Mixing
Solve $x+y+z = x^3 + y^3 + z^3 = 8$ in $mathbb{Z}$
$begingroup$
Solve $x+y+z = x^3 + y^3 + z^3 = 8$ in $mathbb{Z}$
First I tried to transform this equation, substituting $x = 8-y-z$. So I end up with:
$$x^3 + y^3 + z^3 = 8$$
$$(8-y-z)^3 + y^3 + z^3 = 8$$
Using Wolfram Alpha I expanded this equation and tried to factorize it so finally I got:
$$(z-8)(y^2 + y(z-8) - 8z) = 168$$
$z in mathbb{Z}$, which implies $z-8$ is an integer implying that the second term is also an integer.
168 has 16 positive divisors and 16 negative, which means there are 32 distinct cases to be looked, which is a painful work and made me abandon this method in search of a better and easier method.
Second method I tried is Newton Identities.
We want to find the values for the roots $x,y,z$ of the cubic function: $f(x) = x^3 + bx^2 + cx + d$
From the condition we have the following equation:
$$x + y + z = 8 = s_1text{ and } x^3 + y^3 + z^3 = 8 = s_3$$
Using Newton Identities we obtain the following relations:
$$s_1 + b = 0 implies b = -8$$
$$s_2 + bs_1 + 2c = 0$$
$$s_2 - 64 + 2c = 0 implies c = frac{64 - s_2}{2}$$
$$s_3 + bs_2 + cs_1 + 3d = 0$$
$$8 - 8s_2 + frac{64 - s_2}{2} times 8 + 3d = 0$$
$$8 - 8s_2 + 256 - 4s_2 + 3d = 0$$
$$3d - 12s_2 + 264 = 0$$
$$d - 4s_2 + 88 = 0 implies s_2 = frac{88+d}{4}$$
Substituting back we have:
$$c = frac{64 - s_2}{2} = frac{64 - frac{88+d}{4}}{2} = frac{frac{256 - 88 - d}{4}}{2} = frac{168 - d}{8}$$
And I'm stuck here, because I know only the ratio between $c$ and $d$, which is a consequence of not knowing the value $s_2 = x^2 + y^2 + z^2$
number-theory functions roots diophantine-equations factoring
edited Jan 8 at 8:06
Martin Sleziak
44.7k9117272
asked Aug 31 '13 at 14:30
Stefan4024Stefan4024
30.3k63478
$endgroup$
|
show 2 more comments
$begingroup$
Solve $x+y+z = x^3 + y^3 + z^3 = 8$ in $mathbb{Z}$
First I tried to transform this equation, substituting $x = 8-y-z$. So I end up with:
$$x^3 + y^3 + z^3 = 8$$
$$(8-y-z)^3 + y^3 + z^3 = 8$$
Using Wolfram Alpha I expanded this equation and tried to factorize it so finally I got:
$$(z-8)(y^2 + y(z-8) - 8z) = 168$$
$z in mathbb{Z}$, which implies $z-8$ is an integer implying that the second term is also an integer.
168 has 16 positive divisors and 16 negative, which means there are 32 distinct cases to be looked, which is a painful work and made me abandon this method in search of a better and easier method.
Second method I tried is Newton Identities.
We want to find the values for the roots $x,y,z$ of the cubic function: $f(x) = x^3 + bx^2 + cx + d$
From the condition we have the following equation:
$$x + y + z = 8 = s_1text{ and } x^3 + y^3 + z^3 = 8 = s_3$$
Using Newton Identities we obtain the following relations:
$$s_1 + b = 0 implies b = -8$$
$$s_2 + bs_1 + 2c = 0$$
$$s_2 - 64 + 2c = 0 implies c = frac{64 - s_2}{2}$$
$$s_3 + bs_2 + cs_1 + 3d = 0$$
$$8 - 8s_2 + frac{64 - s_2}{2} times 8 + 3d = 0$$
$$8 - 8s_2 + 256 - 4s_2 + 3d = 0$$
$$3d - 12s_2 + 264 = 0$$
$$d - 4s_2 + 88 = 0 implies s_2 = frac{88+d}{4}$$
Substituting back we have:
$$c = frac{64 - s_2}{2} = frac{64 - frac{88+d}{4}}{2} = frac{frac{256 - 88 - d}{4}}{2} = frac{168 - d}{8}$$
And I'm stuck here, because I know only the ratio between $c$ and $d$, which is a consequence of not knowing the value $s_2 = x^2 + y^2 + z^2$
number-theory functions roots diophantine-equations factoring
edited Jan 8 at 8:06
Martin Sleziak
44.7k9117272
asked Aug 31 '13 at 14:30
Stefan4024Stefan4024
30.3k63478
$endgroup$
1
$begingroup$
Why abandon the first path for only 32 cases? The discriminant of the other factor to be a square $(z-8)^2+32z=(z+8)^2$. So you get for each value for $z$ solutions for $y=(2z-16pm(z+8))/2$. These are integers only for even $z$. That cuts down the cases a bit.
$endgroup$
– OR.
Aug 31 '13 at 14:52
1
$begingroup$
What about $x=-16, y=9, z=15$?
$endgroup$
– Axel Kemper
Aug 31 '13 at 15:05
$begingroup$
@ABC Let's say this is a contest problem, would you spend 30 minutes to check every combination? Note that the there are 64 values for y, because of the $pm$ sign in front of the square root.
$endgroup$
– Stefan4024
Aug 31 '13 at 15:09
$begingroup$
@AxelKemper According to Wolfram Alpha (9,15,-16) and its six permutation are the only integer solution, but any idea how to get those values?
$endgroup$
– Stefan4024
Aug 31 '13 at 15:09
$begingroup$
I have used Microsoft Solver Foundation to get this solution. It took less than a second.
$endgroup$
– Axel Kemper
Aug 31 '13 at 15:11
|
show 2 more comments
$begingroup$
Solve $x+y+z = x^3 + y^3 + z^3 = 8$ in $mathbb{Z}$
First I tried to transform this equation, substituting $x = 8-y-z$. So I end up with:
$$x^3 + y^3 + z^3 = 8$$
$$(8-y-z)^3 + y^3 + z^3 = 8$$
Using Wolfram Alpha I expanded this equation and tried to factorize it so finally I got:
$$(z-8)(y^2 + y(z-8) - 8z) = 168$$
$z in mathbb{Z}$, which implies $z-8$ is an integer implying that the second term is also an integer.
168 has 16 positive divisors and 16 negative, which means there are 32 distinct cases to be looked, which is a painful work and made me abandon this method in search of a better and easier method.
Second method I tried is Newton Identities.
We want to find the values for the roots $x,y,z$ of the cubic function: $f(x) = x^3 + bx^2 + cx + d$
From the condition we have the following equation:
$$x + y + z = 8 = s_1text{ and } x^3 + y^3 + z^3 = 8 = s_3$$
Using Newton Identities we obtain the following relations:
$$s_1 + b = 0 implies b = -8$$
$$s_2 + bs_1 + 2c = 0$$
$$s_2 - 64 + 2c = 0 implies c = frac{64 - s_2}{2}$$
$$s_3 + bs_2 + cs_1 + 3d = 0$$
$$8 - 8s_2 + frac{64 - s_2}{2} times 8 + 3d = 0$$
$$8 - 8s_2 + 256 - 4s_2 + 3d = 0$$
$$3d - 12s_2 + 264 = 0$$
$$d - 4s_2 + 88 = 0 implies s_2 = frac{88+d}{4}$$
Substituting back we have:
$$c = frac{64 - s_2}{2} = frac{64 - frac{88+d}{4}}{2} = frac{frac{256 - 88 - d}{4}}{2} = frac{168 - d}{8}$$
And I'm stuck here, because I know only the ratio between $c$ and $d$, which is a consequence of not knowing the value $s_2 = x^2 + y^2 + z^2$
number-theory functions roots diophantine-equations factoring
edited Jan 8 at 8:06
Martin Sleziak
44.7k9117272
asked Aug 31 '13 at 14:30
Stefan4024Stefan4024
30.3k63478
$endgroup$
Solve $x+y+z = x^3 + y^3 + z^3 = 8$ in $mathbb{Z}$
First I tried to transform this equation, substituting $x = 8-y-z$. So I end up with:
$$x^3 + y^3 + z^3 = 8$$
$$(8-y-z)^3 + y^3 + z^3 = 8$$
Using Wolfram Alpha I expanded this equation and tried to factorize it so finally I got:
$$(z-8)(y^2 + y(z-8) - 8z) = 168$$
$z in mathbb{Z}$, which implies $z-8$ is an integer implying that the second term is also an integer.
168 has 16 positive divisors and 16 negative, which means there are 32 distinct cases to be looked, which is a painful work and made me abandon this method in search of a better and easier method.
Second method I tried is Newton Identities.
We want to find the values for the roots $x,y,z$ of the cubic function: $f(x) = x^3 + bx^2 + cx + d$
From the condition we have the following equation:
$$x + y + z = 8 = s_1text{ and } x^3 + y^3 + z^3 = 8 = s_3$$
Using Newton Identities we obtain the following relations:
$$s_1 + b = 0 implies b = -8$$
$$s_2 + bs_1 + 2c = 0$$
$$s_2 - 64 + 2c = 0 implies c = frac{64 - s_2}{2}$$
$$s_3 + bs_2 + cs_1 + 3d = 0$$
$$8 - 8s_2 + frac{64 - s_2}{2} times 8 + 3d = 0$$
$$8 - 8s_2 + 256 - 4s_2 + 3d = 0$$
$$3d - 12s_2 + 264 = 0$$
$$d - 4s_2 + 88 = 0 implies s_2 = frac{88+d}{4}$$
Substituting back we have:
$$c = frac{64 - s_2}{2} = frac{64 - frac{88+d}{4}}{2} = frac{frac{256 - 88 - d}{4}}{2} = frac{168 - d}{8}$$
And I'm stuck here, because I know only the ratio between $c$ and $d$, which is a consequence of not knowing the value $s_2 = x^2 + y^2 + z^2$
number-theory functions roots diophantine-equations factoring
number-theory functions roots diophantine-equations factoring
edited Jan 8 at 8:06
Martin Sleziak
44.7k9117272
asked Aug 31 '13 at 14:30
Stefan4024Stefan4024
30.3k63478
edited Jan 8 at 8:06
Martin Sleziak
44.7k9117272
asked Aug 31 '13 at 14:30
Stefan4024Stefan4024
30.3k63478
edited Jan 8 at 8:06
Martin Sleziak
44.7k9117272
edited Jan 8 at 8:06
Martin Sleziak
44.7k9117272
edited Jan 8 at 8:06
Martin Sleziak
44.7k9117272
44.7k9117272
asked Aug 31 '13 at 14:30
Stefan4024Stefan4024
30.3k63478
asked Aug 31 '13 at 14:30
Stefan4024Stefan4024
30.3k63478
asked Aug 31 '13 at 14:30
Stefan4024Stefan4024
30.3k63478
30.3k63478
1
$begingroup$
Why abandon the first path for only 32 cases? The discriminant of the other factor to be a square $(z-8)^2+32z=(z+8)^2$. So you get for each value for $z$ solutions for $y=(2z-16pm(z+8))/2$. These are integers only for even $z$. That cuts down the cases a bit.
$endgroup$
– OR.
Aug 31 '13 at 14:52
1
$begingroup$
What about $x=-16, y=9, z=15$?
$endgroup$
– Axel Kemper
Aug 31 '13 at 15:05
$begingroup$
@ABC Let's say this is a contest problem, would you spend 30 minutes to check every combination? Note that the there are 64 values for y, because of the $pm$ sign in front of the square root.
$endgroup$
– Stefan4024
Aug 31 '13 at 15:09
$begingroup$
@AxelKemper According to Wolfram Alpha (9,15,-16) and its six permutation are the only integer solution, but any idea how to get those values?
$endgroup$
– Stefan4024
Aug 31 '13 at 15:09
$begingroup$
I have used Microsoft Solver Foundation to get this solution. It took less than a second.
$endgroup$
– Axel Kemper
Aug 31 '13 at 15:11
|
show 2 more comments
1
$begingroup$
Why abandon the first path for only 32 cases? The discriminant of the other factor to be a square $(z-8)^2+32z=(z+8)^2$. So you get for each value for $z$ solutions for $y=(2z-16pm(z+8))/2$. These are integers only for even $z$. That cuts down the cases a bit.
$endgroup$
– OR.
Aug 31 '13 at 14:52
1
$begingroup$
What about $x=-16, y=9, z=15$?
$endgroup$
– Axel Kemper
Aug 31 '13 at 15:05
$begingroup$
@ABC Let's say this is a contest problem, would you spend 30 minutes to check every combination? Note that the there are 64 values for y, because of the $pm$ sign in front of the square root.
$endgroup$
– Stefan4024
Aug 31 '13 at 15:09
$begingroup$
@AxelKemper According to Wolfram Alpha (9,15,-16) and its six permutation are the only integer solution, but any idea how to get those values?
$endgroup$
– Stefan4024
Aug 31 '13 at 15:09
$begingroup$
I have used Microsoft Solver Foundation to get this solution. It took less than a second.
$endgroup$
– Axel Kemper
Aug 31 '13 at 15:11
1
1
$begingroup$
Why abandon the first path for only 32 cases? The discriminant of the other factor to be a square $(z-8)^2+32z=(z+8)^2$. So you get for each value for $z$ solutions for $y=(2z-16pm(z+8))/2$. These are integers only for even $z$. That cuts down the cases a bit.
$endgroup$
– OR.
Aug 31 '13 at 14:52
$begingroup$
Why abandon the first path for only 32 cases? The discriminant of the other factor to be a square $(z-8)^2+32z=(z+8)^2$. So you get for each value for $z$ solutions for $y=(2z-16pm(z+8))/2$. These are integers only for even $z$. That cuts down the cases a bit.
$endgroup$
– OR.
Aug 31 '13 at 14:52
1
1
$begingroup$
What about $x=-16, y=9, z=15$?
$endgroup$
– Axel Kemper
Aug 31 '13 at 15:05
$begingroup$
What about $x=-16, y=9, z=15$?
$endgroup$
– Axel Kemper
Aug 31 '13 at 15:05
$begingroup$
@ABC Let's say this is a contest problem, would you spend 30 minutes to check every combination? Note that the there are 64 values for y, because of the $pm$ sign in front of the square root.
$endgroup$
– Stefan4024
Aug 31 '13 at 15:09
$begingroup$
@ABC Let's say this is a contest problem, would you spend 30 minutes to check every combination? Note that the there are 64 values for y, because of the $pm$ sign in front of the square root.
$endgroup$
– Stefan4024
Aug 31 '13 at 15:09
$begingroup$
@AxelKemper According to Wolfram Alpha (9,15,-16) and its six permutation are the only integer solution, but any idea how to get those values?
$endgroup$
– Stefan4024
Aug 31 '13 at 15:09
$begingroup$
@AxelKemper According to Wolfram Alpha (9,15,-16) and its six permutation are the only integer solution, but any idea how to get those values?
$endgroup$
– Stefan4024
Aug 31 '13 at 15:09
$begingroup$
I have used Microsoft Solver Foundation to get this solution. It took less than a second.
$endgroup$
– Axel Kemper
Aug 31 '13 at 15:11
$begingroup$
I have used Microsoft Solver Foundation to get this solution. It took less than a second.
$endgroup$
– Axel Kemper
Aug 31 '13 at 15:11
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Calculate $(x-8)(y-8)(z-8)$.
There are several ways to do the calculation. Working from the last equation in the first method, observe that the second term can be factorized into $(y-8)(y+z)= -(y-8)(x-8)$, which gives us
$(x-8)(y-8)(z-8) = -168 $
Use this, along with the factors of 168 and that $x-8+y-8+z-8 = -16$ to greatly restrict the cases you have to look at. For example, as Mark comments, since the sum of 3 numbers is even, so either one of them is even, all all three of them are.
Now, if you don't want an 'Wolfram Alpha' expansion of the terms, you could do the expansion yourself, it's not that hard. The way that I arrived at the calculation used the fact that $a^3 + b^3 + c^3 - 3abc = (a+b+c)( (a+b+c)^2 - 3ab -3bc - 3ca) $
Note that since you want to solve this as a Diophantine equation, using algebraic methods can only get you so far. At some point in time, you must change it into a 'number theoretic way', like analyzing the factors, or gcd of terms.
answered Aug 31 '13 at 16:02
Calvin LinCalvin Lin
36.3k349114
$endgroup$
2
$begingroup$
Note that $168=XYZ=8times 3 times 7$ and $X+Y+Z=-16$ implies either three even numbers, or two odd, one even. There is only one possibility for three even. If two factors are odd, the even factor is divisible by $8$ so the sum or difference of the odds has to be a multiple of $8$.
$endgroup$
– Mark Bennet
Aug 31 '13 at 16:38
$begingroup$
Very good answer, significantly cutting the number of cases into a reasonable amount. Just one mistake probably a typo: $(y-8)(y+z) = -(y-8)(x-8)$, not $(y-8)(y+z) = -(y-z)(x-8)$,
$endgroup$
– Stefan4024
Aug 31 '13 at 16:48
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
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$begingroup$
Hint: Calculate $(x-8)(y-8)(z-8)$.
There are several ways to do the calculation. Working from the last equation in the first method, observe that the second term can be factorized into $(y-8)(y+z)= -(y-8)(x-8)$, which gives us
$(x-8)(y-8)(z-8) = -168 $
Use this, along with the factors of 168 and that $x-8+y-8+z-8 = -16$ to greatly restrict the cases you have to look at. For example, as Mark comments, since the sum of 3 numbers is even, so either one of them is even, all all three of them are.
Now, if you don't want an 'Wolfram Alpha' expansion of the terms, you could do the expansion yourself, it's not that hard. The way that I arrived at the calculation used the fact that $a^3 + b^3 + c^3 - 3abc = (a+b+c)( (a+b+c)^2 - 3ab -3bc - 3ca) $
Note that since you want to solve this as a Diophantine equation, using algebraic methods can only get you so far. At some point in time, you must change it into a 'number theoretic way', like analyzing the factors, or gcd of terms.
answered Aug 31 '13 at 16:02
Calvin LinCalvin Lin
36.3k349114
$endgroup$
2
$begingroup$
Note that $168=XYZ=8times 3 times 7$ and $X+Y+Z=-16$ implies either three even numbers, or two odd, one even. There is only one possibility for three even. If two factors are odd, the even factor is divisible by $8$ so the sum or difference of the odds has to be a multiple of $8$.
$endgroup$
– Mark Bennet
Aug 31 '13 at 16:38
$begingroup$
Very good answer, significantly cutting the number of cases into a reasonable amount. Just one mistake probably a typo: $(y-8)(y+z) = -(y-8)(x-8)$, not $(y-8)(y+z) = -(y-z)(x-8)$,
$endgroup$
– Stefan4024
Aug 31 '13 at 16:48
add a comment |
$begingroup$
Hint: Calculate $(x-8)(y-8)(z-8)$.
There are several ways to do the calculation. Working from the last equation in the first method, observe that the second term can be factorized into $(y-8)(y+z)= -(y-8)(x-8)$, which gives us
$(x-8)(y-8)(z-8) = -168 $
Use this, along with the factors of 168 and that $x-8+y-8+z-8 = -16$ to greatly restrict the cases you have to look at. For example, as Mark comments, since the sum of 3 numbers is even, so either one of them is even, all all three of them are.
Now, if you don't want an 'Wolfram Alpha' expansion of the terms, you could do the expansion yourself, it's not that hard. The way that I arrived at the calculation used the fact that $a^3 + b^3 + c^3 - 3abc = (a+b+c)( (a+b+c)^2 - 3ab -3bc - 3ca) $
Note that since you want to solve this as a Diophantine equation, using algebraic methods can only get you so far. At some point in time, you must change it into a 'number theoretic way', like analyzing the factors, or gcd of terms.
answered Aug 31 '13 at 16:02
Calvin LinCalvin Lin
36.3k349114
$endgroup$
2
$begingroup$
Note that $168=XYZ=8times 3 times 7$ and $X+Y+Z=-16$ implies either three even numbers, or two odd, one even. There is only one possibility for three even. If two factors are odd, the even factor is divisible by $8$ so the sum or difference of the odds has to be a multiple of $8$.
$endgroup$
– Mark Bennet
Aug 31 '13 at 16:38
$begingroup$
Very good answer, significantly cutting the number of cases into a reasonable amount. Just one mistake probably a typo: $(y-8)(y+z) = -(y-8)(x-8)$, not $(y-8)(y+z) = -(y-z)(x-8)$,
$endgroup$
– Stefan4024
Aug 31 '13 at 16:48
add a comment |
$begingroup$
Hint: Calculate $(x-8)(y-8)(z-8)$.
There are several ways to do the calculation. Working from the last equation in the first method, observe that the second term can be factorized into $(y-8)(y+z)= -(y-8)(x-8)$, which gives us
$(x-8)(y-8)(z-8) = -168 $
Use this, along with the factors of 168 and that $x-8+y-8+z-8 = -16$ to greatly restrict the cases you have to look at. For example, as Mark comments, since the sum of 3 numbers is even, so either one of them is even, all all three of them are.
Now, if you don't want an 'Wolfram Alpha' expansion of the terms, you could do the expansion yourself, it's not that hard. The way that I arrived at the calculation used the fact that $a^3 + b^3 + c^3 - 3abc = (a+b+c)( (a+b+c)^2 - 3ab -3bc - 3ca) $
Note that since you want to solve this as a Diophantine equation, using algebraic methods can only get you so far. At some point in time, you must change it into a 'number theoretic way', like analyzing the factors, or gcd of terms.
answered Aug 31 '13 at 16:02
Calvin LinCalvin Lin
36.3k349114
$endgroup$
Hint: Calculate $(x-8)(y-8)(z-8)$.
There are several ways to do the calculation. Working from the last equation in the first method, observe that the second term can be factorized into $(y-8)(y+z)= -(y-8)(x-8)$, which gives us
$(x-8)(y-8)(z-8) = -168 $
Use this, along with the factors of 168 and that $x-8+y-8+z-8 = -16$ to greatly restrict the cases you have to look at. For example, as Mark comments, since the sum of 3 numbers is even, so either one of them is even, all all three of them are.
Now, if you don't want an 'Wolfram Alpha' expansion of the terms, you could do the expansion yourself, it's not that hard. The way that I arrived at the calculation used the fact that $a^3 + b^3 + c^3 - 3abc = (a+b+c)( (a+b+c)^2 - 3ab -3bc - 3ca) $
Note that since you want to solve this as a Diophantine equation, using algebraic methods can only get you so far. At some point in time, you must change it into a 'number theoretic way', like analyzing the factors, or gcd of terms.
answered Aug 31 '13 at 16:02
Calvin LinCalvin Lin
36.3k349114
edited Aug 31 '13 at 16:58
answered Aug 31 '13 at 16:02
Calvin LinCalvin Lin
36.3k349114
answered Aug 31 '13 at 16:02
Calvin LinCalvin Lin
36.3k349114
answered Aug 31 '13 at 16:02
Calvin LinCalvin Lin
36.3k349114
36.3k349114
2
$begingroup$
Note that $168=XYZ=8times 3 times 7$ and $X+Y+Z=-16$ implies either three even numbers, or two odd, one even. There is only one possibility for three even. If two factors are odd, the even factor is divisible by $8$ so the sum or difference of the odds has to be a multiple of $8$.
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– Mark Bennet
Aug 31 '13 at 16:38
$begingroup$
Very good answer, significantly cutting the number of cases into a reasonable amount. Just one mistake probably a typo: $(y-8)(y+z) = -(y-8)(x-8)$, not $(y-8)(y+z) = -(y-z)(x-8)$,
$endgroup$
– Stefan4024
Aug 31 '13 at 16:48
add a comment |
2
$begingroup$
Note that $168=XYZ=8times 3 times 7$ and $X+Y+Z=-16$ implies either three even numbers, or two odd, one even. There is only one possibility for three even. If two factors are odd, the even factor is divisible by $8$ so the sum or difference of the odds has to be a multiple of $8$.
$endgroup$
– Mark Bennet
Aug 31 '13 at 16:38
$begingroup$
Very good answer, significantly cutting the number of cases into a reasonable amount. Just one mistake probably a typo: $(y-8)(y+z) = -(y-8)(x-8)$, not $(y-8)(y+z) = -(y-z)(x-8)$,
$endgroup$
– Stefan4024
Aug 31 '13 at 16:48
2
2
$begingroup$
Note that $168=XYZ=8times 3 times 7$ and $X+Y+Z=-16$ implies either three even numbers, or two odd, one even. There is only one possibility for three even. If two factors are odd, the even factor is divisible by $8$ so the sum or difference of the odds has to be a multiple of $8$.
$endgroup$
– Mark Bennet
Aug 31 '13 at 16:38
$begingroup$
Note that $168=XYZ=8times 3 times 7$ and $X+Y+Z=-16$ implies either three even numbers, or two odd, one even. There is only one possibility for three even. If two factors are odd, the even factor is divisible by $8$ so the sum or difference of the odds has to be a multiple of $8$.
$endgroup$
– Mark Bennet
Aug 31 '13 at 16:38
$begingroup$
Very good answer, significantly cutting the number of cases into a reasonable amount. Just one mistake probably a typo: $(y-8)(y+z) = -(y-8)(x-8)$, not $(y-8)(y+z) = -(y-z)(x-8)$,
$endgroup$
– Stefan4024
Aug 31 '13 at 16:48
$begingroup$
Very good answer, significantly cutting the number of cases into a reasonable amount. Just one mistake probably a typo: $(y-8)(y+z) = -(y-8)(x-8)$, not $(y-8)(y+z) = -(y-z)(x-8)$,
$endgroup$
– Stefan4024
Aug 31 '13 at 16:48
add a comment |
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Why abandon the first path for only 32 cases? The discriminant of the other factor to be a square $(z-8)^2+32z=(z+8)^2$. So you get for each value for $z$ solutions for $y=(2z-16pm(z+8))/2$. These are integers only for even $z$. That cuts down the cases a bit.
$endgroup$
– OR.
Aug 31 '13 at 14:52
1
$begingroup$
What about $x=-16, y=9, z=15$?
$endgroup$
– Axel Kemper
Aug 31 '13 at 15:05
$begingroup$
@ABC Let's say this is a contest problem, would you spend 30 minutes to check every combination? Note that the there are 64 values for y, because of the $pm$ sign in front of the square root.
$endgroup$
– Stefan4024
Aug 31 '13 at 15:09
$begingroup$
@AxelKemper According to Wolfram Alpha (9,15,-16) and its six permutation are the only integer solution, but any idea how to get those values?
$endgroup$
– Stefan4024
Aug 31 '13 at 15:09
$begingroup$
I have used Microsoft Solver Foundation to get this solution. It took less than a second.
$endgroup$
– Axel Kemper
Aug 31 '13 at 15:11