Repunits and Goormaghtigh's conjecture
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On the Wikipedia page of Goormaghtigh 's conjecture, there is the following claim:
The Goormaghtigh conjecture may be expressed as saying that $31$ ($111$ in base $5$, $11111$ in base $2$) and $8191$ ($111$ in base $90$, $1111111111111$ in base $2$) are the only two numbers that are repunits with at least 3 digits in two different bases.
I don't see how this follows from the original conjecture, i.e., that the only non-trivial integer solutions of
$$
{frac {x^{m}-1}{x-1}}={frac {y^{n}-1}{y-1}}
$$
satisfying ${displaystyle x>y>1}$ and ${displaystyle n,m>2}$ are
$$
{displaystyle {frac {5^{3}-1}{5-1}}={frac {2^{5}-1}{2-1}}=31}
$$
and
$$
{displaystyle {frac {90^{3}-1}{90-1}}={frac {2^{13}-1}{2-1}}=8191.}
$$
Can someone show a proof that these two statements are equivalent?
elementary-number-theory proof-writing
asked Dec 25 '18 at 10:54
KlangenKlangen
1,76211334
$endgroup$
add a comment |
$begingroup$
On the Wikipedia page of Goormaghtigh 's conjecture, there is the following claim:
The Goormaghtigh conjecture may be expressed as saying that $31$ ($111$ in base $5$, $11111$ in base $2$) and $8191$ ($111$ in base $90$, $1111111111111$ in base $2$) are the only two numbers that are repunits with at least 3 digits in two different bases.
I don't see how this follows from the original conjecture, i.e., that the only non-trivial integer solutions of
$$
{frac {x^{m}-1}{x-1}}={frac {y^{n}-1}{y-1}}
$$
satisfying ${displaystyle x>y>1}$ and ${displaystyle n,m>2}$ are
$$
{displaystyle {frac {5^{3}-1}{5-1}}={frac {2^{5}-1}{2-1}}=31}
$$
and
$$
{displaystyle {frac {90^{3}-1}{90-1}}={frac {2^{13}-1}{2-1}}=8191.}
$$
Can someone show a proof that these two statements are equivalent?
elementary-number-theory proof-writing
asked Dec 25 '18 at 10:54
KlangenKlangen
1,76211334
$endgroup$
add a comment |
$begingroup$
On the Wikipedia page of Goormaghtigh 's conjecture, there is the following claim:
The Goormaghtigh conjecture may be expressed as saying that $31$ ($111$ in base $5$, $11111$ in base $2$) and $8191$ ($111$ in base $90$, $1111111111111$ in base $2$) are the only two numbers that are repunits with at least 3 digits in two different bases.
I don't see how this follows from the original conjecture, i.e., that the only non-trivial integer solutions of
$$
{frac {x^{m}-1}{x-1}}={frac {y^{n}-1}{y-1}}
$$
satisfying ${displaystyle x>y>1}$ and ${displaystyle n,m>2}$ are
$$
{displaystyle {frac {5^{3}-1}{5-1}}={frac {2^{5}-1}{2-1}}=31}
$$
and
$$
{displaystyle {frac {90^{3}-1}{90-1}}={frac {2^{13}-1}{2-1}}=8191.}
$$
Can someone show a proof that these two statements are equivalent?
elementary-number-theory proof-writing
asked Dec 25 '18 at 10:54
KlangenKlangen
1,76211334
$endgroup$
On the Wikipedia page of Goormaghtigh 's conjecture, there is the following claim:
The Goormaghtigh conjecture may be expressed as saying that $31$ ($111$ in base $5$, $11111$ in base $2$) and $8191$ ($111$ in base $90$, $1111111111111$ in base $2$) are the only two numbers that are repunits with at least 3 digits in two different bases.
I don't see how this follows from the original conjecture, i.e., that the only non-trivial integer solutions of
$$
{frac {x^{m}-1}{x-1}}={frac {y^{n}-1}{y-1}}
$$
satisfying ${displaystyle x>y>1}$ and ${displaystyle n,m>2}$ are
$$
{displaystyle {frac {5^{3}-1}{5-1}}={frac {2^{5}-1}{2-1}}=31}
$$
and
$$
{displaystyle {frac {90^{3}-1}{90-1}}={frac {2^{13}-1}{2-1}}=8191.}
$$
Can someone show a proof that these two statements are equivalent?
elementary-number-theory proof-writing
elementary-number-theory proof-writing
asked Dec 25 '18 at 10:54
KlangenKlangen
1,76211334
asked Dec 25 '18 at 10:54
KlangenKlangen
1,76211334
asked Dec 25 '18 at 10:54
KlangenKlangen
1,76211334
asked Dec 25 '18 at 10:54
KlangenKlangen
1,76211334
asked Dec 25 '18 at 10:54
KlangenKlangen
1,76211334
1,76211334
add a comment |
add a comment |
1 Answer
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This follows from the fact that $frac{x^m - 1}{x-1} = 1 + x + x^2 + cdots + x^{m-1}$ is the sum of a geometric series, and hence in base $x$, $frac{x^m - 1}{x-1} = 111ldots1$ is a repunit with $m$ digits.
answered Dec 25 '18 at 10:59
ODFODF
1,486510
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
This follows from the fact that $frac{x^m - 1}{x-1} = 1 + x + x^2 + cdots + x^{m-1}$ is the sum of a geometric series, and hence in base $x$, $frac{x^m - 1}{x-1} = 111ldots1$ is a repunit with $m$ digits.
answered Dec 25 '18 at 10:59
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
This follows from the fact that $frac{x^m - 1}{x-1} = 1 + x + x^2 + cdots + x^{m-1}$ is the sum of a geometric series, and hence in base $x$, $frac{x^m - 1}{x-1} = 111ldots1$ is a repunit with $m$ digits.
answered Dec 25 '18 at 10:59
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
This follows from the fact that $frac{x^m - 1}{x-1} = 1 + x + x^2 + cdots + x^{m-1}$ is the sum of a geometric series, and hence in base $x$, $frac{x^m - 1}{x-1} = 111ldots1$ is a repunit with $m$ digits.
answered Dec 25 '18 at 10:59
ODFODF
1,486510
$endgroup$
This follows from the fact that $frac{x^m - 1}{x-1} = 1 + x + x^2 + cdots + x^{m-1}$ is the sum of a geometric series, and hence in base $x$, $frac{x^m - 1}{x-1} = 111ldots1$ is a repunit with $m$ digits.
answered Dec 25 '18 at 10:59
ODFODF
1,486510
answered Dec 25 '18 at 10:59
ODFODF
1,486510
answered Dec 25 '18 at 10:59
ODFODF
1,486510
answered Dec 25 '18 at 10:59
ODFODF
1,486510
1,486510
add a comment |
add a comment |
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