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To what sets must $a,b,c$ belong?
Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizablewhat type of math is this?Difference between fields $mathbbQ[sqrt2+sqrt3]$ and $mathbbQ[sqrt2,sqrt3]$?Compute all the sets of 87248 into 10 partsWhat are some strong algebraic number theory PhD programs?Number Sets and Harmonics?Can Number Theory be visualized?Difference and Quotient of roots of a quadratic equationWhat is exactly number theory?How to express “elements that belong to two or more sets but not to their intersection”?Searching for intuition in number theorie
$begingroup$
I just thought of kind of a cool number theory/algebra problem.
Given that $$sqrtb^2-4acinBbb N$$
To which sets must $a,b,c$ belong?
It is obvious that $$b^2-4acinleftx^2$$
But beyond that, I do not know what to do. May I have some help?
Edit:
To make things more interesting, what if $a,b,cinBbb N$?
algebra-precalculus number-theory elementary-number-theory quadratics
asked Mar 10 at 20:08
clathratusclathratus
4,9551338
$endgroup$
add a comment |
$begingroup$
I just thought of kind of a cool number theory/algebra problem.
Given that $$sqrtb^2-4acinBbb N$$
To which sets must $a,b,c$ belong?
It is obvious that $$b^2-4acinleftx^2$$
But beyond that, I do not know what to do. May I have some help?
Edit:
To make things more interesting, what if $a,b,cinBbb N$?
algebra-precalculus number-theory elementary-number-theory quadratics
asked Mar 10 at 20:08
clathratusclathratus
4,9551338
$endgroup$
3
$begingroup$
Note that $a,b,c$ need not be integers. So this is no Diophantine equation. So it is a little bit "less cool". We just write $b^2-4ac=x^2$, so that $a=(b^2-d^2)/(4c)$ for $c$ nonzero. Then choose $b,cin Bbb R$ and $d in Bbb Z$.
$endgroup$
– Dietrich Burde
Mar 10 at 20:21
$begingroup$
But what if $a,b,c$ were integers?
$endgroup$
– clathratus
Mar 10 at 20:25
1
$begingroup$
So, should we let $a,b,cinmathbb N$?
$endgroup$
– Dr. Mathva
Mar 10 at 20:38
add a comment |
$begingroup$
I just thought of kind of a cool number theory/algebra problem.
Given that $$sqrtb^2-4acinBbb N$$
To which sets must $a,b,c$ belong?
It is obvious that $$b^2-4acinleftx^2$$
But beyond that, I do not know what to do. May I have some help?
Edit:
To make things more interesting, what if $a,b,cinBbb N$?
algebra-precalculus number-theory elementary-number-theory quadratics
asked Mar 10 at 20:08
clathratusclathratus
4,9551338
$endgroup$
I just thought of kind of a cool number theory/algebra problem.
Given that $$sqrtb^2-4acinBbb N$$
To which sets must $a,b,c$ belong?
It is obvious that $$b^2-4acinleftx^2$$
But beyond that, I do not know what to do. May I have some help?
Edit:
To make things more interesting, what if $a,b,cinBbb N$?
algebra-precalculus number-theory elementary-number-theory quadratics
algebra-precalculus number-theory elementary-number-theory quadratics
asked Mar 10 at 20:08
clathratusclathratus
4,9551338
asked Mar 10 at 20:08
clathratusclathratus
4,9551338
edited Mar 10 at 21:09
clathratus
asked Mar 10 at 20:08
clathratusclathratus
4,9551338
asked Mar 10 at 20:08
clathratusclathratus
4,9551338
asked Mar 10 at 20:08
clathratusclathratus
4,9551338
4,9551338
3
$begingroup$
Note that $a,b,c$ need not be integers. So this is no Diophantine equation. So it is a little bit "less cool". We just write $b^2-4ac=x^2$, so that $a=(b^2-d^2)/(4c)$ for $c$ nonzero. Then choose $b,cin Bbb R$ and $d in Bbb Z$.
$endgroup$
– Dietrich Burde
Mar 10 at 20:21
$begingroup$
But what if $a,b,c$ were integers?
$endgroup$
– clathratus
Mar 10 at 20:25
1
$begingroup$
So, should we let $a,b,cinmathbb N$?
$endgroup$
– Dr. Mathva
Mar 10 at 20:38
add a comment |
3
$begingroup$
Note that $a,b,c$ need not be integers. So this is no Diophantine equation. So it is a little bit "less cool". We just write $b^2-4ac=x^2$, so that $a=(b^2-d^2)/(4c)$ for $c$ nonzero. Then choose $b,cin Bbb R$ and $d in Bbb Z$.
$endgroup$
– Dietrich Burde
Mar 10 at 20:21
$begingroup$
But what if $a,b,c$ were integers?
$endgroup$
– clathratus
Mar 10 at 20:25
1
$begingroup$
So, should we let $a,b,cinmathbb N$?
$endgroup$
– Dr. Mathva
Mar 10 at 20:38
3
3
$begingroup$
Note that $a,b,c$ need not be integers. So this is no Diophantine equation. So it is a little bit "less cool". We just write $b^2-4ac=x^2$, so that $a=(b^2-d^2)/(4c)$ for $c$ nonzero. Then choose $b,cin Bbb R$ and $d in Bbb Z$.
$endgroup$
– Dietrich Burde
Mar 10 at 20:21
$begingroup$
Note that $a,b,c$ need not be integers. So this is no Diophantine equation. So it is a little bit "less cool". We just write $b^2-4ac=x^2$, so that $a=(b^2-d^2)/(4c)$ for $c$ nonzero. Then choose $b,cin Bbb R$ and $d in Bbb Z$.
$endgroup$
– Dietrich Burde
Mar 10 at 20:21
$begingroup$
But what if $a,b,c$ were integers?
$endgroup$
– clathratus
Mar 10 at 20:25
$begingroup$
But what if $a,b,c$ were integers?
$endgroup$
– clathratus
Mar 10 at 20:25
1
1
$begingroup$
So, should we let $a,b,cinmathbb N$?
$endgroup$
– Dr. Mathva
Mar 10 at 20:38
$begingroup$
So, should we let $a,b,cinmathbb N$?
$endgroup$
– Dr. Mathva
Mar 10 at 20:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
all triples $a,b,c$ that work are given by
$$ ax^2 + b x y + c y^2 = (sx+ty)(ux+vy) ; , ; $$
$$ a = su, $$
$$ b = sv+tu, $$
$$ c = tv . $$
When those happen,
$$ b^2 - 4ac = s^2 v^2 + 2 stuv + t^2 u^2 - 4 stuv =s^2 v^2 - 2 stuv + t^2 u^2 = (sv-tu)^2 $$
see Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizable
answered Mar 10 at 21:45
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
Okay, so to which sets do $s,v,t,u$ belong?
$endgroup$
– clathratus
Mar 11 at 1:11
$begingroup$
@clathratus integers
$endgroup$
– Will Jagy
Mar 11 at 1:18
add a comment |
Your Answer
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1 Answer
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1 Answer
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votes
$begingroup$
all triples $a,b,c$ that work are given by
$$ ax^2 + b x y + c y^2 = (sx+ty)(ux+vy) ; , ; $$
$$ a = su, $$
$$ b = sv+tu, $$
$$ c = tv . $$
When those happen,
$$ b^2 - 4ac = s^2 v^2 + 2 stuv + t^2 u^2 - 4 stuv =s^2 v^2 - 2 stuv + t^2 u^2 = (sv-tu)^2 $$
see Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizable
answered Mar 10 at 21:45
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
Okay, so to which sets do $s,v,t,u$ belong?
$endgroup$
– clathratus
Mar 11 at 1:11
$begingroup$
@clathratus integers
$endgroup$
– Will Jagy
Mar 11 at 1:18
add a comment |
$begingroup$
all triples $a,b,c$ that work are given by
$$ ax^2 + b x y + c y^2 = (sx+ty)(ux+vy) ; , ; $$
$$ a = su, $$
$$ b = sv+tu, $$
$$ c = tv . $$
When those happen,
$$ b^2 - 4ac = s^2 v^2 + 2 stuv + t^2 u^2 - 4 stuv =s^2 v^2 - 2 stuv + t^2 u^2 = (sv-tu)^2 $$
see Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizable
answered Mar 10 at 21:45
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
Okay, so to which sets do $s,v,t,u$ belong?
$endgroup$
– clathratus
Mar 11 at 1:11
$begingroup$
@clathratus integers
$endgroup$
– Will Jagy
Mar 11 at 1:18
add a comment |
$begingroup$
all triples $a,b,c$ that work are given by
$$ ax^2 + b x y + c y^2 = (sx+ty)(ux+vy) ; , ; $$
$$ a = su, $$
$$ b = sv+tu, $$
$$ c = tv . $$
When those happen,
$$ b^2 - 4ac = s^2 v^2 + 2 stuv + t^2 u^2 - 4 stuv =s^2 v^2 - 2 stuv + t^2 u^2 = (sv-tu)^2 $$
see Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizable
answered Mar 10 at 21:45
Will JagyWill Jagy
104k5102201
$endgroup$
all triples $a,b,c$ that work are given by
$$ ax^2 + b x y + c y^2 = (sx+ty)(ux+vy) ; , ; $$
$$ a = su, $$
$$ b = sv+tu, $$
$$ c = tv . $$
When those happen,
$$ b^2 - 4ac = s^2 v^2 + 2 stuv + t^2 u^2 - 4 stuv =s^2 v^2 - 2 stuv + t^2 u^2 = (sv-tu)^2 $$
see Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizable
answered Mar 10 at 21:45
Will JagyWill Jagy
104k5102201
edited Mar 10 at 21:50
answered Mar 10 at 21:45
Will JagyWill Jagy
104k5102201
answered Mar 10 at 21:45
Will JagyWill Jagy
104k5102201
answered Mar 10 at 21:45
Will JagyWill Jagy
104k5102201
104k5102201
$begingroup$
Okay, so to which sets do $s,v,t,u$ belong?
$endgroup$
– clathratus
Mar 11 at 1:11
$begingroup$
@clathratus integers
$endgroup$
– Will Jagy
Mar 11 at 1:18
add a comment |
$begingroup$
Okay, so to which sets do $s,v,t,u$ belong?
$endgroup$
– clathratus
Mar 11 at 1:11
$begingroup$
@clathratus integers
$endgroup$
– Will Jagy
Mar 11 at 1:18
$begingroup$
Okay, so to which sets do $s,v,t,u$ belong?
$endgroup$
– clathratus
Mar 11 at 1:11
$begingroup$
Okay, so to which sets do $s,v,t,u$ belong?
$endgroup$
– clathratus
Mar 11 at 1:11
$begingroup$
@clathratus integers
$endgroup$
– Will Jagy
Mar 11 at 1:18
$begingroup$
@clathratus integers
$endgroup$
– Will Jagy
Mar 11 at 1:18
add a comment |
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3
$begingroup$
Note that $a,b,c$ need not be integers. So this is no Diophantine equation. So it is a little bit "less cool". We just write $b^2-4ac=x^2$, so that $a=(b^2-d^2)/(4c)$ for $c$ nonzero. Then choose $b,cin Bbb R$ and $d in Bbb Z$.
$endgroup$
– Dietrich Burde
Mar 10 at 20:21
$begingroup$
But what if $a,b,c$ were integers?
$endgroup$
– clathratus
Mar 10 at 20:25
1
$begingroup$
So, should we let $a,b,cinmathbb N$?
$endgroup$
– Dr. Mathva
Mar 10 at 20:38