Integer equations
$begingroup$
I have $2$ following problems. Find integer roots of
$$begin{align}
&1)~frac{x+y}{x^2-xy+y^2}=frac3z \
&2)~x^3y^3-4xy^3+y^2+x^2-2y-3=0
end{align}$$
I have no idea to solve them. I try to guess roots of the second, they are $left( -2, 1right), left( 0, -1right), left( 0, 3right), left( 2, 1right) $. Please help me. Thank you very much.
number-theory elementary-number-theory arithmetic diophantine-equations
edited Dec 8 '18 at 15:22
mrtaurho
5,21141237
asked Dec 8 '18 at 9:38
CglkttcaCglkttca
826
$endgroup$
|
show 2 more comments
$begingroup$
I have $2$ following problems. Find integer roots of
$$begin{align}
&1)~frac{x+y}{x^2-xy+y^2}=frac3z \
&2)~x^3y^3-4xy^3+y^2+x^2-2y-3=0
end{align}$$
I have no idea to solve them. I try to guess roots of the second, they are $left( -2, 1right), left( 0, -1right), left( 0, 3right), left( 2, 1right) $. Please help me. Thank you very much.
number-theory elementary-number-theory arithmetic diophantine-equations
edited Dec 8 '18 at 15:22
mrtaurho
5,21141237
asked Dec 8 '18 at 9:38
CglkttcaCglkttca
826
$endgroup$
$begingroup$
These problems seem very interesting. Where did you get them from?
$endgroup$
– Toby Mak
Dec 8 '18 at 9:39
$begingroup$
@TobyMak These are my homework.
$endgroup$
– Cglkttca
Dec 8 '18 at 9:42
$begingroup$
What class are you taking?
$endgroup$
– Toby Mak
Dec 8 '18 at 9:42
$begingroup$
@TobyMak I'm in grade 9.
$endgroup$
– Cglkttca
Dec 8 '18 at 9:43
1
$begingroup$
What lesson is this that you were given that exercise ?
$endgroup$
– Rebellos
Dec 8 '18 at 10:03
|
show 2 more comments
$begingroup$
I have $2$ following problems. Find integer roots of
$$begin{align}
&1)~frac{x+y}{x^2-xy+y^2}=frac3z \
&2)~x^3y^3-4xy^3+y^2+x^2-2y-3=0
end{align}$$
I have no idea to solve them. I try to guess roots of the second, they are $left( -2, 1right), left( 0, -1right), left( 0, 3right), left( 2, 1right) $. Please help me. Thank you very much.
number-theory elementary-number-theory arithmetic diophantine-equations
edited Dec 8 '18 at 15:22
mrtaurho
5,21141237
asked Dec 8 '18 at 9:38
CglkttcaCglkttca
826
$endgroup$
I have $2$ following problems. Find integer roots of
$$begin{align}
&1)~frac{x+y}{x^2-xy+y^2}=frac3z \
&2)~x^3y^3-4xy^3+y^2+x^2-2y-3=0
end{align}$$
I have no idea to solve them. I try to guess roots of the second, they are $left( -2, 1right), left( 0, -1right), left( 0, 3right), left( 2, 1right) $. Please help me. Thank you very much.
number-theory elementary-number-theory arithmetic diophantine-equations
number-theory elementary-number-theory arithmetic diophantine-equations
edited Dec 8 '18 at 15:22
mrtaurho
5,21141237
asked Dec 8 '18 at 9:38
CglkttcaCglkttca
826
edited Dec 8 '18 at 15:22
mrtaurho
5,21141237
asked Dec 8 '18 at 9:38
CglkttcaCglkttca
826
edited Dec 8 '18 at 15:22
mrtaurho
5,21141237
edited Dec 8 '18 at 15:22
mrtaurho
5,21141237
edited Dec 8 '18 at 15:22
mrtaurho
5,21141237
5,21141237
asked Dec 8 '18 at 9:38
CglkttcaCglkttca
826
asked Dec 8 '18 at 9:38
CglkttcaCglkttca
826
asked Dec 8 '18 at 9:38
CglkttcaCglkttca
826
826
$begingroup$
These problems seem very interesting. Where did you get them from?
$endgroup$
– Toby Mak
Dec 8 '18 at 9:39
$begingroup$
@TobyMak These are my homework.
$endgroup$
– Cglkttca
Dec 8 '18 at 9:42
$begingroup$
What class are you taking?
$endgroup$
– Toby Mak
Dec 8 '18 at 9:42
$begingroup$
@TobyMak I'm in grade 9.
$endgroup$
– Cglkttca
Dec 8 '18 at 9:43
1
$begingroup$
What lesson is this that you were given that exercise ?
$endgroup$
– Rebellos
Dec 8 '18 at 10:03
|
show 2 more comments
$begingroup$
These problems seem very interesting. Where did you get them from?
$endgroup$
– Toby Mak
Dec 8 '18 at 9:39
$begingroup$
@TobyMak These are my homework.
$endgroup$
– Cglkttca
Dec 8 '18 at 9:42
$begingroup$
What class are you taking?
$endgroup$
– Toby Mak
Dec 8 '18 at 9:42
$begingroup$
@TobyMak I'm in grade 9.
$endgroup$
– Cglkttca
Dec 8 '18 at 9:43
1
$begingroup$
What lesson is this that you were given that exercise ?
$endgroup$
– Rebellos
Dec 8 '18 at 10:03
$begingroup$
These problems seem very interesting. Where did you get them from?
$endgroup$
– Toby Mak
Dec 8 '18 at 9:39
$begingroup$
These problems seem very interesting. Where did you get them from?
$endgroup$
– Toby Mak
Dec 8 '18 at 9:39
$begingroup$
@TobyMak These are my homework.
$endgroup$
– Cglkttca
Dec 8 '18 at 9:42
$begingroup$
@TobyMak These are my homework.
$endgroup$
– Cglkttca
Dec 8 '18 at 9:42
$begingroup$
What class are you taking?
$endgroup$
– Toby Mak
Dec 8 '18 at 9:42
$begingroup$
What class are you taking?
$endgroup$
– Toby Mak
Dec 8 '18 at 9:42
$begingroup$
@TobyMak I'm in grade 9.
$endgroup$
– Cglkttca
Dec 8 '18 at 9:43
$begingroup$
@TobyMak I'm in grade 9.
$endgroup$
– Cglkttca
Dec 8 '18 at 9:43
1
1
$begingroup$
What lesson is this that you were given that exercise ?
$endgroup$
– Rebellos
Dec 8 '18 at 10:03
$begingroup$
What lesson is this that you were given that exercise ?
$endgroup$
– Rebellos
Dec 8 '18 at 10:03
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Question 1:
begin{align}
& frac{x+y}{x^2-xy+y^2} = frac 3z \
iff & frac{3(x^2-xy+y^2)}{x+y} = z\
iff & 3(x+y) - frac{9xy}{x+y} = z \
iff & frac{9xy}{x+y} = 3x+3y-z \
implies & frac{9}{frac 1x + frac 1y} = 3x+3y-z
end{align}
From the LHS we see that $frac 1x + frac 1y$ must be equal to $pm 1, pm 3, pm 9$ since the RHS is an integer. But of course, since $x,y$ are integers, $frac 1x + frac 1y in [-2,2]$ and it follows that $frac 1x + frac 1y = pm 1$.
Moreover, the only way this can happen is if $x = y = pm 2$.
Hence, the only solutions are $(x,y,z) = (2,2,3)$ and $(x,y,z) = (-2,-2,-3)$.
EDIT:
As pointed out in the comments, I have made the mistake of dividing by $0$, so I have changed the last $iff$ to an $implies$.
If $x=0$, then $z=3y$ so we get the solutions $(x,y,z) = (0,t,3t)$
Similarly when $y=0$ we get the solutions $(x,y,z) = (t,0,3t)$ for any $t in Bbb Z$ with $t neq 0$.
answered Dec 8 '18 at 10:16
glowstonetreesglowstonetrees
2,368418
$endgroup$
$begingroup$
I got it. Thank you very much.
$endgroup$
– Cglkttca
Dec 8 '18 at 10:20
1
$begingroup$
It's a nice answer, however, you forgot the cases when x and y are 0.
$endgroup$
– Ankit Kumar
Dec 8 '18 at 10:29
$begingroup$
Yes, in this case, we get more solutions
$endgroup$
– Cglkttca
Dec 8 '18 at 10:34
add a comment |
Your Answer
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1 Answer
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votes
1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
Question 1:
begin{align}
& frac{x+y}{x^2-xy+y^2} = frac 3z \
iff & frac{3(x^2-xy+y^2)}{x+y} = z\
iff & 3(x+y) - frac{9xy}{x+y} = z \
iff & frac{9xy}{x+y} = 3x+3y-z \
implies & frac{9}{frac 1x + frac 1y} = 3x+3y-z
end{align}
From the LHS we see that $frac 1x + frac 1y$ must be equal to $pm 1, pm 3, pm 9$ since the RHS is an integer. But of course, since $x,y$ are integers, $frac 1x + frac 1y in [-2,2]$ and it follows that $frac 1x + frac 1y = pm 1$.
Moreover, the only way this can happen is if $x = y = pm 2$.
Hence, the only solutions are $(x,y,z) = (2,2,3)$ and $(x,y,z) = (-2,-2,-3)$.
EDIT:
As pointed out in the comments, I have made the mistake of dividing by $0$, so I have changed the last $iff$ to an $implies$.
If $x=0$, then $z=3y$ so we get the solutions $(x,y,z) = (0,t,3t)$
Similarly when $y=0$ we get the solutions $(x,y,z) = (t,0,3t)$ for any $t in Bbb Z$ with $t neq 0$.
answered Dec 8 '18 at 10:16
glowstonetreesglowstonetrees
2,368418
$endgroup$
$begingroup$
I got it. Thank you very much.
$endgroup$
– Cglkttca
Dec 8 '18 at 10:20
1
$begingroup$
It's a nice answer, however, you forgot the cases when x and y are 0.
$endgroup$
– Ankit Kumar
Dec 8 '18 at 10:29
$begingroup$
Yes, in this case, we get more solutions
$endgroup$
– Cglkttca
Dec 8 '18 at 10:34
add a comment |
$begingroup$
Question 1:
begin{align}
& frac{x+y}{x^2-xy+y^2} = frac 3z \
iff & frac{3(x^2-xy+y^2)}{x+y} = z\
iff & 3(x+y) - frac{9xy}{x+y} = z \
iff & frac{9xy}{x+y} = 3x+3y-z \
implies & frac{9}{frac 1x + frac 1y} = 3x+3y-z
end{align}
From the LHS we see that $frac 1x + frac 1y$ must be equal to $pm 1, pm 3, pm 9$ since the RHS is an integer. But of course, since $x,y$ are integers, $frac 1x + frac 1y in [-2,2]$ and it follows that $frac 1x + frac 1y = pm 1$.
Moreover, the only way this can happen is if $x = y = pm 2$.
Hence, the only solutions are $(x,y,z) = (2,2,3)$ and $(x,y,z) = (-2,-2,-3)$.
EDIT:
As pointed out in the comments, I have made the mistake of dividing by $0$, so I have changed the last $iff$ to an $implies$.
If $x=0$, then $z=3y$ so we get the solutions $(x,y,z) = (0,t,3t)$
Similarly when $y=0$ we get the solutions $(x,y,z) = (t,0,3t)$ for any $t in Bbb Z$ with $t neq 0$.
answered Dec 8 '18 at 10:16
glowstonetreesglowstonetrees
2,368418
$endgroup$
$begingroup$
I got it. Thank you very much.
$endgroup$
– Cglkttca
Dec 8 '18 at 10:20
1
$begingroup$
It's a nice answer, however, you forgot the cases when x and y are 0.
$endgroup$
– Ankit Kumar
Dec 8 '18 at 10:29
$begingroup$
Yes, in this case, we get more solutions
$endgroup$
– Cglkttca
Dec 8 '18 at 10:34
add a comment |
$begingroup$
Question 1:
begin{align}
& frac{x+y}{x^2-xy+y^2} = frac 3z \
iff & frac{3(x^2-xy+y^2)}{x+y} = z\
iff & 3(x+y) - frac{9xy}{x+y} = z \
iff & frac{9xy}{x+y} = 3x+3y-z \
implies & frac{9}{frac 1x + frac 1y} = 3x+3y-z
end{align}
From the LHS we see that $frac 1x + frac 1y$ must be equal to $pm 1, pm 3, pm 9$ since the RHS is an integer. But of course, since $x,y$ are integers, $frac 1x + frac 1y in [-2,2]$ and it follows that $frac 1x + frac 1y = pm 1$.
Moreover, the only way this can happen is if $x = y = pm 2$.
Hence, the only solutions are $(x,y,z) = (2,2,3)$ and $(x,y,z) = (-2,-2,-3)$.
EDIT:
As pointed out in the comments, I have made the mistake of dividing by $0$, so I have changed the last $iff$ to an $implies$.
If $x=0$, then $z=3y$ so we get the solutions $(x,y,z) = (0,t,3t)$
Similarly when $y=0$ we get the solutions $(x,y,z) = (t,0,3t)$ for any $t in Bbb Z$ with $t neq 0$.
answered Dec 8 '18 at 10:16
glowstonetreesglowstonetrees
2,368418
$endgroup$
Question 1:
begin{align}
& frac{x+y}{x^2-xy+y^2} = frac 3z \
iff & frac{3(x^2-xy+y^2)}{x+y} = z\
iff & 3(x+y) - frac{9xy}{x+y} = z \
iff & frac{9xy}{x+y} = 3x+3y-z \
implies & frac{9}{frac 1x + frac 1y} = 3x+3y-z
end{align}
From the LHS we see that $frac 1x + frac 1y$ must be equal to $pm 1, pm 3, pm 9$ since the RHS is an integer. But of course, since $x,y$ are integers, $frac 1x + frac 1y in [-2,2]$ and it follows that $frac 1x + frac 1y = pm 1$.
Moreover, the only way this can happen is if $x = y = pm 2$.
Hence, the only solutions are $(x,y,z) = (2,2,3)$ and $(x,y,z) = (-2,-2,-3)$.
EDIT:
As pointed out in the comments, I have made the mistake of dividing by $0$, so I have changed the last $iff$ to an $implies$.
If $x=0$, then $z=3y$ so we get the solutions $(x,y,z) = (0,t,3t)$
Similarly when $y=0$ we get the solutions $(x,y,z) = (t,0,3t)$ for any $t in Bbb Z$ with $t neq 0$.
answered Dec 8 '18 at 10:16
glowstonetreesglowstonetrees
2,368418
edited Dec 8 '18 at 10:37
answered Dec 8 '18 at 10:16
glowstonetreesglowstonetrees
2,368418
answered Dec 8 '18 at 10:16
glowstonetreesglowstonetrees
2,368418
answered Dec 8 '18 at 10:16
glowstonetreesglowstonetrees
2,368418
2,368418
$begingroup$
I got it. Thank you very much.
$endgroup$
– Cglkttca
Dec 8 '18 at 10:20
1
$begingroup$
It's a nice answer, however, you forgot the cases when x and y are 0.
$endgroup$
– Ankit Kumar
Dec 8 '18 at 10:29
$begingroup$
Yes, in this case, we get more solutions
$endgroup$
– Cglkttca
Dec 8 '18 at 10:34
add a comment |
$begingroup$
I got it. Thank you very much.
$endgroup$
– Cglkttca
Dec 8 '18 at 10:20
1
$begingroup$
It's a nice answer, however, you forgot the cases when x and y are 0.
$endgroup$
– Ankit Kumar
Dec 8 '18 at 10:29
$begingroup$
Yes, in this case, we get more solutions
$endgroup$
– Cglkttca
Dec 8 '18 at 10:34
$begingroup$
I got it. Thank you very much.
$endgroup$
– Cglkttca
Dec 8 '18 at 10:20
$begingroup$
I got it. Thank you very much.
$endgroup$
– Cglkttca
Dec 8 '18 at 10:20
1
1
$begingroup$
It's a nice answer, however, you forgot the cases when x and y are 0.
$endgroup$
– Ankit Kumar
Dec 8 '18 at 10:29
$begingroup$
It's a nice answer, however, you forgot the cases when x and y are 0.
$endgroup$
– Ankit Kumar
Dec 8 '18 at 10:29
$begingroup$
Yes, in this case, we get more solutions
$endgroup$
– Cglkttca
Dec 8 '18 at 10:34
$begingroup$
Yes, in this case, we get more solutions
$endgroup$
– Cglkttca
Dec 8 '18 at 10:34
add a comment |
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$begingroup$
These problems seem very interesting. Where did you get them from?
$endgroup$
– Toby Mak
Dec 8 '18 at 9:39
$begingroup$
@TobyMak These are my homework.
$endgroup$
– Cglkttca
Dec 8 '18 at 9:42
$begingroup$
What class are you taking?
$endgroup$
– Toby Mak
Dec 8 '18 at 9:42
$begingroup$
@TobyMak I'm in grade 9.
$endgroup$
– Cglkttca
Dec 8 '18 at 9:43
1
$begingroup$
What lesson is this that you were given that exercise ?
$endgroup$
– Rebellos
Dec 8 '18 at 10:03