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For a given $Din mathbbN$ are there infinite solutions to $D=b^2-4ac$ with $a,b,c in mathbbN$
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I am wondering if for given $Din mathbbN$ we can find an infinite amount of solutions to $D=b^2-4ac$. Obvious $D$ is the discriminant of a binary quadratic form or a polynomial of degree $2$. Maybe the answer can be given with the use of those, but I do not know. Thanks for your answers.
abstract-algebra quadratic-forms
asked Jul 29 at 11:47
Deavor
568513
add a comment |Â
up vote
3
down vote
favorite
I am wondering if for given $Din mathbbN$ we can find an infinite amount of solutions to $D=b^2-4ac$. Obvious $D$ is the discriminant of a binary quadratic form or a polynomial of degree $2$. Maybe the answer can be given with the use of those, but I do not know. Thanks for your answers.
abstract-algebra quadratic-forms
asked Jul 29 at 11:47
Deavor
568513
3
If $Dequiv0$ or $1pmod 4$, yes. If $Dequiv2$ or $3pmod 4$, no.
– Lord Shark the Unknown
Jul 29 at 11:52
Thanks for your answer. Could you outline why this is a fact, or maybe give me a reference.
– Deavor
Jul 29 at 11:56
For $D = 4$, it can be shown that there are infinite solutions. $1 + ac = left(dfracb2right)^2$. For every $b$ s.t. $dfracb2 $ is odd, clearly atleast one solution exists $(b > 2)$. And there are infinite such $b$, so infinite solutions.
– Rahul Goswami
Jul 29 at 11:59
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am wondering if for given $Din mathbbN$ we can find an infinite amount of solutions to $D=b^2-4ac$. Obvious $D$ is the discriminant of a binary quadratic form or a polynomial of degree $2$. Maybe the answer can be given with the use of those, but I do not know. Thanks for your answers.
abstract-algebra quadratic-forms
asked Jul 29 at 11:47
Deavor
568513
I am wondering if for given $Din mathbbN$ we can find an infinite amount of solutions to $D=b^2-4ac$. Obvious $D$ is the discriminant of a binary quadratic form or a polynomial of degree $2$. Maybe the answer can be given with the use of those, but I do not know. Thanks for your answers.
abstract-algebra quadratic-forms
asked Jul 29 at 11:47
Deavor
568513
asked Jul 29 at 11:47
Deavor
568513
asked Jul 29 at 11:47
Deavor
568513
asked Jul 29 at 11:47
Deavor
568513
568513
3
If $Dequiv0$ or $1pmod 4$, yes. If $Dequiv2$ or $3pmod 4$, no.
– Lord Shark the Unknown
Jul 29 at 11:52
Thanks for your answer. Could you outline why this is a fact, or maybe give me a reference.
– Deavor
Jul 29 at 11:56
For $D = 4$, it can be shown that there are infinite solutions. $1 + ac = left(dfracb2right)^2$. For every $b$ s.t. $dfracb2 $ is odd, clearly atleast one solution exists $(b > 2)$. And there are infinite such $b$, so infinite solutions.
– Rahul Goswami
Jul 29 at 11:59
add a comment |Â
3
If $Dequiv0$ or $1pmod 4$, yes. If $Dequiv2$ or $3pmod 4$, no.
– Lord Shark the Unknown
Jul 29 at 11:52
Thanks for your answer. Could you outline why this is a fact, or maybe give me a reference.
– Deavor
Jul 29 at 11:56
For $D = 4$, it can be shown that there are infinite solutions. $1 + ac = left(dfracb2right)^2$. For every $b$ s.t. $dfracb2 $ is odd, clearly atleast one solution exists $(b > 2)$. And there are infinite such $b$, so infinite solutions.
– Rahul Goswami
Jul 29 at 11:59
3
3
If $Dequiv0$ or $1pmod 4$, yes. If $Dequiv2$ or $3pmod 4$, no.
– Lord Shark the Unknown
Jul 29 at 11:52
If $Dequiv0$ or $1pmod 4$, yes. If $Dequiv2$ or $3pmod 4$, no.
– Lord Shark the Unknown
Jul 29 at 11:52
Thanks for your answer. Could you outline why this is a fact, or maybe give me a reference.
– Deavor
Jul 29 at 11:56
Thanks for your answer. Could you outline why this is a fact, or maybe give me a reference.
– Deavor
Jul 29 at 11:56
For $D = 4$, it can be shown that there are infinite solutions. $1 + ac = left(dfracb2right)^2$. For every $b$ s.t. $dfracb2 $ is odd, clearly atleast one solution exists $(b > 2)$. And there are infinite such $b$, so infinite solutions.
– Rahul Goswami
Jul 29 at 11:59
For $D = 4$, it can be shown that there are infinite solutions. $1 + ac = left(dfracb2right)^2$. For every $b$ s.t. $dfracb2 $ is odd, clearly atleast one solution exists $(b > 2)$. And there are infinite such $b$, so infinite solutions.
– Rahul Goswami
Jul 29 at 11:59
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
I am wondering if for given $Din mathbbN$ we can find an infinite amount of solutions to $D=b^2-4ac$
No, not in general.
Consider $D=1$. Then $b^2-1=4ac$. Hence $(b+1)(b-1)=4ac$.
This obviously has four immediate solutions. $b=pm 1$ and $a=0vee c=0$.
Since $4|(b+1)(b-1)$, $b$ has to be odd. Which gives $b=2k+1$ for some $kinmathbbN$ and we get:
$(2k+2)2k=4k^2+4k=4k(k+1)$
Hence we have an infinte amount of solutions. With $k=a$ and $k+1=c$
Consider $D=3$. Then $3=b^2-4ac$. If $a=c$ we get $3=b^2-4a^2=(b-2a)(b+2a)$ but $3$ is prime. Hence there are only (at most) four solutions, for that case, which result from $b-2a=1 wedge b+2a=3$ and $b-2a=3wedge b+2a=1$.
If $aneq c$.
We get, after adding $1$ to the equation:
$b^2+1=4ac+4=4(ac+1)$
Hence $4|b^2+1$ as above $b$ has to be odd.
$(2k+1)^2+1=4k^2+4k+1+1=4k^2+4k+2=2(2k^2+2k+1)$
$2k^2+2k+1=2(ac+1)$
But $2nmid 2(k^2+k)+1$
As Lord Shark the Unknwon suggested in the comments, it is enough to view the problem $mod 4$. Because we can always just factor out the 4 of $D$ and reduce to $0, 1,2, 3$ that way.
The case $Dequiv 0mod 4$ is trivial and has infinite solutions, since then $b^2=4ac$ and if $b=2k$ we get $4k^2=4ac$. Now $a=c=k$ gives infinite solutions.
The last case is $Dequiv 2mod 4$.
This fails for a similar argument like the shown $D=3$ case.
$2=b^2-4ac$.
$b^2=4ac+2=2(2ac+1)$
Yet again, $b$ has to be even. This gives us:
$4k^2=2(2ac+1)Leftrightarrow 2k^2=2ac+1$
But now the LHS is even, while the RHS is odd.
answered Jul 29 at 12:06
Cornman
2,30021027
Thank you very much :). This answers the questions pretty well.
– Deavor
Jul 29 at 13:04
You are welcome. :)
– Cornman
Jul 29 at 13:17
Could you maybe explain why it is sufficient to reduce the equation modulo $4$? For example, If I have infinite solutions for $D=1$. How can I lift them to solutions for $D=5,9,13$ respectively $D=4,8,12$. Thank you very much.
– Deavor
Aug 1 at 13:59
@Deavor You can always write $D=r+4q$ where $rin0,1,2,3$, and then you have to solve $r+4q=b^2-4acLeftrightarrow r=b^2-4ac-4q=b^2-4underbrace(ac+q)_a'c'$. Thats why it is sufficient to reduce the equation modulo 4.
– Cornman
Aug 1 at 15:04
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I am wondering if for given $Din mathbbN$ we can find an infinite amount of solutions to $D=b^2-4ac$
No, not in general.
Consider $D=1$. Then $b^2-1=4ac$. Hence $(b+1)(b-1)=4ac$.
This obviously has four immediate solutions. $b=pm 1$ and $a=0vee c=0$.
Since $4|(b+1)(b-1)$, $b$ has to be odd. Which gives $b=2k+1$ for some $kinmathbbN$ and we get:
$(2k+2)2k=4k^2+4k=4k(k+1)$
Hence we have an infinte amount of solutions. With $k=a$ and $k+1=c$
Consider $D=3$. Then $3=b^2-4ac$. If $a=c$ we get $3=b^2-4a^2=(b-2a)(b+2a)$ but $3$ is prime. Hence there are only (at most) four solutions, for that case, which result from $b-2a=1 wedge b+2a=3$ and $b-2a=3wedge b+2a=1$.
If $aneq c$.
We get, after adding $1$ to the equation:
$b^2+1=4ac+4=4(ac+1)$
Hence $4|b^2+1$ as above $b$ has to be odd.
$(2k+1)^2+1=4k^2+4k+1+1=4k^2+4k+2=2(2k^2+2k+1)$
$2k^2+2k+1=2(ac+1)$
But $2nmid 2(k^2+k)+1$
As Lord Shark the Unknwon suggested in the comments, it is enough to view the problem $mod 4$. Because we can always just factor out the 4 of $D$ and reduce to $0, 1,2, 3$ that way.
The case $Dequiv 0mod 4$ is trivial and has infinite solutions, since then $b^2=4ac$ and if $b=2k$ we get $4k^2=4ac$. Now $a=c=k$ gives infinite solutions.
The last case is $Dequiv 2mod 4$.
This fails for a similar argument like the shown $D=3$ case.
$2=b^2-4ac$.
$b^2=4ac+2=2(2ac+1)$
Yet again, $b$ has to be even. This gives us:
$4k^2=2(2ac+1)Leftrightarrow 2k^2=2ac+1$
But now the LHS is even, while the RHS is odd.
answered Jul 29 at 12:06
Cornman
2,30021027
Thank you very much :). This answers the questions pretty well.
– Deavor
Jul 29 at 13:04
You are welcome. :)
– Cornman
Jul 29 at 13:17
Could you maybe explain why it is sufficient to reduce the equation modulo $4$? For example, If I have infinite solutions for $D=1$. How can I lift them to solutions for $D=5,9,13$ respectively $D=4,8,12$. Thank you very much.
– Deavor
Aug 1 at 13:59
@Deavor You can always write $D=r+4q$ where $rin0,1,2,3$, and then you have to solve $r+4q=b^2-4acLeftrightarrow r=b^2-4ac-4q=b^2-4underbrace(ac+q)_a'c'$. Thats why it is sufficient to reduce the equation modulo 4.
– Cornman
Aug 1 at 15:04
add a comment |Â
up vote
2
down vote
accepted
I am wondering if for given $Din mathbbN$ we can find an infinite amount of solutions to $D=b^2-4ac$
No, not in general.
Consider $D=1$. Then $b^2-1=4ac$. Hence $(b+1)(b-1)=4ac$.
This obviously has four immediate solutions. $b=pm 1$ and $a=0vee c=0$.
Since $4|(b+1)(b-1)$, $b$ has to be odd. Which gives $b=2k+1$ for some $kinmathbbN$ and we get:
$(2k+2)2k=4k^2+4k=4k(k+1)$
Hence we have an infinte amount of solutions. With $k=a$ and $k+1=c$
Consider $D=3$. Then $3=b^2-4ac$. If $a=c$ we get $3=b^2-4a^2=(b-2a)(b+2a)$ but $3$ is prime. Hence there are only (at most) four solutions, for that case, which result from $b-2a=1 wedge b+2a=3$ and $b-2a=3wedge b+2a=1$.
If $aneq c$.
We get, after adding $1$ to the equation:
$b^2+1=4ac+4=4(ac+1)$
Hence $4|b^2+1$ as above $b$ has to be odd.
$(2k+1)^2+1=4k^2+4k+1+1=4k^2+4k+2=2(2k^2+2k+1)$
$2k^2+2k+1=2(ac+1)$
But $2nmid 2(k^2+k)+1$
As Lord Shark the Unknwon suggested in the comments, it is enough to view the problem $mod 4$. Because we can always just factor out the 4 of $D$ and reduce to $0, 1,2, 3$ that way.
The case $Dequiv 0mod 4$ is trivial and has infinite solutions, since then $b^2=4ac$ and if $b=2k$ we get $4k^2=4ac$. Now $a=c=k$ gives infinite solutions.
The last case is $Dequiv 2mod 4$.
This fails for a similar argument like the shown $D=3$ case.
$2=b^2-4ac$.
$b^2=4ac+2=2(2ac+1)$
Yet again, $b$ has to be even. This gives us:
$4k^2=2(2ac+1)Leftrightarrow 2k^2=2ac+1$
But now the LHS is even, while the RHS is odd.
answered Jul 29 at 12:06
Cornman
2,30021027
Thank you very much :). This answers the questions pretty well.
– Deavor
Jul 29 at 13:04
You are welcome. :)
– Cornman
Jul 29 at 13:17
Could you maybe explain why it is sufficient to reduce the equation modulo $4$? For example, If I have infinite solutions for $D=1$. How can I lift them to solutions for $D=5,9,13$ respectively $D=4,8,12$. Thank you very much.
– Deavor
Aug 1 at 13:59
@Deavor You can always write $D=r+4q$ where $rin0,1,2,3$, and then you have to solve $r+4q=b^2-4acLeftrightarrow r=b^2-4ac-4q=b^2-4underbrace(ac+q)_a'c'$. Thats why it is sufficient to reduce the equation modulo 4.
– Cornman
Aug 1 at 15:04
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I am wondering if for given $Din mathbbN$ we can find an infinite amount of solutions to $D=b^2-4ac$
No, not in general.
Consider $D=1$. Then $b^2-1=4ac$. Hence $(b+1)(b-1)=4ac$.
This obviously has four immediate solutions. $b=pm 1$ and $a=0vee c=0$.
Since $4|(b+1)(b-1)$, $b$ has to be odd. Which gives $b=2k+1$ for some $kinmathbbN$ and we get:
$(2k+2)2k=4k^2+4k=4k(k+1)$
Hence we have an infinte amount of solutions. With $k=a$ and $k+1=c$
Consider $D=3$. Then $3=b^2-4ac$. If $a=c$ we get $3=b^2-4a^2=(b-2a)(b+2a)$ but $3$ is prime. Hence there are only (at most) four solutions, for that case, which result from $b-2a=1 wedge b+2a=3$ and $b-2a=3wedge b+2a=1$.
If $aneq c$.
We get, after adding $1$ to the equation:
$b^2+1=4ac+4=4(ac+1)$
Hence $4|b^2+1$ as above $b$ has to be odd.
$(2k+1)^2+1=4k^2+4k+1+1=4k^2+4k+2=2(2k^2+2k+1)$
$2k^2+2k+1=2(ac+1)$
But $2nmid 2(k^2+k)+1$
As Lord Shark the Unknwon suggested in the comments, it is enough to view the problem $mod 4$. Because we can always just factor out the 4 of $D$ and reduce to $0, 1,2, 3$ that way.
The case $Dequiv 0mod 4$ is trivial and has infinite solutions, since then $b^2=4ac$ and if $b=2k$ we get $4k^2=4ac$. Now $a=c=k$ gives infinite solutions.
The last case is $Dequiv 2mod 4$.
This fails for a similar argument like the shown $D=3$ case.
$2=b^2-4ac$.
$b^2=4ac+2=2(2ac+1)$
Yet again, $b$ has to be even. This gives us:
$4k^2=2(2ac+1)Leftrightarrow 2k^2=2ac+1$
But now the LHS is even, while the RHS is odd.
answered Jul 29 at 12:06
Cornman
2,30021027
I am wondering if for given $Din mathbbN$ we can find an infinite amount of solutions to $D=b^2-4ac$
No, not in general.
Consider $D=1$. Then $b^2-1=4ac$. Hence $(b+1)(b-1)=4ac$.
This obviously has four immediate solutions. $b=pm 1$ and $a=0vee c=0$.
Since $4|(b+1)(b-1)$, $b$ has to be odd. Which gives $b=2k+1$ for some $kinmathbbN$ and we get:
$(2k+2)2k=4k^2+4k=4k(k+1)$
Hence we have an infinte amount of solutions. With $k=a$ and $k+1=c$
Consider $D=3$. Then $3=b^2-4ac$. If $a=c$ we get $3=b^2-4a^2=(b-2a)(b+2a)$ but $3$ is prime. Hence there are only (at most) four solutions, for that case, which result from $b-2a=1 wedge b+2a=3$ and $b-2a=3wedge b+2a=1$.
If $aneq c$.
We get, after adding $1$ to the equation:
$b^2+1=4ac+4=4(ac+1)$
Hence $4|b^2+1$ as above $b$ has to be odd.
$(2k+1)^2+1=4k^2+4k+1+1=4k^2+4k+2=2(2k^2+2k+1)$
$2k^2+2k+1=2(ac+1)$
But $2nmid 2(k^2+k)+1$
As Lord Shark the Unknwon suggested in the comments, it is enough to view the problem $mod 4$. Because we can always just factor out the 4 of $D$ and reduce to $0, 1,2, 3$ that way.
The case $Dequiv 0mod 4$ is trivial and has infinite solutions, since then $b^2=4ac$ and if $b=2k$ we get $4k^2=4ac$. Now $a=c=k$ gives infinite solutions.
The last case is $Dequiv 2mod 4$.
This fails for a similar argument like the shown $D=3$ case.
$2=b^2-4ac$.
$b^2=4ac+2=2(2ac+1)$
Yet again, $b$ has to be even. This gives us:
$4k^2=2(2ac+1)Leftrightarrow 2k^2=2ac+1$
But now the LHS is even, while the RHS is odd.
answered Jul 29 at 12:06
Cornman
2,30021027
edited Jul 31 at 15:22
answered Jul 29 at 12:06
Cornman
2,30021027
answered Jul 29 at 12:06
Cornman
2,30021027
answered Jul 29 at 12:06
Cornman
2,30021027
2,30021027
Thank you very much :). This answers the questions pretty well.
– Deavor
Jul 29 at 13:04
You are welcome. :)
– Cornman
Jul 29 at 13:17
Could you maybe explain why it is sufficient to reduce the equation modulo $4$? For example, If I have infinite solutions for $D=1$. How can I lift them to solutions for $D=5,9,13$ respectively $D=4,8,12$. Thank you very much.
– Deavor
Aug 1 at 13:59
@Deavor You can always write $D=r+4q$ where $rin0,1,2,3$, and then you have to solve $r+4q=b^2-4acLeftrightarrow r=b^2-4ac-4q=b^2-4underbrace(ac+q)_a'c'$. Thats why it is sufficient to reduce the equation modulo 4.
– Cornman
Aug 1 at 15:04
add a comment |Â
Thank you very much :). This answers the questions pretty well.
– Deavor
Jul 29 at 13:04
You are welcome. :)
– Cornman
Jul 29 at 13:17
Could you maybe explain why it is sufficient to reduce the equation modulo $4$? For example, If I have infinite solutions for $D=1$. How can I lift them to solutions for $D=5,9,13$ respectively $D=4,8,12$. Thank you very much.
– Deavor
Aug 1 at 13:59
@Deavor You can always write $D=r+4q$ where $rin0,1,2,3$, and then you have to solve $r+4q=b^2-4acLeftrightarrow r=b^2-4ac-4q=b^2-4underbrace(ac+q)_a'c'$. Thats why it is sufficient to reduce the equation modulo 4.
– Cornman
Aug 1 at 15:04
Thank you very much :). This answers the questions pretty well.
– Deavor
Jul 29 at 13:04
Thank you very much :). This answers the questions pretty well.
– Deavor
Jul 29 at 13:04
You are welcome. :)
– Cornman
Jul 29 at 13:17
You are welcome. :)
– Cornman
Jul 29 at 13:17
Could you maybe explain why it is sufficient to reduce the equation modulo $4$? For example, If I have infinite solutions for $D=1$. How can I lift them to solutions for $D=5,9,13$ respectively $D=4,8,12$. Thank you very much.
– Deavor
Aug 1 at 13:59
Could you maybe explain why it is sufficient to reduce the equation modulo $4$? For example, If I have infinite solutions for $D=1$. How can I lift them to solutions for $D=5,9,13$ respectively $D=4,8,12$. Thank you very much.
– Deavor
Aug 1 at 13:59
@Deavor You can always write $D=r+4q$ where $rin0,1,2,3$, and then you have to solve $r+4q=b^2-4acLeftrightarrow r=b^2-4ac-4q=b^2-4underbrace(ac+q)_a'c'$. Thats why it is sufficient to reduce the equation modulo 4.
– Cornman
Aug 1 at 15:04
@Deavor You can always write $D=r+4q$ where $rin0,1,2,3$, and then you have to solve $r+4q=b^2-4acLeftrightarrow r=b^2-4ac-4q=b^2-4underbrace(ac+q)_a'c'$. Thats why it is sufficient to reduce the equation modulo 4.
– Cornman
Aug 1 at 15:04
add a comment |Â
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3
If $Dequiv0$ or $1pmod 4$, yes. If $Dequiv2$ or $3pmod 4$, no.
– Lord Shark the Unknown
Jul 29 at 11:52
Thanks for your answer. Could you outline why this is a fact, or maybe give me a reference.
– Deavor
Jul 29 at 11:56
For $D = 4$, it can be shown that there are infinite solutions. $1 + ac = left(dfracb2right)^2$. For every $b$ s.t. $dfracb2 $ is odd, clearly atleast one solution exists $(b > 2)$. And there are infinite such $b$, so infinite solutions.
– Rahul Goswami
Jul 29 at 11:59