Mixing
Find all integer solutions to $ x^3 - y^3 = 3(x^2 - y^2) $
$begingroup$
The objective is to find all solutions to
$$ x^3 - y^3 = 3(x^2 - y^2) $$
where $x,y in mathbb{Z}$.
So far I've got one pair of solution. Try $(x, y)=(0,0)$:
$$ 0^3-0^3=3(0^2-0^2) \
0=0 qquad text{equation satisfied}$$
Another try $ (x, y) = (x, x)$ then
$$ x^3 - x^3 = 3(x^2 -x^2) \
0 = 0 qquad text{equation satisfied}$$
But the above are just particular solutions. To find all $(x, y)$ pairs, I tried:
$$begin{aligned}
(x-y)(x^2+xy+y^2) &= 3(x+y)(x-y) \
x^2+xy+y^2 &= 3(x+y) end{aligned}\
begin{aligned}
x^2+xy+y^2-3x-3y &= 0 \
x^2+x(y-3)+y(y-3) &=0 \
x^2 + (y-3)(x+y) &= 0 end{aligned}$$
What now? Am I on the right track?
algebra-precalculus polynomials
edited Jan 29 at 18:12
J. W. Tanner
3,8971320
asked Jan 28 at 1:40
Bruce_MoustacheBruce_Moustache
83
$endgroup$
add a comment |
$begingroup$
The objective is to find all solutions to
$$ x^3 - y^3 = 3(x^2 - y^2) $$
where $x,y in mathbb{Z}$.
So far I've got one pair of solution. Try $(x, y)=(0,0)$:
$$ 0^3-0^3=3(0^2-0^2) \
0=0 qquad text{equation satisfied}$$
Another try $ (x, y) = (x, x)$ then
$$ x^3 - x^3 = 3(x^2 -x^2) \
0 = 0 qquad text{equation satisfied}$$
But the above are just particular solutions. To find all $(x, y)$ pairs, I tried:
$$begin{aligned}
(x-y)(x^2+xy+y^2) &= 3(x+y)(x-y) \
x^2+xy+y^2 &= 3(x+y) end{aligned}\
begin{aligned}
x^2+xy+y^2-3x-3y &= 0 \
x^2+x(y-3)+y(y-3) &=0 \
x^2 + (y-3)(x+y) &= 0 end{aligned}$$
What now? Am I on the right track?
algebra-precalculus polynomials
edited Jan 29 at 18:12
J. W. Tanner
3,8971320
asked Jan 28 at 1:40
Bruce_MoustacheBruce_Moustache
83
$endgroup$
$begingroup$
Would be baskara a good way?
$endgroup$
– Bruce_Moustache
Jan 28 at 2:02
add a comment |
$begingroup$
The objective is to find all solutions to
$$ x^3 - y^3 = 3(x^2 - y^2) $$
where $x,y in mathbb{Z}$.
So far I've got one pair of solution. Try $(x, y)=(0,0)$:
$$ 0^3-0^3=3(0^2-0^2) \
0=0 qquad text{equation satisfied}$$
Another try $ (x, y) = (x, x)$ then
$$ x^3 - x^3 = 3(x^2 -x^2) \
0 = 0 qquad text{equation satisfied}$$
But the above are just particular solutions. To find all $(x, y)$ pairs, I tried:
$$begin{aligned}
(x-y)(x^2+xy+y^2) &= 3(x+y)(x-y) \
x^2+xy+y^2 &= 3(x+y) end{aligned}\
begin{aligned}
x^2+xy+y^2-3x-3y &= 0 \
x^2+x(y-3)+y(y-3) &=0 \
x^2 + (y-3)(x+y) &= 0 end{aligned}$$
What now? Am I on the right track?
algebra-precalculus polynomials
edited Jan 29 at 18:12
J. W. Tanner
3,8971320
asked Jan 28 at 1:40
Bruce_MoustacheBruce_Moustache
83
$endgroup$
The objective is to find all solutions to
$$ x^3 - y^3 = 3(x^2 - y^2) $$
where $x,y in mathbb{Z}$.
So far I've got one pair of solution. Try $(x, y)=(0,0)$:
$$ 0^3-0^3=3(0^2-0^2) \
0=0 qquad text{equation satisfied}$$
Another try $ (x, y) = (x, x)$ then
$$ x^3 - x^3 = 3(x^2 -x^2) \
0 = 0 qquad text{equation satisfied}$$
But the above are just particular solutions. To find all $(x, y)$ pairs, I tried:
$$begin{aligned}
(x-y)(x^2+xy+y^2) &= 3(x+y)(x-y) \
x^2+xy+y^2 &= 3(x+y) end{aligned}\
begin{aligned}
x^2+xy+y^2-3x-3y &= 0 \
x^2+x(y-3)+y(y-3) &=0 \
x^2 + (y-3)(x+y) &= 0 end{aligned}$$
What now? Am I on the right track?
algebra-precalculus polynomials
algebra-precalculus polynomials
edited Jan 29 at 18:12
J. W. Tanner
3,8971320
asked Jan 28 at 1:40
Bruce_MoustacheBruce_Moustache
83
edited Jan 29 at 18:12
J. W. Tanner
3,8971320
asked Jan 28 at 1:40
Bruce_MoustacheBruce_Moustache
83
edited Jan 29 at 18:12
J. W. Tanner
3,8971320
edited Jan 29 at 18:12
J. W. Tanner
3,8971320
edited Jan 29 at 18:12
J. W. Tanner
3,8971320
3,8971320
asked Jan 28 at 1:40
Bruce_MoustacheBruce_Moustache
83
asked Jan 28 at 1:40
Bruce_MoustacheBruce_Moustache
83
asked Jan 28 at 1:40
Bruce_MoustacheBruce_Moustache
83
83
$begingroup$
Would be baskara a good way?
$endgroup$
– Bruce_Moustache
Jan 28 at 2:02
add a comment |
$begingroup$
Would be baskara a good way?
$endgroup$
– Bruce_Moustache
Jan 28 at 2:02
$begingroup$
Would be baskara a good way?
$endgroup$
– Bruce_Moustache
Jan 28 at 2:02
$begingroup$
Would be baskara a good way?
$endgroup$
– Bruce_Moustache
Jan 28 at 2:02
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
By symmetry, we can say that whenever $(x,y)$ is a solution, $(y,x)$ is also a solution. Also, one of them must be positive integer, as $x+y=frac{x^2+xy+y^2}{3}>0$
Apply quadratic formula in $3$ specific cases, when $y=1,2,3.$ We get $(2,2), (-1,2), (0,3), (3,0), (2,-1)$ as the integer solutions in these cases. There is no other case, as otherwise $y>3$ or $x>3$ and Left Hand Side of the last equation you wrote will be strictly positive.
Hope it is helpful:)
answered Jan 28 at 2:21
MartundMartund
1,667213
$endgroup$
add a comment |
$begingroup$
$x=y$ and $x=3$ with $y=0$ (or $x=0$ with $y=3$) are trivial answers. To show there are no others for $xne y$, let $n=x+y$, then the expression can be written $x^2+xy+y^2=(x+y)^2 +(3-n)n$ or $x(n-x)+3n-n^2=0$. Solve the quadratic for $x$ to get $x=frac{n+sqrt{12n-3n^2}}{2}$. Since the discriminant has to be real, $nle 4$, and a perfect square to get an integer solution. The only possibilities are $n=1$ giving $(2,-1)$ (not allowed) and $n=3$ giving $(3,0)$.
answered Jan 28 at 2:49
herb steinbergherb steinberg
3,0982311
$endgroup$
add a comment |
$begingroup$
So first you factorize the expression:
$$(x-y)(x^2+xy+y^2-3x-3y)=0$$
$$(x-y)(x^2+x(y-3)+(y^2-3y))=0$$
When the product of two expressions are $0$ either of them is $0$
So first factor is $(x-y)=0$ giving the solution pair $x=y$ so for all such pairs the equation will be 0.
Second factor can be treated as a quadratic equation in terms of either $x$ or $y$,
$$x^2+x(y-3)+(y^2-3y)=0$$
Where you consider each $y$ as a constant and solve the equation to get:
$$x=frac{-(y-3)pmsqrt{(y-3)^2-4(y^2-3y)}}{2}$$
$$x=frac{-(y-3)pmsqrt{3(9-y^2)}}{2}$$
For $x$ to be real the quantity inside the square root should be positive, moreover you need the integer solutions for $x$ and $y$ so the only possibility that remains is if the discriminant becomes a perfect square, So $y=0,pm1,pm2,pm3$ but you also want it to be a perfect square so only $y=pm3$ which gives:
$$x = 0,3 $$
Final solutions:
All integers for which $x=y$,
$(0,3)$,
$(3,-3)$
You can also solve the equation for $y$, hope this helps .....
answered Jan 28 at 6:27
SNEHIL SANYALSNEHIL SANYAL
654110
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By symmetry, we can say that whenever $(x,y)$ is a solution, $(y,x)$ is also a solution. Also, one of them must be positive integer, as $x+y=frac{x^2+xy+y^2}{3}>0$
Apply quadratic formula in $3$ specific cases, when $y=1,2,3.$ We get $(2,2), (-1,2), (0,3), (3,0), (2,-1)$ as the integer solutions in these cases. There is no other case, as otherwise $y>3$ or $x>3$ and Left Hand Side of the last equation you wrote will be strictly positive.
Hope it is helpful:)
answered Jan 28 at 2:21
MartundMartund
1,667213
$endgroup$
add a comment |
$begingroup$
By symmetry, we can say that whenever $(x,y)$ is a solution, $(y,x)$ is also a solution. Also, one of them must be positive integer, as $x+y=frac{x^2+xy+y^2}{3}>0$
Apply quadratic formula in $3$ specific cases, when $y=1,2,3.$ We get $(2,2), (-1,2), (0,3), (3,0), (2,-1)$ as the integer solutions in these cases. There is no other case, as otherwise $y>3$ or $x>3$ and Left Hand Side of the last equation you wrote will be strictly positive.
Hope it is helpful:)
answered Jan 28 at 2:21
MartundMartund
1,667213
$endgroup$
add a comment |
$begingroup$
By symmetry, we can say that whenever $(x,y)$ is a solution, $(y,x)$ is also a solution. Also, one of them must be positive integer, as $x+y=frac{x^2+xy+y^2}{3}>0$
Apply quadratic formula in $3$ specific cases, when $y=1,2,3.$ We get $(2,2), (-1,2), (0,3), (3,0), (2,-1)$ as the integer solutions in these cases. There is no other case, as otherwise $y>3$ or $x>3$ and Left Hand Side of the last equation you wrote will be strictly positive.
Hope it is helpful:)
answered Jan 28 at 2:21
MartundMartund
1,667213
$endgroup$
By symmetry, we can say that whenever $(x,y)$ is a solution, $(y,x)$ is also a solution. Also, one of them must be positive integer, as $x+y=frac{x^2+xy+y^2}{3}>0$
Apply quadratic formula in $3$ specific cases, when $y=1,2,3.$ We get $(2,2), (-1,2), (0,3), (3,0), (2,-1)$ as the integer solutions in these cases. There is no other case, as otherwise $y>3$ or $x>3$ and Left Hand Side of the last equation you wrote will be strictly positive.
Hope it is helpful:)
answered Jan 28 at 2:21
MartundMartund
1,667213
answered Jan 28 at 2:21
MartundMartund
1,667213
answered Jan 28 at 2:21
MartundMartund
1,667213
answered Jan 28 at 2:21
MartundMartund
1,667213
1,667213
add a comment |
add a comment |
$begingroup$
$x=y$ and $x=3$ with $y=0$ (or $x=0$ with $y=3$) are trivial answers. To show there are no others for $xne y$, let $n=x+y$, then the expression can be written $x^2+xy+y^2=(x+y)^2 +(3-n)n$ or $x(n-x)+3n-n^2=0$. Solve the quadratic for $x$ to get $x=frac{n+sqrt{12n-3n^2}}{2}$. Since the discriminant has to be real, $nle 4$, and a perfect square to get an integer solution. The only possibilities are $n=1$ giving $(2,-1)$ (not allowed) and $n=3$ giving $(3,0)$.
answered Jan 28 at 2:49
herb steinbergherb steinberg
3,0982311
$endgroup$
add a comment |
$begingroup$
$x=y$ and $x=3$ with $y=0$ (or $x=0$ with $y=3$) are trivial answers. To show there are no others for $xne y$, let $n=x+y$, then the expression can be written $x^2+xy+y^2=(x+y)^2 +(3-n)n$ or $x(n-x)+3n-n^2=0$. Solve the quadratic for $x$ to get $x=frac{n+sqrt{12n-3n^2}}{2}$. Since the discriminant has to be real, $nle 4$, and a perfect square to get an integer solution. The only possibilities are $n=1$ giving $(2,-1)$ (not allowed) and $n=3$ giving $(3,0)$.
answered Jan 28 at 2:49
herb steinbergherb steinberg
3,0982311
$endgroup$
add a comment |
$begingroup$
$x=y$ and $x=3$ with $y=0$ (or $x=0$ with $y=3$) are trivial answers. To show there are no others for $xne y$, let $n=x+y$, then the expression can be written $x^2+xy+y^2=(x+y)^2 +(3-n)n$ or $x(n-x)+3n-n^2=0$. Solve the quadratic for $x$ to get $x=frac{n+sqrt{12n-3n^2}}{2}$. Since the discriminant has to be real, $nle 4$, and a perfect square to get an integer solution. The only possibilities are $n=1$ giving $(2,-1)$ (not allowed) and $n=3$ giving $(3,0)$.
answered Jan 28 at 2:49
herb steinbergherb steinberg
3,0982311
$endgroup$
$x=y$ and $x=3$ with $y=0$ (or $x=0$ with $y=3$) are trivial answers. To show there are no others for $xne y$, let $n=x+y$, then the expression can be written $x^2+xy+y^2=(x+y)^2 +(3-n)n$ or $x(n-x)+3n-n^2=0$. Solve the quadratic for $x$ to get $x=frac{n+sqrt{12n-3n^2}}{2}$. Since the discriminant has to be real, $nle 4$, and a perfect square to get an integer solution. The only possibilities are $n=1$ giving $(2,-1)$ (not allowed) and $n=3$ giving $(3,0)$.
answered Jan 28 at 2:49
herb steinbergherb steinberg
3,0982311
answered Jan 28 at 2:49
herb steinbergherb steinberg
3,0982311
answered Jan 28 at 2:49
herb steinbergherb steinberg
3,0982311
answered Jan 28 at 2:49
herb steinbergherb steinberg
3,0982311
3,0982311
add a comment |
add a comment |
$begingroup$
So first you factorize the expression:
$$(x-y)(x^2+xy+y^2-3x-3y)=0$$
$$(x-y)(x^2+x(y-3)+(y^2-3y))=0$$
When the product of two expressions are $0$ either of them is $0$
So first factor is $(x-y)=0$ giving the solution pair $x=y$ so for all such pairs the equation will be 0.
Second factor can be treated as a quadratic equation in terms of either $x$ or $y$,
$$x^2+x(y-3)+(y^2-3y)=0$$
Where you consider each $y$ as a constant and solve the equation to get:
$$x=frac{-(y-3)pmsqrt{(y-3)^2-4(y^2-3y)}}{2}$$
$$x=frac{-(y-3)pmsqrt{3(9-y^2)}}{2}$$
For $x$ to be real the quantity inside the square root should be positive, moreover you need the integer solutions for $x$ and $y$ so the only possibility that remains is if the discriminant becomes a perfect square, So $y=0,pm1,pm2,pm3$ but you also want it to be a perfect square so only $y=pm3$ which gives:
$$x = 0,3 $$
Final solutions:
All integers for which $x=y$,
$(0,3)$,
$(3,-3)$
You can also solve the equation for $y$, hope this helps .....
answered Jan 28 at 6:27
SNEHIL SANYALSNEHIL SANYAL
654110
$endgroup$
add a comment |
$begingroup$
So first you factorize the expression:
$$(x-y)(x^2+xy+y^2-3x-3y)=0$$
$$(x-y)(x^2+x(y-3)+(y^2-3y))=0$$
When the product of two expressions are $0$ either of them is $0$
So first factor is $(x-y)=0$ giving the solution pair $x=y$ so for all such pairs the equation will be 0.
Second factor can be treated as a quadratic equation in terms of either $x$ or $y$,
$$x^2+x(y-3)+(y^2-3y)=0$$
Where you consider each $y$ as a constant and solve the equation to get:
$$x=frac{-(y-3)pmsqrt{(y-3)^2-4(y^2-3y)}}{2}$$
$$x=frac{-(y-3)pmsqrt{3(9-y^2)}}{2}$$
For $x$ to be real the quantity inside the square root should be positive, moreover you need the integer solutions for $x$ and $y$ so the only possibility that remains is if the discriminant becomes a perfect square, So $y=0,pm1,pm2,pm3$ but you also want it to be a perfect square so only $y=pm3$ which gives:
$$x = 0,3 $$
Final solutions:
All integers for which $x=y$,
$(0,3)$,
$(3,-3)$
You can also solve the equation for $y$, hope this helps .....
answered Jan 28 at 6:27
SNEHIL SANYALSNEHIL SANYAL
654110
$endgroup$
add a comment |
$begingroup$
So first you factorize the expression:
$$(x-y)(x^2+xy+y^2-3x-3y)=0$$
$$(x-y)(x^2+x(y-3)+(y^2-3y))=0$$
When the product of two expressions are $0$ either of them is $0$
So first factor is $(x-y)=0$ giving the solution pair $x=y$ so for all such pairs the equation will be 0.
Second factor can be treated as a quadratic equation in terms of either $x$ or $y$,
$$x^2+x(y-3)+(y^2-3y)=0$$
Where you consider each $y$ as a constant and solve the equation to get:
$$x=frac{-(y-3)pmsqrt{(y-3)^2-4(y^2-3y)}}{2}$$
$$x=frac{-(y-3)pmsqrt{3(9-y^2)}}{2}$$
For $x$ to be real the quantity inside the square root should be positive, moreover you need the integer solutions for $x$ and $y$ so the only possibility that remains is if the discriminant becomes a perfect square, So $y=0,pm1,pm2,pm3$ but you also want it to be a perfect square so only $y=pm3$ which gives:
$$x = 0,3 $$
Final solutions:
All integers for which $x=y$,
$(0,3)$,
$(3,-3)$
You can also solve the equation for $y$, hope this helps .....
answered Jan 28 at 6:27
SNEHIL SANYALSNEHIL SANYAL
654110
$endgroup$
So first you factorize the expression:
$$(x-y)(x^2+xy+y^2-3x-3y)=0$$
$$(x-y)(x^2+x(y-3)+(y^2-3y))=0$$
When the product of two expressions are $0$ either of them is $0$
So first factor is $(x-y)=0$ giving the solution pair $x=y$ so for all such pairs the equation will be 0.
Second factor can be treated as a quadratic equation in terms of either $x$ or $y$,
$$x^2+x(y-3)+(y^2-3y)=0$$
Where you consider each $y$ as a constant and solve the equation to get:
$$x=frac{-(y-3)pmsqrt{(y-3)^2-4(y^2-3y)}}{2}$$
$$x=frac{-(y-3)pmsqrt{3(9-y^2)}}{2}$$
For $x$ to be real the quantity inside the square root should be positive, moreover you need the integer solutions for $x$ and $y$ so the only possibility that remains is if the discriminant becomes a perfect square, So $y=0,pm1,pm2,pm3$ but you also want it to be a perfect square so only $y=pm3$ which gives:
$$x = 0,3 $$
Final solutions:
All integers for which $x=y$,
$(0,3)$,
$(3,-3)$
You can also solve the equation for $y$, hope this helps .....
answered Jan 28 at 6:27
SNEHIL SANYALSNEHIL SANYAL
654110
answered Jan 28 at 6:27
SNEHIL SANYALSNEHIL SANYAL
654110
answered Jan 28 at 6:27
SNEHIL SANYALSNEHIL SANYAL
654110
answered Jan 28 at 6:27
SNEHIL SANYALSNEHIL SANYAL
654110
654110
add a comment |
add a comment |
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$begingroup$
Would be baskara a good way?
$endgroup$
– Bruce_Moustache
Jan 28 at 2:02