Proving that, for $l$, $m$, $n$ odd integers, $sqrt{l^2-4mn}$ is never rational
$begingroup$
So the question I was solving was:
Prove that if $l$,$m$ & $n$ are odd then the line $lx+my+n=0$ will never intersect the parabola $y=x^2$ at a rational point.
I tried to solve the question and arrived at the condition that for this to be true $sqrt{l^2-4mn}$ should never be rational. How do I prove this?
number-theory conic-sections
edited Jan 3 at 18:56
Blue
49.3k870157
asked Nov 23 '18 at 16:55
harshit54harshit54
346113
$endgroup$
add a comment |
$begingroup$
So the question I was solving was:
Prove that if $l$,$m$ & $n$ are odd then the line $lx+my+n=0$ will never intersect the parabola $y=x^2$ at a rational point.
I tried to solve the question and arrived at the condition that for this to be true $sqrt{l^2-4mn}$ should never be rational. How do I prove this?
number-theory conic-sections
edited Jan 3 at 18:56
Blue
49.3k870157
asked Nov 23 '18 at 16:55
harshit54harshit54
346113
$endgroup$
$begingroup$
@Anguepa But then $l$ isn't odd.
$endgroup$
– saulspatz
Nov 23 '18 at 16:59
$begingroup$
You re right sorry. The square root of a natural number can only be an integer or irrational. Try to prove that the expression inside the square root can never be a perfect square.
$endgroup$
– Anguepa
Nov 23 '18 at 17:07
add a comment |
$begingroup$
So the question I was solving was:
Prove that if $l$,$m$ & $n$ are odd then the line $lx+my+n=0$ will never intersect the parabola $y=x^2$ at a rational point.
I tried to solve the question and arrived at the condition that for this to be true $sqrt{l^2-4mn}$ should never be rational. How do I prove this?
number-theory conic-sections
edited Jan 3 at 18:56
Blue
49.3k870157
asked Nov 23 '18 at 16:55
harshit54harshit54
346113
$endgroup$
So the question I was solving was:
Prove that if $l$,$m$ & $n$ are odd then the line $lx+my+n=0$ will never intersect the parabola $y=x^2$ at a rational point.
I tried to solve the question and arrived at the condition that for this to be true $sqrt{l^2-4mn}$ should never be rational. How do I prove this?
number-theory conic-sections
number-theory conic-sections
edited Jan 3 at 18:56
Blue
49.3k870157
asked Nov 23 '18 at 16:55
harshit54harshit54
346113
edited Jan 3 at 18:56
Blue
49.3k870157
asked Nov 23 '18 at 16:55
harshit54harshit54
346113
edited Jan 3 at 18:56
Blue
49.3k870157
edited Jan 3 at 18:56
Blue
49.3k870157
edited Jan 3 at 18:56
Blue
49.3k870157
49.3k870157
asked Nov 23 '18 at 16:55
harshit54harshit54
346113
asked Nov 23 '18 at 16:55
harshit54harshit54
346113
asked Nov 23 '18 at 16:55
harshit54harshit54
346113
346113
$begingroup$
@Anguepa But then $l$ isn't odd.
$endgroup$
– saulspatz
Nov 23 '18 at 16:59
$begingroup$
You re right sorry. The square root of a natural number can only be an integer or irrational. Try to prove that the expression inside the square root can never be a perfect square.
$endgroup$
– Anguepa
Nov 23 '18 at 17:07
add a comment |
$begingroup$
@Anguepa But then $l$ isn't odd.
$endgroup$
– saulspatz
Nov 23 '18 at 16:59
$begingroup$
You re right sorry. The square root of a natural number can only be an integer or irrational. Try to prove that the expression inside the square root can never be a perfect square.
$endgroup$
– Anguepa
Nov 23 '18 at 17:07
$begingroup$
@Anguepa But then $l$ isn't odd.
$endgroup$
– saulspatz
Nov 23 '18 at 16:59
$begingroup$
@Anguepa But then $l$ isn't odd.
$endgroup$
– saulspatz
Nov 23 '18 at 16:59
$begingroup$
You re right sorry. The square root of a natural number can only be an integer or irrational. Try to prove that the expression inside the square root can never be a perfect square.
$endgroup$
– Anguepa
Nov 23 '18 at 17:07
$begingroup$
You re right sorry. The square root of a natural number can only be an integer or irrational. Try to prove that the expression inside the square root can never be a perfect square.
$endgroup$
– Anguepa
Nov 23 '18 at 17:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since $l^2-4mn$ would have to be an odd square, i.e. $1$ more than a multiple of $8$ just like $l^2$, $mn$ would be even, a contradiction.
answered Nov 23 '18 at 17:00
J.G.J.G.
32.6k23250
$endgroup$
$begingroup$
Wow. Thanks for the quick response.
$endgroup$
– harshit54
Nov 23 '18 at 17:05
add a comment |
$begingroup$
The system of equations
$$begin{cases} my + lx + n = 0 \ y = x^2 end{cases}$$
Can be simplified as
$$mx^2 + lx + n = 0$$
As the right hand side is divisible by 2, the left hand side must be too.
But
$$mx^2 + lx + n equiv x^2 + x + 1 equiv x + x + 1 equiv 1 pmod{2}$$ as $m,l,n$ are odd.
answered Jan 3 at 19:10
Jonas De SchouwerJonas De Schouwer
3769
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $l^2-4mn$ would have to be an odd square, i.e. $1$ more than a multiple of $8$ just like $l^2$, $mn$ would be even, a contradiction.
answered Nov 23 '18 at 17:00
J.G.J.G.
32.6k23250
$endgroup$
$begingroup$
Wow. Thanks for the quick response.
$endgroup$
– harshit54
Nov 23 '18 at 17:05
add a comment |
$begingroup$
Since $l^2-4mn$ would have to be an odd square, i.e. $1$ more than a multiple of $8$ just like $l^2$, $mn$ would be even, a contradiction.
answered Nov 23 '18 at 17:00
J.G.J.G.
32.6k23250
$endgroup$
$begingroup$
Wow. Thanks for the quick response.
$endgroup$
– harshit54
Nov 23 '18 at 17:05
add a comment |
$begingroup$
Since $l^2-4mn$ would have to be an odd square, i.e. $1$ more than a multiple of $8$ just like $l^2$, $mn$ would be even, a contradiction.
answered Nov 23 '18 at 17:00
J.G.J.G.
32.6k23250
$endgroup$
Since $l^2-4mn$ would have to be an odd square, i.e. $1$ more than a multiple of $8$ just like $l^2$, $mn$ would be even, a contradiction.
answered Nov 23 '18 at 17:00
J.G.J.G.
32.6k23250
answered Nov 23 '18 at 17:00
J.G.J.G.
32.6k23250
answered Nov 23 '18 at 17:00
J.G.J.G.
32.6k23250
answered Nov 23 '18 at 17:00
J.G.J.G.
32.6k23250
32.6k23250
$begingroup$
Wow. Thanks for the quick response.
$endgroup$
– harshit54
Nov 23 '18 at 17:05
add a comment |
$begingroup$
Wow. Thanks for the quick response.
$endgroup$
– harshit54
Nov 23 '18 at 17:05
$begingroup$
Wow. Thanks for the quick response.
$endgroup$
– harshit54
Nov 23 '18 at 17:05
$begingroup$
Wow. Thanks for the quick response.
$endgroup$
– harshit54
Nov 23 '18 at 17:05
add a comment |
$begingroup$
The system of equations
$$begin{cases} my + lx + n = 0 \ y = x^2 end{cases}$$
Can be simplified as
$$mx^2 + lx + n = 0$$
As the right hand side is divisible by 2, the left hand side must be too.
But
$$mx^2 + lx + n equiv x^2 + x + 1 equiv x + x + 1 equiv 1 pmod{2}$$ as $m,l,n$ are odd.
answered Jan 3 at 19:10
Jonas De SchouwerJonas De Schouwer
3769
$endgroup$
add a comment |
$begingroup$
The system of equations
$$begin{cases} my + lx + n = 0 \ y = x^2 end{cases}$$
Can be simplified as
$$mx^2 + lx + n = 0$$
As the right hand side is divisible by 2, the left hand side must be too.
But
$$mx^2 + lx + n equiv x^2 + x + 1 equiv x + x + 1 equiv 1 pmod{2}$$ as $m,l,n$ are odd.
answered Jan 3 at 19:10
Jonas De SchouwerJonas De Schouwer
3769
$endgroup$
add a comment |
$begingroup$
The system of equations
$$begin{cases} my + lx + n = 0 \ y = x^2 end{cases}$$
Can be simplified as
$$mx^2 + lx + n = 0$$
As the right hand side is divisible by 2, the left hand side must be too.
But
$$mx^2 + lx + n equiv x^2 + x + 1 equiv x + x + 1 equiv 1 pmod{2}$$ as $m,l,n$ are odd.
answered Jan 3 at 19:10
Jonas De SchouwerJonas De Schouwer
3769
$endgroup$
The system of equations
$$begin{cases} my + lx + n = 0 \ y = x^2 end{cases}$$
Can be simplified as
$$mx^2 + lx + n = 0$$
As the right hand side is divisible by 2, the left hand side must be too.
But
$$mx^2 + lx + n equiv x^2 + x + 1 equiv x + x + 1 equiv 1 pmod{2}$$ as $m,l,n$ are odd.
answered Jan 3 at 19:10
Jonas De SchouwerJonas De Schouwer
3769
answered Jan 3 at 19:10
Jonas De SchouwerJonas De Schouwer
3769
answered Jan 3 at 19:10
Jonas De SchouwerJonas De Schouwer
3769
answered Jan 3 at 19:10
Jonas De SchouwerJonas De Schouwer
3769
3769
add a comment |
add a comment |
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$begingroup$
@Anguepa But then $l$ isn't odd.
$endgroup$
– saulspatz
Nov 23 '18 at 16:59
$begingroup$
You re right sorry. The square root of a natural number can only be an integer or irrational. Try to prove that the expression inside the square root can never be a perfect square.
$endgroup$
– Anguepa
Nov 23 '18 at 17:07