Show that the curve $2Y^2 = X^4 - 17$ has no points in $mathbb{Q}$
$begingroup$
There is a hint to show that if there were points in $mathbb{Q}$, then there would exist $r,s, t in mathbb{Z}$ with $gcd(r, t) = 1$ such that $2s^2 = t^4 - 17r^4$
, and then show that any
prime dividing
$s$
is a quadratic residue modulo $17$.
I'm stuck on the first bit - if you suppose $left(frac{a}{b}, frac{c}{d}right)$ is a point on the curve in $mathbb{Q}$ (so $a, b, c, d in mathbb{Z}$) then you get to $2a^2d^4 = b^2c^4 - 17b^2d^4$, but I'm not sure where to go from here since you don't know that $b$ is a square?
number-theory elliptic-curves
asked Feb 24 '14 at 9:48
jemimajemima
49929
$endgroup$
add a comment |
$begingroup$
There is a hint to show that if there were points in $mathbb{Q}$, then there would exist $r,s, t in mathbb{Z}$ with $gcd(r, t) = 1$ such that $2s^2 = t^4 - 17r^4$
, and then show that any
prime dividing
$s$
is a quadratic residue modulo $17$.
I'm stuck on the first bit - if you suppose $left(frac{a}{b}, frac{c}{d}right)$ is a point on the curve in $mathbb{Q}$ (so $a, b, c, d in mathbb{Z}$) then you get to $2a^2d^4 = b^2c^4 - 17b^2d^4$, but I'm not sure where to go from here since you don't know that $b$ is a square?
number-theory elliptic-curves
asked Feb 24 '14 at 9:48
jemimajemima
49929
$endgroup$
add a comment |
$begingroup$
There is a hint to show that if there were points in $mathbb{Q}$, then there would exist $r,s, t in mathbb{Z}$ with $gcd(r, t) = 1$ such that $2s^2 = t^4 - 17r^4$
, and then show that any
prime dividing
$s$
is a quadratic residue modulo $17$.
I'm stuck on the first bit - if you suppose $left(frac{a}{b}, frac{c}{d}right)$ is a point on the curve in $mathbb{Q}$ (so $a, b, c, d in mathbb{Z}$) then you get to $2a^2d^4 = b^2c^4 - 17b^2d^4$, but I'm not sure where to go from here since you don't know that $b$ is a square?
number-theory elliptic-curves
asked Feb 24 '14 at 9:48
jemimajemima
49929
$endgroup$
There is a hint to show that if there were points in $mathbb{Q}$, then there would exist $r,s, t in mathbb{Z}$ with $gcd(r, t) = 1$ such that $2s^2 = t^4 - 17r^4$
, and then show that any
prime dividing
$s$
is a quadratic residue modulo $17$.
I'm stuck on the first bit - if you suppose $left(frac{a}{b}, frac{c}{d}right)$ is a point on the curve in $mathbb{Q}$ (so $a, b, c, d in mathbb{Z}$) then you get to $2a^2d^4 = b^2c^4 - 17b^2d^4$, but I'm not sure where to go from here since you don't know that $b$ is a square?
number-theory elliptic-curves
number-theory elliptic-curves
asked Feb 24 '14 at 9:48
jemimajemima
49929
asked Feb 24 '14 at 9:48
jemimajemima
49929
asked Feb 24 '14 at 9:48
jemimajemima
49929
asked Feb 24 '14 at 9:48
jemimajemima
49929
asked Feb 24 '14 at 9:48
jemimajemima
49929
49929
add a comment |
add a comment |
1 Answer
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$begingroup$
$b$ need not be a square. Instead, the trick is to show that you can actually divide the equation by $b^2$.
Let $p$ be a prime that divides $b$ and is not 2. Let $p^n$ be the highest power of $p$ that divides $b$.
Since $gcd(a,b) = 1$, and we know that $ 2a^2 d^4 equiv 0 pmod{p^{2n}}$, thus $ d^4 equiv 0 pmod{p^{2n}}$ and this implies that $ p^n mid d^2 $.
If $ 2^n mid mid b$ (highest power of 2 that divides $b$), then we have $ 2 not mid a$. Similarly, we get $ 2 d^4 equiv 0 pmod{2^{2n}}$. Let $2^m mid mid d^2$, then we know that $ 1 + 2m geq 2n$. Since $m$ and $n$ are integers, we can conclude that $m geq n$. Thus, we have $ 2^n mid d^2$.
Hence, $b mid d^2$, and so we may divide throughout by $b^2$, to obtain $ 2 left( a timesfrac{d^2}{b} right)^2 = c^4 - 17 d^4$, and $gcd (c, d) = 1 $ by assumption.
In general, you see that this will work, as long as the coefficient (2) is square-free.
answered Feb 24 '14 at 10:01
Calvin LinCalvin Lin
36.3k349114
$endgroup$
add a comment |
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$begingroup$
$b$ need not be a square. Instead, the trick is to show that you can actually divide the equation by $b^2$.
Let $p$ be a prime that divides $b$ and is not 2. Let $p^n$ be the highest power of $p$ that divides $b$.
Since $gcd(a,b) = 1$, and we know that $ 2a^2 d^4 equiv 0 pmod{p^{2n}}$, thus $ d^4 equiv 0 pmod{p^{2n}}$ and this implies that $ p^n mid d^2 $.
If $ 2^n mid mid b$ (highest power of 2 that divides $b$), then we have $ 2 not mid a$. Similarly, we get $ 2 d^4 equiv 0 pmod{2^{2n}}$. Let $2^m mid mid d^2$, then we know that $ 1 + 2m geq 2n$. Since $m$ and $n$ are integers, we can conclude that $m geq n$. Thus, we have $ 2^n mid d^2$.
Hence, $b mid d^2$, and so we may divide throughout by $b^2$, to obtain $ 2 left( a timesfrac{d^2}{b} right)^2 = c^4 - 17 d^4$, and $gcd (c, d) = 1 $ by assumption.
In general, you see that this will work, as long as the coefficient (2) is square-free.
answered Feb 24 '14 at 10:01
Calvin LinCalvin Lin
36.3k349114
$endgroup$
add a comment |
$begingroup$
$b$ need not be a square. Instead, the trick is to show that you can actually divide the equation by $b^2$.
Let $p$ be a prime that divides $b$ and is not 2. Let $p^n$ be the highest power of $p$ that divides $b$.
Since $gcd(a,b) = 1$, and we know that $ 2a^2 d^4 equiv 0 pmod{p^{2n}}$, thus $ d^4 equiv 0 pmod{p^{2n}}$ and this implies that $ p^n mid d^2 $.
If $ 2^n mid mid b$ (highest power of 2 that divides $b$), then we have $ 2 not mid a$. Similarly, we get $ 2 d^4 equiv 0 pmod{2^{2n}}$. Let $2^m mid mid d^2$, then we know that $ 1 + 2m geq 2n$. Since $m$ and $n$ are integers, we can conclude that $m geq n$. Thus, we have $ 2^n mid d^2$.
Hence, $b mid d^2$, and so we may divide throughout by $b^2$, to obtain $ 2 left( a timesfrac{d^2}{b} right)^2 = c^4 - 17 d^4$, and $gcd (c, d) = 1 $ by assumption.
In general, you see that this will work, as long as the coefficient (2) is square-free.
answered Feb 24 '14 at 10:01
Calvin LinCalvin Lin
36.3k349114
$endgroup$
add a comment |
$begingroup$
$b$ need not be a square. Instead, the trick is to show that you can actually divide the equation by $b^2$.
Let $p$ be a prime that divides $b$ and is not 2. Let $p^n$ be the highest power of $p$ that divides $b$.
Since $gcd(a,b) = 1$, and we know that $ 2a^2 d^4 equiv 0 pmod{p^{2n}}$, thus $ d^4 equiv 0 pmod{p^{2n}}$ and this implies that $ p^n mid d^2 $.
If $ 2^n mid mid b$ (highest power of 2 that divides $b$), then we have $ 2 not mid a$. Similarly, we get $ 2 d^4 equiv 0 pmod{2^{2n}}$. Let $2^m mid mid d^2$, then we know that $ 1 + 2m geq 2n$. Since $m$ and $n$ are integers, we can conclude that $m geq n$. Thus, we have $ 2^n mid d^2$.
Hence, $b mid d^2$, and so we may divide throughout by $b^2$, to obtain $ 2 left( a timesfrac{d^2}{b} right)^2 = c^4 - 17 d^4$, and $gcd (c, d) = 1 $ by assumption.
In general, you see that this will work, as long as the coefficient (2) is square-free.
answered Feb 24 '14 at 10:01
Calvin LinCalvin Lin
36.3k349114
$endgroup$
$b$ need not be a square. Instead, the trick is to show that you can actually divide the equation by $b^2$.
Let $p$ be a prime that divides $b$ and is not 2. Let $p^n$ be the highest power of $p$ that divides $b$.
Since $gcd(a,b) = 1$, and we know that $ 2a^2 d^4 equiv 0 pmod{p^{2n}}$, thus $ d^4 equiv 0 pmod{p^{2n}}$ and this implies that $ p^n mid d^2 $.
If $ 2^n mid mid b$ (highest power of 2 that divides $b$), then we have $ 2 not mid a$. Similarly, we get $ 2 d^4 equiv 0 pmod{2^{2n}}$. Let $2^m mid mid d^2$, then we know that $ 1 + 2m geq 2n$. Since $m$ and $n$ are integers, we can conclude that $m geq n$. Thus, we have $ 2^n mid d^2$.
Hence, $b mid d^2$, and so we may divide throughout by $b^2$, to obtain $ 2 left( a timesfrac{d^2}{b} right)^2 = c^4 - 17 d^4$, and $gcd (c, d) = 1 $ by assumption.
In general, you see that this will work, as long as the coefficient (2) is square-free.
answered Feb 24 '14 at 10:01
Calvin LinCalvin Lin
36.3k349114
edited Feb 24 '14 at 10:14
answered Feb 24 '14 at 10:01
Calvin LinCalvin Lin
36.3k349114
answered Feb 24 '14 at 10:01
Calvin LinCalvin Lin
36.3k349114
answered Feb 24 '14 at 10:01
Calvin LinCalvin Lin
36.3k349114
36.3k349114
add a comment |
add a comment |
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