Mixing
$ a^{n}+b^{n} $ prime. How to show that n is a power of two?
$begingroup$
Let $ a,b in mathbb{N} setminus {0,1}$ and $n in mathbb{N^{ast }}. $
Let $a^{n}+b^{n} $ prime. How to show that n is a power of two ? I know it works if we suppose that $n=2k(2p+1)$ and then show that it implies $ a^{n}+b^{n} = 0 $ but can't do it.
Sorry, I have a little bit of trouble with LaTex too.
elementary-number-theory arithmetic
edited Jul 8 '17 at 14:14
MathOverview
8,54343163
asked Jul 8 '17 at 13:56
ClipperClipper
192
$endgroup$
add a comment |
$begingroup$
Let $ a,b in mathbb{N} setminus {0,1}$ and $n in mathbb{N^{ast }}. $
Let $a^{n}+b^{n} $ prime. How to show that n is a power of two ? I know it works if we suppose that $n=2k(2p+1)$ and then show that it implies $ a^{n}+b^{n} = 0 $ but can't do it.
Sorry, I have a little bit of trouble with LaTex too.
elementary-number-theory arithmetic
edited Jul 8 '17 at 14:14
MathOverview
8,54343163
asked Jul 8 '17 at 13:56
ClipperClipper
192
$endgroup$
$begingroup$
Hint: if $n=2^km$ where $m$ is odd then $a^n+b^n$ is divisible by $a^{2^k}+b^{2^k}$. Of course you have to handle the case where the quotient is $1$, as in $2=1^3+1^3$.
$endgroup$
– lulu
Jul 8 '17 at 14:05
add a comment |
$begingroup$
Let $ a,b in mathbb{N} setminus {0,1}$ and $n in mathbb{N^{ast }}. $
Let $a^{n}+b^{n} $ prime. How to show that n is a power of two ? I know it works if we suppose that $n=2k(2p+1)$ and then show that it implies $ a^{n}+b^{n} = 0 $ but can't do it.
Sorry, I have a little bit of trouble with LaTex too.
elementary-number-theory arithmetic
edited Jul 8 '17 at 14:14
MathOverview
8,54343163
asked Jul 8 '17 at 13:56
ClipperClipper
192
$endgroup$
Let $ a,b in mathbb{N} setminus {0,1}$ and $n in mathbb{N^{ast }}. $
Let $a^{n}+b^{n} $ prime. How to show that n is a power of two ? I know it works if we suppose that $n=2k(2p+1)$ and then show that it implies $ a^{n}+b^{n} = 0 $ but can't do it.
Sorry, I have a little bit of trouble with LaTex too.
elementary-number-theory arithmetic
elementary-number-theory arithmetic
edited Jul 8 '17 at 14:14
MathOverview
8,54343163
asked Jul 8 '17 at 13:56
ClipperClipper
192
edited Jul 8 '17 at 14:14
MathOverview
8,54343163
asked Jul 8 '17 at 13:56
ClipperClipper
192
edited Jul 8 '17 at 14:14
MathOverview
8,54343163
edited Jul 8 '17 at 14:14
MathOverview
8,54343163
edited Jul 8 '17 at 14:14
MathOverview
8,54343163
8,54343163
asked Jul 8 '17 at 13:56
ClipperClipper
192
asked Jul 8 '17 at 13:56
ClipperClipper
192
asked Jul 8 '17 at 13:56
ClipperClipper
192
192
$begingroup$
Hint: if $n=2^km$ where $m$ is odd then $a^n+b^n$ is divisible by $a^{2^k}+b^{2^k}$. Of course you have to handle the case where the quotient is $1$, as in $2=1^3+1^3$.
$endgroup$
– lulu
Jul 8 '17 at 14:05
add a comment |
$begingroup$
Hint: if $n=2^km$ where $m$ is odd then $a^n+b^n$ is divisible by $a^{2^k}+b^{2^k}$. Of course you have to handle the case where the quotient is $1$, as in $2=1^3+1^3$.
$endgroup$
– lulu
Jul 8 '17 at 14:05
$begingroup$
Hint: if $n=2^km$ where $m$ is odd then $a^n+b^n$ is divisible by $a^{2^k}+b^{2^k}$. Of course you have to handle the case where the quotient is $1$, as in $2=1^3+1^3$.
$endgroup$
– lulu
Jul 8 '17 at 14:05
$begingroup$
Hint: if $n=2^km$ where $m$ is odd then $a^n+b^n$ is divisible by $a^{2^k}+b^{2^k}$. Of course you have to handle the case where the quotient is $1$, as in $2=1^3+1^3$.
$endgroup$
– lulu
Jul 8 '17 at 14:05
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Try to show the following:
Let $n=l2^k$, where $l$ is co prime to $2$ (in other words, it's odd). Then, it is infact true that $a^n+b^n$ is a multiple of $a^{2^k} + b^{2^k}$.
If you can't then look at this box:
Let $d = a^{2^k}$ and $e=b^{2^k}$, then this is basically saying that $d^l + e^l$ is a multiple of $d+e$ whenever $l$ is odd.
Which is true, since $dfrac{d^l+e^l}{d+e} = displaystylesum_{k=0}^{l-1} (-1)^ka^kb^{l-1-k}$. Put back $d$ and $e$ in terms of $a$ and $b$ to get the proof.
Therefore, unless $l=1$, the expression $a^n+b^n$ always has a non-trivial factor. Infact, we can even figure out what it is from the above formula.
For example, $7^6+2^6 = 117713 = 53 times 2221$, and $53 = 7^2+2^2$.
This proves your statement.
answered Jul 8 '17 at 14:21
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.7k33376
$endgroup$
add a comment |
$begingroup$
Hint If $n=(2k+1)m$ with $kgeq 1$ odd then
$$a^n+b^n=(a^m)^{2k+1}+(b^m)^{2k+1}=(a^m+b^m)(a^{2km}-a^{(2k-1)m}b^m+...+b^{2km})$$
answered Jul 8 '17 at 14:53
N. S.N. S.
103k6111208
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Try to show the following:
Let $n=l2^k$, where $l$ is co prime to $2$ (in other words, it's odd). Then, it is infact true that $a^n+b^n$ is a multiple of $a^{2^k} + b^{2^k}$.
If you can't then look at this box:
Let $d = a^{2^k}$ and $e=b^{2^k}$, then this is basically saying that $d^l + e^l$ is a multiple of $d+e$ whenever $l$ is odd.
Which is true, since $dfrac{d^l+e^l}{d+e} = displaystylesum_{k=0}^{l-1} (-1)^ka^kb^{l-1-k}$. Put back $d$ and $e$ in terms of $a$ and $b$ to get the proof.
Therefore, unless $l=1$, the expression $a^n+b^n$ always has a non-trivial factor. Infact, we can even figure out what it is from the above formula.
For example, $7^6+2^6 = 117713 = 53 times 2221$, and $53 = 7^2+2^2$.
This proves your statement.
answered Jul 8 '17 at 14:21
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.7k33376
$endgroup$
add a comment |
$begingroup$
Try to show the following:
Let $n=l2^k$, where $l$ is co prime to $2$ (in other words, it's odd). Then, it is infact true that $a^n+b^n$ is a multiple of $a^{2^k} + b^{2^k}$.
If you can't then look at this box:
Let $d = a^{2^k}$ and $e=b^{2^k}$, then this is basically saying that $d^l + e^l$ is a multiple of $d+e$ whenever $l$ is odd.
Which is true, since $dfrac{d^l+e^l}{d+e} = displaystylesum_{k=0}^{l-1} (-1)^ka^kb^{l-1-k}$. Put back $d$ and $e$ in terms of $a$ and $b$ to get the proof.
Therefore, unless $l=1$, the expression $a^n+b^n$ always has a non-trivial factor. Infact, we can even figure out what it is from the above formula.
For example, $7^6+2^6 = 117713 = 53 times 2221$, and $53 = 7^2+2^2$.
This proves your statement.
answered Jul 8 '17 at 14:21
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.7k33376
$endgroup$
add a comment |
$begingroup$
Try to show the following:
Let $n=l2^k$, where $l$ is co prime to $2$ (in other words, it's odd). Then, it is infact true that $a^n+b^n$ is a multiple of $a^{2^k} + b^{2^k}$.
If you can't then look at this box:
Let $d = a^{2^k}$ and $e=b^{2^k}$, then this is basically saying that $d^l + e^l$ is a multiple of $d+e$ whenever $l$ is odd.
Which is true, since $dfrac{d^l+e^l}{d+e} = displaystylesum_{k=0}^{l-1} (-1)^ka^kb^{l-1-k}$. Put back $d$ and $e$ in terms of $a$ and $b$ to get the proof.
Therefore, unless $l=1$, the expression $a^n+b^n$ always has a non-trivial factor. Infact, we can even figure out what it is from the above formula.
For example, $7^6+2^6 = 117713 = 53 times 2221$, and $53 = 7^2+2^2$.
This proves your statement.
answered Jul 8 '17 at 14:21
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.7k33376
$endgroup$
Try to show the following:
Let $n=l2^k$, where $l$ is co prime to $2$ (in other words, it's odd). Then, it is infact true that $a^n+b^n$ is a multiple of $a^{2^k} + b^{2^k}$.
If you can't then look at this box:
Let $d = a^{2^k}$ and $e=b^{2^k}$, then this is basically saying that $d^l + e^l$ is a multiple of $d+e$ whenever $l$ is odd.
Which is true, since $dfrac{d^l+e^l}{d+e} = displaystylesum_{k=0}^{l-1} (-1)^ka^kb^{l-1-k}$. Put back $d$ and $e$ in terms of $a$ and $b$ to get the proof.
Therefore, unless $l=1$, the expression $a^n+b^n$ always has a non-trivial factor. Infact, we can even figure out what it is from the above formula.
For example, $7^6+2^6 = 117713 = 53 times 2221$, and $53 = 7^2+2^2$.
This proves your statement.
answered Jul 8 '17 at 14:21
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.7k33376
edited Jan 4 at 5:50
answered Jul 8 '17 at 14:21
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.7k33376
answered Jul 8 '17 at 14:21
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.7k33376
answered Jul 8 '17 at 14:21
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.7k33376
37.7k33376
add a comment |
add a comment |
$begingroup$
Hint If $n=(2k+1)m$ with $kgeq 1$ odd then
$$a^n+b^n=(a^m)^{2k+1}+(b^m)^{2k+1}=(a^m+b^m)(a^{2km}-a^{(2k-1)m}b^m+...+b^{2km})$$
answered Jul 8 '17 at 14:53
N. S.N. S.
103k6111208
$endgroup$
add a comment |
$begingroup$
Hint If $n=(2k+1)m$ with $kgeq 1$ odd then
$$a^n+b^n=(a^m)^{2k+1}+(b^m)^{2k+1}=(a^m+b^m)(a^{2km}-a^{(2k-1)m}b^m+...+b^{2km})$$
answered Jul 8 '17 at 14:53
N. S.N. S.
103k6111208
$endgroup$
add a comment |
$begingroup$
Hint If $n=(2k+1)m$ with $kgeq 1$ odd then
$$a^n+b^n=(a^m)^{2k+1}+(b^m)^{2k+1}=(a^m+b^m)(a^{2km}-a^{(2k-1)m}b^m+...+b^{2km})$$
answered Jul 8 '17 at 14:53
N. S.N. S.
103k6111208
$endgroup$
Hint If $n=(2k+1)m$ with $kgeq 1$ odd then
$$a^n+b^n=(a^m)^{2k+1}+(b^m)^{2k+1}=(a^m+b^m)(a^{2km}-a^{(2k-1)m}b^m+...+b^{2km})$$
answered Jul 8 '17 at 14:53
N. S.N. S.
103k6111208
answered Jul 8 '17 at 14:53
N. S.N. S.
103k6111208
answered Jul 8 '17 at 14:53
N. S.N. S.
103k6111208
answered Jul 8 '17 at 14:53
N. S.N. S.
103k6111208
103k6111208
add a comment |
add a comment |
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$begingroup$
Hint: if $n=2^km$ where $m$ is odd then $a^n+b^n$ is divisible by $a^{2^k}+b^{2^k}$. Of course you have to handle the case where the quotient is $1$, as in $2=1^3+1^3$.
$endgroup$
– lulu
Jul 8 '17 at 14:05