Mixing
Proof that $x^4 - qy^4 = az^2$ has no integral solution
$begingroup$
This is a question from Takashi Ono's book, Problem 1.45 to be exact.
The question is
Let $q$ be a prime such that $q = 1 mod 8$ and $a$ be an integer such that $p^2notmid a$ for any prime $p$ and that $x^4 = a mod q$
has no solution in $Z$. Prove that the equation $x^4 - qy^4 = az^2$ has no
integral solution other than x = y = z = 0.
I'm really stuck, this question comes after the section on quadratic reciprocity but not sure what I can do with it
number-theory quadratic-reciprocity
asked Jan 11 at 1:50
john takeuchijohn takeuchi
313
$endgroup$
add a comment |
$begingroup$
This is a question from Takashi Ono's book, Problem 1.45 to be exact.
The question is
Let $q$ be a prime such that $q = 1 mod 8$ and $a$ be an integer such that $p^2notmid a$ for any prime $p$ and that $x^4 = a mod q$
has no solution in $Z$. Prove that the equation $x^4 - qy^4 = az^2$ has no
integral solution other than x = y = z = 0.
I'm really stuck, this question comes after the section on quadratic reciprocity but not sure what I can do with it
number-theory quadratic-reciprocity
asked Jan 11 at 1:50
john takeuchijohn takeuchi
313
$endgroup$
$begingroup$
If one replaces the fourth powers with squares, you can solve this only assuming $q$ is $1$ modulo $4$. Start with the case $a$ is prime and consider the equation modulo $a$. This easily generalizes to $a$ not prime. I don't know if this helps with the original problem.
$endgroup$
– RghtHndSd
Jan 11 at 2:40
add a comment |
$begingroup$
This is a question from Takashi Ono's book, Problem 1.45 to be exact.
The question is
Let $q$ be a prime such that $q = 1 mod 8$ and $a$ be an integer such that $p^2notmid a$ for any prime $p$ and that $x^4 = a mod q$
has no solution in $Z$. Prove that the equation $x^4 - qy^4 = az^2$ has no
integral solution other than x = y = z = 0.
I'm really stuck, this question comes after the section on quadratic reciprocity but not sure what I can do with it
number-theory quadratic-reciprocity
asked Jan 11 at 1:50
john takeuchijohn takeuchi
313
$endgroup$
This is a question from Takashi Ono's book, Problem 1.45 to be exact.
The question is
Let $q$ be a prime such that $q = 1 mod 8$ and $a$ be an integer such that $p^2notmid a$ for any prime $p$ and that $x^4 = a mod q$
has no solution in $Z$. Prove that the equation $x^4 - qy^4 = az^2$ has no
integral solution other than x = y = z = 0.
I'm really stuck, this question comes after the section on quadratic reciprocity but not sure what I can do with it
number-theory quadratic-reciprocity
number-theory quadratic-reciprocity
asked Jan 11 at 1:50
john takeuchijohn takeuchi
313
asked Jan 11 at 1:50
john takeuchijohn takeuchi
313
asked Jan 11 at 1:50
john takeuchijohn takeuchi
313
asked Jan 11 at 1:50
john takeuchijohn takeuchi
313
asked Jan 11 at 1:50
john takeuchijohn takeuchi
313
313
$begingroup$
If one replaces the fourth powers with squares, you can solve this only assuming $q$ is $1$ modulo $4$. Start with the case $a$ is prime and consider the equation modulo $a$. This easily generalizes to $a$ not prime. I don't know if this helps with the original problem.
$endgroup$
– RghtHndSd
Jan 11 at 2:40
add a comment |
$begingroup$
If one replaces the fourth powers with squares, you can solve this only assuming $q$ is $1$ modulo $4$. Start with the case $a$ is prime and consider the equation modulo $a$. This easily generalizes to $a$ not prime. I don't know if this helps with the original problem.
$endgroup$
– RghtHndSd
Jan 11 at 2:40
$begingroup$
If one replaces the fourth powers with squares, you can solve this only assuming $q$ is $1$ modulo $4$. Start with the case $a$ is prime and consider the equation modulo $a$. This easily generalizes to $a$ not prime. I don't know if this helps with the original problem.
$endgroup$
– RghtHndSd
Jan 11 at 2:40
$begingroup$
If one replaces the fourth powers with squares, you can solve this only assuming $q$ is $1$ modulo $4$. Start with the case $a$ is prime and consider the equation modulo $a$. This easily generalizes to $a$ not prime. I don't know if this helps with the original problem.
$endgroup$
– RghtHndSd
Jan 11 at 2:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If there are solutions, then there are solutions with $x$, $y$, $z$ pairwise
coprime. If $pmid x$, $y$, then $p^4mid az^2$ and as $p^2nmid a$ then
$p^2mid z$. Then $(x/p,y/p,z/p^2)$ is also a solution. Similar
arguments work when $pmid x,z$ and when $pmid y,z$.
Suppose that $(x,y,z)$ form a
pairwise coprime solution to the equation.
Let $p$ be an odd prime dividing $z$. Then as $x^4equiv qy^4pmod p$
we get $left(frac qpright)=1$ and then
$left(frac pqright)=1$ by quadratic reciprocity.
As $left(frac {-1}qright)=left(frac 2qright)=1$ then
$left(frac zqright)=1$.
There is $u$ with $zequiv u^2pmod q$. Then $x^4equiv au^4pmod q$,
and as both sides are nonzero modulo $q$, then $a$ is a quartic residue,
a contradiction.
The example $x^4-17y^2=2z^2$ is a counterexample to the Hasse principle
to quartics and is due to Reichardt.
answered Jan 11 at 4:48
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
Thank you for the response :) I was stuck because I was only thinking of mod q and mod a when what I need to do was mod z thanks again
$endgroup$
– john takeuchi
Jan 11 at 19:15
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If there are solutions, then there are solutions with $x$, $y$, $z$ pairwise
coprime. If $pmid x$, $y$, then $p^4mid az^2$ and as $p^2nmid a$ then
$p^2mid z$. Then $(x/p,y/p,z/p^2)$ is also a solution. Similar
arguments work when $pmid x,z$ and when $pmid y,z$.
Suppose that $(x,y,z)$ form a
pairwise coprime solution to the equation.
Let $p$ be an odd prime dividing $z$. Then as $x^4equiv qy^4pmod p$
we get $left(frac qpright)=1$ and then
$left(frac pqright)=1$ by quadratic reciprocity.
As $left(frac {-1}qright)=left(frac 2qright)=1$ then
$left(frac zqright)=1$.
There is $u$ with $zequiv u^2pmod q$. Then $x^4equiv au^4pmod q$,
and as both sides are nonzero modulo $q$, then $a$ is a quartic residue,
a contradiction.
The example $x^4-17y^2=2z^2$ is a counterexample to the Hasse principle
to quartics and is due to Reichardt.
answered Jan 11 at 4:48
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
Thank you for the response :) I was stuck because I was only thinking of mod q and mod a when what I need to do was mod z thanks again
$endgroup$
– john takeuchi
Jan 11 at 19:15
add a comment |
$begingroup$
If there are solutions, then there are solutions with $x$, $y$, $z$ pairwise
coprime. If $pmid x$, $y$, then $p^4mid az^2$ and as $p^2nmid a$ then
$p^2mid z$. Then $(x/p,y/p,z/p^2)$ is also a solution. Similar
arguments work when $pmid x,z$ and when $pmid y,z$.
Suppose that $(x,y,z)$ form a
pairwise coprime solution to the equation.
Let $p$ be an odd prime dividing $z$. Then as $x^4equiv qy^4pmod p$
we get $left(frac qpright)=1$ and then
$left(frac pqright)=1$ by quadratic reciprocity.
As $left(frac {-1}qright)=left(frac 2qright)=1$ then
$left(frac zqright)=1$.
There is $u$ with $zequiv u^2pmod q$. Then $x^4equiv au^4pmod q$,
and as both sides are nonzero modulo $q$, then $a$ is a quartic residue,
a contradiction.
The example $x^4-17y^2=2z^2$ is a counterexample to the Hasse principle
to quartics and is due to Reichardt.
answered Jan 11 at 4:48
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
Thank you for the response :) I was stuck because I was only thinking of mod q and mod a when what I need to do was mod z thanks again
$endgroup$
– john takeuchi
Jan 11 at 19:15
add a comment |
$begingroup$
If there are solutions, then there are solutions with $x$, $y$, $z$ pairwise
coprime. If $pmid x$, $y$, then $p^4mid az^2$ and as $p^2nmid a$ then
$p^2mid z$. Then $(x/p,y/p,z/p^2)$ is also a solution. Similar
arguments work when $pmid x,z$ and when $pmid y,z$.
Suppose that $(x,y,z)$ form a
pairwise coprime solution to the equation.
Let $p$ be an odd prime dividing $z$. Then as $x^4equiv qy^4pmod p$
we get $left(frac qpright)=1$ and then
$left(frac pqright)=1$ by quadratic reciprocity.
As $left(frac {-1}qright)=left(frac 2qright)=1$ then
$left(frac zqright)=1$.
There is $u$ with $zequiv u^2pmod q$. Then $x^4equiv au^4pmod q$,
and as both sides are nonzero modulo $q$, then $a$ is a quartic residue,
a contradiction.
The example $x^4-17y^2=2z^2$ is a counterexample to the Hasse principle
to quartics and is due to Reichardt.
answered Jan 11 at 4:48
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
If there are solutions, then there are solutions with $x$, $y$, $z$ pairwise
coprime. If $pmid x$, $y$, then $p^4mid az^2$ and as $p^2nmid a$ then
$p^2mid z$. Then $(x/p,y/p,z/p^2)$ is also a solution. Similar
arguments work when $pmid x,z$ and when $pmid y,z$.
Suppose that $(x,y,z)$ form a
pairwise coprime solution to the equation.
Let $p$ be an odd prime dividing $z$. Then as $x^4equiv qy^4pmod p$
we get $left(frac qpright)=1$ and then
$left(frac pqright)=1$ by quadratic reciprocity.
As $left(frac {-1}qright)=left(frac 2qright)=1$ then
$left(frac zqright)=1$.
There is $u$ with $zequiv u^2pmod q$. Then $x^4equiv au^4pmod q$,
and as both sides are nonzero modulo $q$, then $a$ is a quartic residue,
a contradiction.
The example $x^4-17y^2=2z^2$ is a counterexample to the Hasse principle
to quartics and is due to Reichardt.
answered Jan 11 at 4:48
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Jan 11 at 4:48
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Jan 11 at 4:48
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Jan 11 at 4:48
Lord Shark the UnknownLord Shark the Unknown
104k1160132
104k1160132
$begingroup$
Thank you for the response :) I was stuck because I was only thinking of mod q and mod a when what I need to do was mod z thanks again
$endgroup$
– john takeuchi
Jan 11 at 19:15
add a comment |
$begingroup$
Thank you for the response :) I was stuck because I was only thinking of mod q and mod a when what I need to do was mod z thanks again
$endgroup$
– john takeuchi
Jan 11 at 19:15
$begingroup$
Thank you for the response :) I was stuck because I was only thinking of mod q and mod a when what I need to do was mod z thanks again
$endgroup$
– john takeuchi
Jan 11 at 19:15
$begingroup$
Thank you for the response :) I was stuck because I was only thinking of mod q and mod a when what I need to do was mod z thanks again
$endgroup$
– john takeuchi
Jan 11 at 19:15
add a comment |
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$begingroup$
If one replaces the fourth powers with squares, you can solve this only assuming $q$ is $1$ modulo $4$. Start with the case $a$ is prime and consider the equation modulo $a$. This easily generalizes to $a$ not prime. I don't know if this helps with the original problem.
$endgroup$
– RghtHndSd
Jan 11 at 2:40