatest-common-divisor
edited Feb 16 '16 at 21:37
Martin Sleziak
44.9k10122276
asked Nov 18 '13 at 19:06
Joe Johnson 126Joe Johnson 126
13.8k32771
$endgroup$
add a comment |
$begingroup$
A student of mine has been self-studying some elementary number theory. She came by my office today and asked if I had any hints on how to prove the statement
If $m$ is odd then $gcd(2^m-1,2^n+1)=1$.
It's been a while since I took number theory and I'm not sure what to do. She said she is learning about congruences, primitive roots, and power residues. She has not taken any group theory.
elementary-number-theory divisibility greatest-common-divisor
edited Feb 16 '16 at 21:37
Martin Sleziak
44.9k10122276
asked Nov 18 '13 at 19:06
Joe Johnson 126Joe Johnson 126
13.8k32771
$endgroup$
add a comment |
$begingroup$
A student of mine has been self-studying some elementary number theory. She came by my office today and asked if I had any hints on how to prove the statement
If $m$ is odd then $gcd(2^m-1,2^n+1)=1$.
It's been a while since I took number theory and I'm not sure what to do. She said she is learning about congruences, primitive roots, and power residues. She has not taken any group theory.
elementary-number-theory divisibility greatest-common-divisor
edited Feb 16 '16 at 21:37
Martin Sleziak
44.9k10122276
asked Nov 18 '13 at 19:06
Joe Johnson 126Joe Johnson 126
13.8k32771
$endgroup$
A student of mine has been self-studying some elementary number theory. She came by my office today and asked if I had any hints on how to prove the statement
If $m$ is odd then $gcd(2^m-1,2^n+1)=1$.
It's been a while since I took number theory and I'm not sure what to do. She said she is learning about congruences, primitive roots, and power residues. She has not taken any group theory.
elementary-number-theory divisibility greatest-common-divisor
elementary-number-theory divisibility greatest-common-divisor
edited Feb 16 '16 at 21:37
Martin Sleziak
44.9k10122276
asked Nov 18 '13 at 19:06
Joe Johnson 126Joe Johnson 126
13.8k32771
edited Feb 16 '16 at 21:37
Martin Sleziak
44.9k10122276
asked Nov 18 '13 at 19:06
Joe Johnson 126Joe Johnson 126
13.8k32771
edited Feb 16 '16 at 21:37
Martin Sleziak
44.9k10122276
edited Feb 16 '16 at 21:37
Martin Sleziak
44.9k10122276
edited Feb 16 '16 at 21:37
Martin Sleziak
44.9k10122276
44.9k10122276
asked Nov 18 '13 at 19:06
Joe Johnson 126Joe Johnson 126
13.8k32771
asked Nov 18 '13 at 19:06
Joe Johnson 126Joe Johnson 126
13.8k32771
asked Nov 18 '13 at 19:06
Joe Johnson 126Joe Johnson 126
13.8k32771
13.8k32771
add a comment |
add a comment |
1 Answer
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$begingroup$
If an odd prime $p$ divides $2^n+1$, then the order of $2$ modulo $p$ is even (it is a divisor of $2n$, but not of $n$). If an odd prime $q$ divides $2^m-1$ with $m$ odd, then the order of $2$ modulo $q$ is odd (it is a divisor of $m$). Hence $p neq q$. Since $2^m - 1$ is odd for $m > 0$, in particular all odd $m$, the greatest common divisor cannot be even. So no prime divides both, $2^n+1$ and $2^m-1$.
Alternatively, we can use
$$gcd (2^t-1, 2^u-1) = 2^{gcd (t,u)}-1tag{1}$$
to conclude
$$gcd (2^m-1, 2^{2n}-1) = 2^{gcd(m,2n)}-1.$$
But since $m$ is odd, we have $gcd (m,2n) = gcd(m,n)$, and hence
$$2^{gcd(m,2n)}-1 mid 2^n-1,$$
which, since
$$gcd(2^n-1,2^n+1) = gcd(2^n-1,2) mid 2$$
and $2^{gcd(m,2n)}-1$ is odd, implies $gcd (2^{gcd(m,2n)}-1,2^n+1) = 1$ and hence $gcd(2^m-1,2^n+1) = 1$.
To see $(1)$, write $u = qcdot t + r$ with $0 leqslant r < t$, and
$$2^u-1 = 2^rleft(2^{qcdot t}-1right) + left(2^r-1right),$$
which, since $2^t-1 mid (2^t)^q-1$, yields
$$gcd(2^t-1,2^u-1) = gcd(2^t-1,2^r-1),$$
and continuing the Euclidean algorithm for the exponents finally yields $(1)$.
answered Nov 18 '13 at 19:09
Daniel FischerDaniel Fischer
174k17169288
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1 Answer
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1 Answer
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active
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active
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$begingroup$
If an odd prime $p$ divides $2^n+1$, then the order of $2$ modulo $p$ is even (it is a divisor of $2n$, but not of $n$). If an odd prime $q$ divides $2^m-1$ with $m$ odd, then the order of $2$ modulo $q$ is odd (it is a divisor of $m$). Hence $p neq q$. Since $2^m - 1$ is odd for $m > 0$, in particular all odd $m$, the greatest common divisor cannot be even. So no prime divides both, $2^n+1$ and $2^m-1$.
Alternatively, we can use
$$gcd (2^t-1, 2^u-1) = 2^{gcd (t,u)}-1tag{1}$$
to conclude
$$gcd (2^m-1, 2^{2n}-1) = 2^{gcd(m,2n)}-1.$$
But since $m$ is odd, we have $gcd (m,2n) = gcd(m,n)$, and hence
$$2^{gcd(m,2n)}-1 mid 2^n-1,$$
which, since
$$gcd(2^n-1,2^n+1) = gcd(2^n-1,2) mid 2$$
and $2^{gcd(m,2n)}-1$ is odd, implies $gcd (2^{gcd(m,2n)}-1,2^n+1) = 1$ and hence $gcd(2^m-1,2^n+1) = 1$.
To see $(1)$, write $u = qcdot t + r$ with $0 leqslant r < t$, and
$$2^u-1 = 2^rleft(2^{qcdot t}-1right) + left(2^r-1right),$$
which, since $2^t-1 mid (2^t)^q-1$, yields
$$gcd(2^t-1,2^u-1) = gcd(2^t-1,2^r-1),$$
and continuing the Euclidean algorithm for the exponents finally yields $(1)$.
answered Nov 18 '13 at 19:09
Daniel FischerDaniel Fischer
174k17169288
$endgroup$
add a comment |
$begingroup$
If an odd prime $p$ divides $2^n+1$, then the order of $2$ modulo $p$ is even (it is a divisor of $2n$, but not of $n$). If an odd prime $q$ divides $2^m-1$ with $m$ odd, then the order of $2$ modulo $q$ is odd (it is a divisor of $m$). Hence $p neq q$. Since $2^m - 1$ is odd for $m > 0$, in particular all odd $m$, the greatest common divisor cannot be even. So no prime divides both, $2^n+1$ and $2^m-1$.
Alternatively, we can use
$$gcd (2^t-1, 2^u-1) = 2^{gcd (t,u)}-1tag{1}$$
to conclude
$$gcd (2^m-1, 2^{2n}-1) = 2^{gcd(m,2n)}-1.$$
But since $m$ is odd, we have $gcd (m,2n) = gcd(m,n)$, and hence
$$2^{gcd(m,2n)}-1 mid 2^n-1,$$
which, since
$$gcd(2^n-1,2^n+1) = gcd(2^n-1,2) mid 2$$
and $2^{gcd(m,2n)}-1$ is odd, implies $gcd (2^{gcd(m,2n)}-1,2^n+1) = 1$ and hence $gcd(2^m-1,2^n+1) = 1$.
To see $(1)$, write $u = qcdot t + r$ with $0 leqslant r < t$, and
$$2^u-1 = 2^rleft(2^{qcdot t}-1right) + left(2^r-1right),$$
which, since $2^t-1 mid (2^t)^q-1$, yields
$$gcd(2^t-1,2^u-1) = gcd(2^t-1,2^r-1),$$
and continuing the Euclidean algorithm for the exponents finally yields $(1)$.
answered Nov 18 '13 at 19:09
Daniel FischerDaniel Fischer
174k17169288
$endgroup$
add a comment |
$begingroup$
If an odd prime $p$ divides $2^n+1$, then the order of $2$ modulo $p$ is even (it is a divisor of $2n$, but not of $n$). If an odd prime $q$ divides $2^m-1$ with $m$ odd, then the order of $2$ modulo $q$ is odd (it is a divisor of $m$). Hence $p neq q$. Since $2^m - 1$ is odd for $m > 0$, in particular all odd $m$, the greatest common divisor cannot be even. So no prime divides both, $2^n+1$ and $2^m-1$.
Alternatively, we can use
$$gcd (2^t-1, 2^u-1) = 2^{gcd (t,u)}-1tag{1}$$
to conclude
$$gcd (2^m-1, 2^{2n}-1) = 2^{gcd(m,2n)}-1.$$
But since $m$ is odd, we have $gcd (m,2n) = gcd(m,n)$, and hence
$$2^{gcd(m,2n)}-1 mid 2^n-1,$$
which, since
$$gcd(2^n-1,2^n+1) = gcd(2^n-1,2) mid 2$$
and $2^{gcd(m,2n)}-1$ is odd, implies $gcd (2^{gcd(m,2n)}-1,2^n+1) = 1$ and hence $gcd(2^m-1,2^n+1) = 1$.
To see $(1)$, write $u = qcdot t + r$ with $0 leqslant r < t$, and
$$2^u-1 = 2^rleft(2^{qcdot t}-1right) + left(2^r-1right),$$
which, since $2^t-1 mid (2^t)^q-1$, yields
$$gcd(2^t-1,2^u-1) = gcd(2^t-1,2^r-1),$$
and continuing the Euclidean algorithm for the exponents finally yields $(1)$.
answered Nov 18 '13 at 19:09
Daniel FischerDaniel Fischer
174k17169288
$endgroup$
If an odd prime $p$ divides $2^n+1$, then the order of $2$ modulo $p$ is even (it is a divisor of $2n$, but not of $n$). If an odd prime $q$ divides $2^m-1$ with $m$ odd, then the order of $2$ modulo $q$ is odd (it is a divisor of $m$). Hence $p neq q$. Since $2^m - 1$ is odd for $m > 0$, in particular all odd $m$, the greatest common divisor cannot be even. So no prime divides both, $2^n+1$ and $2^m-1$.
Alternatively, we can use
$$gcd (2^t-1, 2^u-1) = 2^{gcd (t,u)}-1tag{1}$$
to conclude
$$gcd (2^m-1, 2^{2n}-1) = 2^{gcd(m,2n)}-1.$$
But since $m$ is odd, we have $gcd (m,2n) = gcd(m,n)$, and hence
$$2^{gcd(m,2n)}-1 mid 2^n-1,$$
which, since
$$gcd(2^n-1,2^n+1) = gcd(2^n-1,2) mid 2$$
and $2^{gcd(m,2n)}-1$ is odd, implies $gcd (2^{gcd(m,2n)}-1,2^n+1) = 1$ and hence $gcd(2^m-1,2^n+1) = 1$.
To see $(1)$, write $u = qcdot t + r$ with $0 leqslant r < t$, and
$$2^u-1 = 2^rleft(2^{qcdot t}-1right) + left(2^r-1right),$$
which, since $2^t-1 mid (2^t)^q-1$, yields
$$gcd(2^t-1,2^u-1) = gcd(2^t-1,2^r-1),$$
and continuing the Euclidean algorithm for the exponents finally yields $(1)$.
answered Nov 18 '13 at 19:09
Daniel FischerDaniel Fischer
174k17169288
edited Nov 18 '13 at 22:11
answered Nov 18 '13 at 19:09
Daniel FischerDaniel Fischer
174k17169288
answered Nov 18 '13 at 19:09
Daniel FischerDaniel Fischer
174k17169288
answered Nov 18 '13 at 19:09
Daniel FischerDaniel Fischer
174k17169288
174k17169288
add a comment |
add a comment |
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