Integer equations












2












$begingroup$


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







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$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2


0



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 15:22









mrtaurho

5,21141237




5,21141237










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















4












$begingroup$

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 $t neq 0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

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4












$begingroup$

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 $t neq 0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















4












$begingroup$

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 $t neq 0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














4












4








4





$begingroup$

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 $t neq 0$.






share|cite|improve this answer











$endgroup$



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 $t neq 0$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 10:37

























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


















  • $begingroup$
    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


















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