Integer equations












2














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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  • These problems seem very interesting. Where did you get them from?
    – Toby Mak
    Dec 8 at 9:39










  • @TobyMak These are my homework.
    – Cglkttca
    Dec 8 at 9:42










  • What class are you taking?
    – Toby Mak
    Dec 8 at 9:42










  • @TobyMak I'm in grade 9.
    – Cglkttca
    Dec 8 at 9:43






  • 1




    What lesson is this that you were given that exercise ?
    – Rebellos
    Dec 8 at 10:03
















2














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
























  • These problems seem very interesting. Where did you get them from?
    – Toby Mak
    Dec 8 at 9:39










  • @TobyMak These are my homework.
    – Cglkttca
    Dec 8 at 9:42










  • What class are you taking?
    – Toby Mak
    Dec 8 at 9:42










  • @TobyMak I'm in grade 9.
    – Cglkttca
    Dec 8 at 9:43






  • 1




    What lesson is this that you were given that exercise ?
    – Rebellos
    Dec 8 at 10:03














2












2








2


1





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















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 at 15:22









mrtaurho

3,3982932




3,3982932










asked Dec 8 at 9:38









Cglkttca

826




826












  • These problems seem very interesting. Where did you get them from?
    – Toby Mak
    Dec 8 at 9:39










  • @TobyMak These are my homework.
    – Cglkttca
    Dec 8 at 9:42










  • What class are you taking?
    – Toby Mak
    Dec 8 at 9:42










  • @TobyMak I'm in grade 9.
    – Cglkttca
    Dec 8 at 9:43






  • 1




    What lesson is this that you were given that exercise ?
    – Rebellos
    Dec 8 at 10:03


















  • These problems seem very interesting. Where did you get them from?
    – Toby Mak
    Dec 8 at 9:39










  • @TobyMak These are my homework.
    – Cglkttca
    Dec 8 at 9:42










  • What class are you taking?
    – Toby Mak
    Dec 8 at 9:42










  • @TobyMak I'm in grade 9.
    – Cglkttca
    Dec 8 at 9:43






  • 1




    What lesson is this that you were given that exercise ?
    – Rebellos
    Dec 8 at 10:03
















These problems seem very interesting. Where did you get them from?
– Toby Mak
Dec 8 at 9:39




These problems seem very interesting. Where did you get them from?
– Toby Mak
Dec 8 at 9:39












@TobyMak These are my homework.
– Cglkttca
Dec 8 at 9:42




@TobyMak These are my homework.
– Cglkttca
Dec 8 at 9:42












What class are you taking?
– Toby Mak
Dec 8 at 9:42




What class are you taking?
– Toby Mak
Dec 8 at 9:42












@TobyMak I'm in grade 9.
– Cglkttca
Dec 8 at 9:43




@TobyMak I'm in grade 9.
– Cglkttca
Dec 8 at 9:43




1




1




What lesson is this that you were given that exercise ?
– Rebellos
Dec 8 at 10:03




What lesson is this that you were given that exercise ?
– Rebellos
Dec 8 at 10:03










1 Answer
1






active

oldest

votes


















4














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























  • I got it. Thank you very much.
    – Cglkttca
    Dec 8 at 10:20






  • 1




    It's a nice answer, however, you forgot the cases when x and y are 0.
    – Ankit Kumar
    Dec 8 at 10:29










  • Yes, in this case, we get more solutions
    – Cglkttca
    Dec 8 at 10:34











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1 Answer
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1 Answer
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active

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active

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4














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























  • I got it. Thank you very much.
    – Cglkttca
    Dec 8 at 10:20






  • 1




    It's a nice answer, however, you forgot the cases when x and y are 0.
    – Ankit Kumar
    Dec 8 at 10:29










  • Yes, in this case, we get more solutions
    – Cglkttca
    Dec 8 at 10:34
















4














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























  • I got it. Thank you very much.
    – Cglkttca
    Dec 8 at 10:20






  • 1




    It's a nice answer, however, you forgot the cases when x and y are 0.
    – Ankit Kumar
    Dec 8 at 10:29










  • Yes, in this case, we get more solutions
    – Cglkttca
    Dec 8 at 10:34














4












4








4






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














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 at 10:37

























answered Dec 8 at 10:16









glowstonetrees

2,285317




2,285317












  • I got it. Thank you very much.
    – Cglkttca
    Dec 8 at 10:20






  • 1




    It's a nice answer, however, you forgot the cases when x and y are 0.
    – Ankit Kumar
    Dec 8 at 10:29










  • Yes, in this case, we get more solutions
    – Cglkttca
    Dec 8 at 10:34


















  • I got it. Thank you very much.
    – Cglkttca
    Dec 8 at 10:20






  • 1




    It's a nice answer, however, you forgot the cases when x and y are 0.
    – Ankit Kumar
    Dec 8 at 10:29










  • Yes, in this case, we get more solutions
    – Cglkttca
    Dec 8 at 10:34
















I got it. Thank you very much.
– Cglkttca
Dec 8 at 10:20




I got it. Thank you very much.
– Cglkttca
Dec 8 at 10:20




1




1




It's a nice answer, however, you forgot the cases when x and y are 0.
– Ankit Kumar
Dec 8 at 10:29




It's a nice answer, however, you forgot the cases when x and y are 0.
– Ankit Kumar
Dec 8 at 10:29












Yes, in this case, we get more solutions
– Cglkttca
Dec 8 at 10:34




Yes, in this case, we get more solutions
– Cglkttca
Dec 8 at 10:34


















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