Mixing
Nontrivial integer solutions of $ a^3+b^3=c^3+d^3$
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
How can we obtain a set of nontrivial solutions of
$$
a^3+b^3=c^3+d^3,
$$
for $a,b,c,din mathbbZ$
where $(a,b)neq (c,d)$ and $(a,b)neq (d,c)$.
Say in the range that $|a|,|b|,|c|,|d| in [0,30]$, whose absolute values small or eqaul to 30?
How many simple but nontrivial solutions are there around this range?
The simple form the solutions are the better.
Thank you!
p.s. When $d=0$, we know it is impossible due to the Fermat's Last Theorem.
number-theory prime-numbers arithmetic pythagorean-triples fermat-numbers
asked Jul 15 at 22:40
wonderich
1,67021226
 |Â
show 3 more comments
up vote
1
down vote
favorite
How can we obtain a set of nontrivial solutions of
$$
a^3+b^3=c^3+d^3,
$$
for $a,b,c,din mathbbZ$
where $(a,b)neq (c,d)$ and $(a,b)neq (d,c)$.
Say in the range that $|a|,|b|,|c|,|d| in [0,30]$, whose absolute values small or eqaul to 30?
How many simple but nontrivial solutions are there around this range?
The simple form the solutions are the better.
Thank you!
p.s. When $d=0$, we know it is impossible due to the Fermat's Last Theorem.
number-theory prime-numbers arithmetic pythagorean-triples fermat-numbers
asked Jul 15 at 22:40
wonderich
1,67021226
3
Here's one such set: $(1,12,9,10),(2,16,9,15)$
– Henning Makholm
Jul 15 at 22:44
1
Have a look at the paper Characterizing the Sum of Two Cubes.
– Théophile
Jul 15 at 22:45
Could you include your own efforts in trying to answer these two questions? As it is, it looks like an exercise or assigned problem and that you're expecting us to answer it entirely for you.
– amWhy
Jul 15 at 22:45
2
See Euler's general solution here for example.
– dxiv
Jul 15 at 22:50
1
One thing I knew is that Plato's number actually works if you want to see my efforts: $$ 3^3 + 4^3 = (-5)^3 + 6^3=91, $$ also formulating a question may be an effort if you contemplate something bigger than the question seems to be. Thanks for all other's useful answers! These are very beneficial (much more than simply complaining or voting down).
– wonderich
Jul 16 at 2:46
 |Â
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How can we obtain a set of nontrivial solutions of
$$
a^3+b^3=c^3+d^3,
$$
for $a,b,c,din mathbbZ$
where $(a,b)neq (c,d)$ and $(a,b)neq (d,c)$.
Say in the range that $|a|,|b|,|c|,|d| in [0,30]$, whose absolute values small or eqaul to 30?
How many simple but nontrivial solutions are there around this range?
The simple form the solutions are the better.
Thank you!
p.s. When $d=0$, we know it is impossible due to the Fermat's Last Theorem.
number-theory prime-numbers arithmetic pythagorean-triples fermat-numbers
asked Jul 15 at 22:40
wonderich
1,67021226
How can we obtain a set of nontrivial solutions of
$$
a^3+b^3=c^3+d^3,
$$
for $a,b,c,din mathbbZ$
where $(a,b)neq (c,d)$ and $(a,b)neq (d,c)$.
Say in the range that $|a|,|b|,|c|,|d| in [0,30]$, whose absolute values small or eqaul to 30?
How many simple but nontrivial solutions are there around this range?
The simple form the solutions are the better.
Thank you!
p.s. When $d=0$, we know it is impossible due to the Fermat's Last Theorem.
number-theory prime-numbers arithmetic pythagorean-triples fermat-numbers
asked Jul 15 at 22:40
wonderich
1,67021226
asked Jul 15 at 22:40
wonderich
1,67021226
asked Jul 15 at 22:40
wonderich
1,67021226
asked Jul 15 at 22:40
wonderich
1,67021226
1,67021226
3
Here's one such set: $(1,12,9,10),(2,16,9,15)$
– Henning Makholm
Jul 15 at 22:44
1
Have a look at the paper Characterizing the Sum of Two Cubes.
– Théophile
Jul 15 at 22:45
Could you include your own efforts in trying to answer these two questions? As it is, it looks like an exercise or assigned problem and that you're expecting us to answer it entirely for you.
– amWhy
Jul 15 at 22:45
2
See Euler's general solution here for example.
– dxiv
Jul 15 at 22:50
1
One thing I knew is that Plato's number actually works if you want to see my efforts: $$ 3^3 + 4^3 = (-5)^3 + 6^3=91, $$ also formulating a question may be an effort if you contemplate something bigger than the question seems to be. Thanks for all other's useful answers! These are very beneficial (much more than simply complaining or voting down).
– wonderich
Jul 16 at 2:46
 |Â
show 3 more comments
3
Here's one such set: $(1,12,9,10),(2,16,9,15)$
– Henning Makholm
Jul 15 at 22:44
1
Have a look at the paper Characterizing the Sum of Two Cubes.
– Théophile
Jul 15 at 22:45
Could you include your own efforts in trying to answer these two questions? As it is, it looks like an exercise or assigned problem and that you're expecting us to answer it entirely for you.
– amWhy
Jul 15 at 22:45
2
See Euler's general solution here for example.
– dxiv
Jul 15 at 22:50
1
One thing I knew is that Plato's number actually works if you want to see my efforts: $$ 3^3 + 4^3 = (-5)^3 + 6^3=91, $$ also formulating a question may be an effort if you contemplate something bigger than the question seems to be. Thanks for all other's useful answers! These are very beneficial (much more than simply complaining or voting down).
– wonderich
Jul 16 at 2:46
3
3
Here's one such set: $(1,12,9,10),(2,16,9,15)$
– Henning Makholm
Jul 15 at 22:44
Here's one such set: $(1,12,9,10),(2,16,9,15)$
– Henning Makholm
Jul 15 at 22:44
1
1
Have a look at the paper Characterizing the Sum of Two Cubes.
– Théophile
Jul 15 at 22:45
Have a look at the paper Characterizing the Sum of Two Cubes.
– Théophile
Jul 15 at 22:45
Could you include your own efforts in trying to answer these two questions? As it is, it looks like an exercise or assigned problem and that you're expecting us to answer it entirely for you.
– amWhy
Jul 15 at 22:45
Could you include your own efforts in trying to answer these two questions? As it is, it looks like an exercise or assigned problem and that you're expecting us to answer it entirely for you.
– amWhy
Jul 15 at 22:45
2
2
See Euler's general solution here for example.
– dxiv
Jul 15 at 22:50
See Euler's general solution here for example.
– dxiv
Jul 15 at 22:50
1
1
One thing I knew is that Plato's number actually works if you want to see my efforts: $$ 3^3 + 4^3 = (-5)^3 + 6^3=91, $$ also formulating a question may be an effort if you contemplate something bigger than the question seems to be. Thanks for all other's useful answers! These are very beneficial (much more than simply complaining or voting down).
– wonderich
Jul 16 at 2:46
One thing I knew is that Plato's number actually works if you want to see my efforts: $$ 3^3 + 4^3 = (-5)^3 + 6^3=91, $$ also formulating a question may be an effort if you contemplate something bigger than the question seems to be. Thanks for all other's useful answers! These are very beneficial (much more than simply complaining or voting down).
– wonderich
Jul 16 at 2:46
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
It is possible to give a partial answer by mimicking Kummer's approach to the first case of FLT. As in your other related question, let us introduce $omega$ = a primitive 3rd root of unity, the quadratic field $mathbf Q(omega)$, its ring of integers $mathbf Z[omega]$. We intend to solve the diophantine equation $a^3+b^3=c^3+d^3$ under the additional hypothesis that $3nmid a+b$ (or equivalently $3nmid c+d$, by Fermat's little theorem). For simplification, let us concentrate on primitive solutions, i.e. suppose that $a, b$ are coprime (and similarly $c,d$ ).To exploit the decomposition $a^3+b^3=(a+b)(a+bomega)(a+bomega^2)$, we must recall that $mathbf Z[omega]$ is a principal ideal ring, whose group of units coincides with its group of roots of unity, generated by $pm omega$.
Lemma: Under our hypotheses, for $mneq n$ mod 3, the factors $(a+bomega^m)$ and $(a+bomega^n)$ are coprime in $mathbf Z[omega]$. Proof: Since $mathbf Z[omega]$ is principal, let us apply Bézout's theorem by showing that the ideal $J$ generated by our two elements contains $1$. Because $omega^k$ is a primitive 3rd root of unity whenever $k neq 0$ mod 3, the quotient $epsilon_k=frac 1-omega^k1-omega$ is a unit. But $(a+bomega^m)-(a+bomega^n)=omega^m(1-omega^n-m)b=epsilon_n-momega^m(1-omega)b$ , and similarly $(a+bomega^m)omega^n-(a+bomega^n)omega^m=-omega^m(1-omega^n-m)a= - epsilon_n-momega^m(1-omega)a$, hence $(1-omega)b$ and $ (1-omega)a in J$. Since $a,b$ are coprime in $mathbf Z$, Bézout asserts the existence of $u,v in mathbf Z$ s.t. $ua+vb=1$, hence $(1-omega)ua+(1-omega)vb=(1-omega) in J$, hence also $3in J$. Moreover, $a+b=(a+bomega^m)+(1-omega^m)b=(a+bomega^ma+bomega^m)+(1-omega)epsilon_mb$, so that $a+b in J$. Finally, our additional hypothesis implies the existence of $s,tin mathbf Z$ s.t.
$(a+b)s+3t=1$, hence $1in J$. OUF
EDIT : Any rational prime $pneq 3$ is unramified in $mathbf Q(omega)$. Let $delta$= gcd $(a+b,c+d)$ and $N$= the norm of $mathbf Q (omega)/mathbf Q$ . Our additional hypothesis and the lemma then imply that $frac a+bdelta=pm N(c+domega)=pm (c^2+d^2-cd)=pm ((c+d)^2-3cd)$, and similarly $frac c+ddelta=pm ((a+b)^2-3ab)$. Besides, since $c,d$ are the roots of the quadratic equation $x^2-(c+d)x+cd=0$, the discriminant $Delta=(c+d)^2-4cd=((c+d)^2-3cd)-cd$ is necessarily a perfect square (=square of an integer), and similarly $((a+b)^2-3ab)-cd$ is a perfect square. Summarizing, $pmfraca+bdelta-cd$ and $pmfracc+ddelta-ab$ must be perfect squares.
EDIT 2: I made a mistake again ! $(a+bomega)$ and $(c+domega)$ could have a common factor. This seems to be a dead end.
answered Jul 18 at 16:13
nguyen quang do
7,4621621
thanks for your nice efforts - +1.
– wonderich
Jul 18 at 17:58
Unfortunately my conclusion was too hasty. There is an obvious contradiction with the Plato number. I think I made an error in the derivation of the equality $a+b=pm(c+d)$. I'll think about it more thoroughly and edit it if possible.
– nguyen quang do
Jul 18 at 19:39
thanks for letting us know -
– wonderich
Jul 18 at 22:01
I edited my first answer. What I get now is a necessary condition, which can be checked numerically in a limited range of absolute values of the variables.
– nguyen quang do
Jul 20 at 12:52
I made a mistake again ! $a+bomega$ and $c+domega$ could have a common factor. This seems to be a dead end.
– nguyen quang do
Jul 21 at 8:06
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
It is possible to give a partial answer by mimicking Kummer's approach to the first case of FLT. As in your other related question, let us introduce $omega$ = a primitive 3rd root of unity, the quadratic field $mathbf Q(omega)$, its ring of integers $mathbf Z[omega]$. We intend to solve the diophantine equation $a^3+b^3=c^3+d^3$ under the additional hypothesis that $3nmid a+b$ (or equivalently $3nmid c+d$, by Fermat's little theorem). For simplification, let us concentrate on primitive solutions, i.e. suppose that $a, b$ are coprime (and similarly $c,d$ ).To exploit the decomposition $a^3+b^3=(a+b)(a+bomega)(a+bomega^2)$, we must recall that $mathbf Z[omega]$ is a principal ideal ring, whose group of units coincides with its group of roots of unity, generated by $pm omega$.
Lemma: Under our hypotheses, for $mneq n$ mod 3, the factors $(a+bomega^m)$ and $(a+bomega^n)$ are coprime in $mathbf Z[omega]$. Proof: Since $mathbf Z[omega]$ is principal, let us apply Bézout's theorem by showing that the ideal $J$ generated by our two elements contains $1$. Because $omega^k$ is a primitive 3rd root of unity whenever $k neq 0$ mod 3, the quotient $epsilon_k=frac 1-omega^k1-omega$ is a unit. But $(a+bomega^m)-(a+bomega^n)=omega^m(1-omega^n-m)b=epsilon_n-momega^m(1-omega)b$ , and similarly $(a+bomega^m)omega^n-(a+bomega^n)omega^m=-omega^m(1-omega^n-m)a= - epsilon_n-momega^m(1-omega)a$, hence $(1-omega)b$ and $ (1-omega)a in J$. Since $a,b$ are coprime in $mathbf Z$, Bézout asserts the existence of $u,v in mathbf Z$ s.t. $ua+vb=1$, hence $(1-omega)ua+(1-omega)vb=(1-omega) in J$, hence also $3in J$. Moreover, $a+b=(a+bomega^m)+(1-omega^m)b=(a+bomega^ma+bomega^m)+(1-omega)epsilon_mb$, so that $a+b in J$. Finally, our additional hypothesis implies the existence of $s,tin mathbf Z$ s.t.
$(a+b)s+3t=1$, hence $1in J$. OUF
EDIT : Any rational prime $pneq 3$ is unramified in $mathbf Q(omega)$. Let $delta$= gcd $(a+b,c+d)$ and $N$= the norm of $mathbf Q (omega)/mathbf Q$ . Our additional hypothesis and the lemma then imply that $frac a+bdelta=pm N(c+domega)=pm (c^2+d^2-cd)=pm ((c+d)^2-3cd)$, and similarly $frac c+ddelta=pm ((a+b)^2-3ab)$. Besides, since $c,d$ are the roots of the quadratic equation $x^2-(c+d)x+cd=0$, the discriminant $Delta=(c+d)^2-4cd=((c+d)^2-3cd)-cd$ is necessarily a perfect square (=square of an integer), and similarly $((a+b)^2-3ab)-cd$ is a perfect square. Summarizing, $pmfraca+bdelta-cd$ and $pmfracc+ddelta-ab$ must be perfect squares.
EDIT 2: I made a mistake again ! $(a+bomega)$ and $(c+domega)$ could have a common factor. This seems to be a dead end.
answered Jul 18 at 16:13
nguyen quang do
7,4621621
thanks for your nice efforts - +1.
– wonderich
Jul 18 at 17:58
Unfortunately my conclusion was too hasty. There is an obvious contradiction with the Plato number. I think I made an error in the derivation of the equality $a+b=pm(c+d)$. I'll think about it more thoroughly and edit it if possible.
– nguyen quang do
Jul 18 at 19:39
thanks for letting us know -
– wonderich
Jul 18 at 22:01
I edited my first answer. What I get now is a necessary condition, which can be checked numerically in a limited range of absolute values of the variables.
– nguyen quang do
Jul 20 at 12:52
I made a mistake again ! $a+bomega$ and $c+domega$ could have a common factor. This seems to be a dead end.
– nguyen quang do
Jul 21 at 8:06
add a comment |Â
up vote
1
down vote
It is possible to give a partial answer by mimicking Kummer's approach to the first case of FLT. As in your other related question, let us introduce $omega$ = a primitive 3rd root of unity, the quadratic field $mathbf Q(omega)$, its ring of integers $mathbf Z[omega]$. We intend to solve the diophantine equation $a^3+b^3=c^3+d^3$ under the additional hypothesis that $3nmid a+b$ (or equivalently $3nmid c+d$, by Fermat's little theorem). For simplification, let us concentrate on primitive solutions, i.e. suppose that $a, b$ are coprime (and similarly $c,d$ ).To exploit the decomposition $a^3+b^3=(a+b)(a+bomega)(a+bomega^2)$, we must recall that $mathbf Z[omega]$ is a principal ideal ring, whose group of units coincides with its group of roots of unity, generated by $pm omega$.
Lemma: Under our hypotheses, for $mneq n$ mod 3, the factors $(a+bomega^m)$ and $(a+bomega^n)$ are coprime in $mathbf Z[omega]$. Proof: Since $mathbf Z[omega]$ is principal, let us apply Bézout's theorem by showing that the ideal $J$ generated by our two elements contains $1$. Because $omega^k$ is a primitive 3rd root of unity whenever $k neq 0$ mod 3, the quotient $epsilon_k=frac 1-omega^k1-omega$ is a unit. But $(a+bomega^m)-(a+bomega^n)=omega^m(1-omega^n-m)b=epsilon_n-momega^m(1-omega)b$ , and similarly $(a+bomega^m)omega^n-(a+bomega^n)omega^m=-omega^m(1-omega^n-m)a= - epsilon_n-momega^m(1-omega)a$, hence $(1-omega)b$ and $ (1-omega)a in J$. Since $a,b$ are coprime in $mathbf Z$, Bézout asserts the existence of $u,v in mathbf Z$ s.t. $ua+vb=1$, hence $(1-omega)ua+(1-omega)vb=(1-omega) in J$, hence also $3in J$. Moreover, $a+b=(a+bomega^m)+(1-omega^m)b=(a+bomega^ma+bomega^m)+(1-omega)epsilon_mb$, so that $a+b in J$. Finally, our additional hypothesis implies the existence of $s,tin mathbf Z$ s.t.
$(a+b)s+3t=1$, hence $1in J$. OUF
EDIT : Any rational prime $pneq 3$ is unramified in $mathbf Q(omega)$. Let $delta$= gcd $(a+b,c+d)$ and $N$= the norm of $mathbf Q (omega)/mathbf Q$ . Our additional hypothesis and the lemma then imply that $frac a+bdelta=pm N(c+domega)=pm (c^2+d^2-cd)=pm ((c+d)^2-3cd)$, and similarly $frac c+ddelta=pm ((a+b)^2-3ab)$. Besides, since $c,d$ are the roots of the quadratic equation $x^2-(c+d)x+cd=0$, the discriminant $Delta=(c+d)^2-4cd=((c+d)^2-3cd)-cd$ is necessarily a perfect square (=square of an integer), and similarly $((a+b)^2-3ab)-cd$ is a perfect square. Summarizing, $pmfraca+bdelta-cd$ and $pmfracc+ddelta-ab$ must be perfect squares.
EDIT 2: I made a mistake again ! $(a+bomega)$ and $(c+domega)$ could have a common factor. This seems to be a dead end.
answered Jul 18 at 16:13
nguyen quang do
7,4621621
thanks for your nice efforts - +1.
– wonderich
Jul 18 at 17:58
Unfortunately my conclusion was too hasty. There is an obvious contradiction with the Plato number. I think I made an error in the derivation of the equality $a+b=pm(c+d)$. I'll think about it more thoroughly and edit it if possible.
– nguyen quang do
Jul 18 at 19:39
thanks for letting us know -
– wonderich
Jul 18 at 22:01
I edited my first answer. What I get now is a necessary condition, which can be checked numerically in a limited range of absolute values of the variables.
– nguyen quang do
Jul 20 at 12:52
I made a mistake again ! $a+bomega$ and $c+domega$ could have a common factor. This seems to be a dead end.
– nguyen quang do
Jul 21 at 8:06
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is possible to give a partial answer by mimicking Kummer's approach to the first case of FLT. As in your other related question, let us introduce $omega$ = a primitive 3rd root of unity, the quadratic field $mathbf Q(omega)$, its ring of integers $mathbf Z[omega]$. We intend to solve the diophantine equation $a^3+b^3=c^3+d^3$ under the additional hypothesis that $3nmid a+b$ (or equivalently $3nmid c+d$, by Fermat's little theorem). For simplification, let us concentrate on primitive solutions, i.e. suppose that $a, b$ are coprime (and similarly $c,d$ ).To exploit the decomposition $a^3+b^3=(a+b)(a+bomega)(a+bomega^2)$, we must recall that $mathbf Z[omega]$ is a principal ideal ring, whose group of units coincides with its group of roots of unity, generated by $pm omega$.
Lemma: Under our hypotheses, for $mneq n$ mod 3, the factors $(a+bomega^m)$ and $(a+bomega^n)$ are coprime in $mathbf Z[omega]$. Proof: Since $mathbf Z[omega]$ is principal, let us apply Bézout's theorem by showing that the ideal $J$ generated by our two elements contains $1$. Because $omega^k$ is a primitive 3rd root of unity whenever $k neq 0$ mod 3, the quotient $epsilon_k=frac 1-omega^k1-omega$ is a unit. But $(a+bomega^m)-(a+bomega^n)=omega^m(1-omega^n-m)b=epsilon_n-momega^m(1-omega)b$ , and similarly $(a+bomega^m)omega^n-(a+bomega^n)omega^m=-omega^m(1-omega^n-m)a= - epsilon_n-momega^m(1-omega)a$, hence $(1-omega)b$ and $ (1-omega)a in J$. Since $a,b$ are coprime in $mathbf Z$, Bézout asserts the existence of $u,v in mathbf Z$ s.t. $ua+vb=1$, hence $(1-omega)ua+(1-omega)vb=(1-omega) in J$, hence also $3in J$. Moreover, $a+b=(a+bomega^m)+(1-omega^m)b=(a+bomega^ma+bomega^m)+(1-omega)epsilon_mb$, so that $a+b in J$. Finally, our additional hypothesis implies the existence of $s,tin mathbf Z$ s.t.
$(a+b)s+3t=1$, hence $1in J$. OUF
EDIT : Any rational prime $pneq 3$ is unramified in $mathbf Q(omega)$. Let $delta$= gcd $(a+b,c+d)$ and $N$= the norm of $mathbf Q (omega)/mathbf Q$ . Our additional hypothesis and the lemma then imply that $frac a+bdelta=pm N(c+domega)=pm (c^2+d^2-cd)=pm ((c+d)^2-3cd)$, and similarly $frac c+ddelta=pm ((a+b)^2-3ab)$. Besides, since $c,d$ are the roots of the quadratic equation $x^2-(c+d)x+cd=0$, the discriminant $Delta=(c+d)^2-4cd=((c+d)^2-3cd)-cd$ is necessarily a perfect square (=square of an integer), and similarly $((a+b)^2-3ab)-cd$ is a perfect square. Summarizing, $pmfraca+bdelta-cd$ and $pmfracc+ddelta-ab$ must be perfect squares.
EDIT 2: I made a mistake again ! $(a+bomega)$ and $(c+domega)$ could have a common factor. This seems to be a dead end.
answered Jul 18 at 16:13
nguyen quang do
7,4621621
It is possible to give a partial answer by mimicking Kummer's approach to the first case of FLT. As in your other related question, let us introduce $omega$ = a primitive 3rd root of unity, the quadratic field $mathbf Q(omega)$, its ring of integers $mathbf Z[omega]$. We intend to solve the diophantine equation $a^3+b^3=c^3+d^3$ under the additional hypothesis that $3nmid a+b$ (or equivalently $3nmid c+d$, by Fermat's little theorem). For simplification, let us concentrate on primitive solutions, i.e. suppose that $a, b$ are coprime (and similarly $c,d$ ).To exploit the decomposition $a^3+b^3=(a+b)(a+bomega)(a+bomega^2)$, we must recall that $mathbf Z[omega]$ is a principal ideal ring, whose group of units coincides with its group of roots of unity, generated by $pm omega$.
Lemma: Under our hypotheses, for $mneq n$ mod 3, the factors $(a+bomega^m)$ and $(a+bomega^n)$ are coprime in $mathbf Z[omega]$. Proof: Since $mathbf Z[omega]$ is principal, let us apply Bézout's theorem by showing that the ideal $J$ generated by our two elements contains $1$. Because $omega^k$ is a primitive 3rd root of unity whenever $k neq 0$ mod 3, the quotient $epsilon_k=frac 1-omega^k1-omega$ is a unit. But $(a+bomega^m)-(a+bomega^n)=omega^m(1-omega^n-m)b=epsilon_n-momega^m(1-omega)b$ , and similarly $(a+bomega^m)omega^n-(a+bomega^n)omega^m=-omega^m(1-omega^n-m)a= - epsilon_n-momega^m(1-omega)a$, hence $(1-omega)b$ and $ (1-omega)a in J$. Since $a,b$ are coprime in $mathbf Z$, Bézout asserts the existence of $u,v in mathbf Z$ s.t. $ua+vb=1$, hence $(1-omega)ua+(1-omega)vb=(1-omega) in J$, hence also $3in J$. Moreover, $a+b=(a+bomega^m)+(1-omega^m)b=(a+bomega^ma+bomega^m)+(1-omega)epsilon_mb$, so that $a+b in J$. Finally, our additional hypothesis implies the existence of $s,tin mathbf Z$ s.t.
$(a+b)s+3t=1$, hence $1in J$. OUF
EDIT : Any rational prime $pneq 3$ is unramified in $mathbf Q(omega)$. Let $delta$= gcd $(a+b,c+d)$ and $N$= the norm of $mathbf Q (omega)/mathbf Q$ . Our additional hypothesis and the lemma then imply that $frac a+bdelta=pm N(c+domega)=pm (c^2+d^2-cd)=pm ((c+d)^2-3cd)$, and similarly $frac c+ddelta=pm ((a+b)^2-3ab)$. Besides, since $c,d$ are the roots of the quadratic equation $x^2-(c+d)x+cd=0$, the discriminant $Delta=(c+d)^2-4cd=((c+d)^2-3cd)-cd$ is necessarily a perfect square (=square of an integer), and similarly $((a+b)^2-3ab)-cd$ is a perfect square. Summarizing, $pmfraca+bdelta-cd$ and $pmfracc+ddelta-ab$ must be perfect squares.
EDIT 2: I made a mistake again ! $(a+bomega)$ and $(c+domega)$ could have a common factor. This seems to be a dead end.
answered Jul 18 at 16:13
nguyen quang do
7,4621621
edited Jul 21 at 8:04
answered Jul 18 at 16:13
nguyen quang do
7,4621621
answered Jul 18 at 16:13
nguyen quang do
7,4621621
answered Jul 18 at 16:13
nguyen quang do
7,4621621
7,4621621
thanks for your nice efforts - +1.
– wonderich
Jul 18 at 17:58
Unfortunately my conclusion was too hasty. There is an obvious contradiction with the Plato number. I think I made an error in the derivation of the equality $a+b=pm(c+d)$. I'll think about it more thoroughly and edit it if possible.
– nguyen quang do
Jul 18 at 19:39
thanks for letting us know -
– wonderich
Jul 18 at 22:01
I edited my first answer. What I get now is a necessary condition, which can be checked numerically in a limited range of absolute values of the variables.
– nguyen quang do
Jul 20 at 12:52
I made a mistake again ! $a+bomega$ and $c+domega$ could have a common factor. This seems to be a dead end.
– nguyen quang do
Jul 21 at 8:06
add a comment |Â
thanks for your nice efforts - +1.
– wonderich
Jul 18 at 17:58
Unfortunately my conclusion was too hasty. There is an obvious contradiction with the Plato number. I think I made an error in the derivation of the equality $a+b=pm(c+d)$. I'll think about it more thoroughly and edit it if possible.
– nguyen quang do
Jul 18 at 19:39
thanks for letting us know -
– wonderich
Jul 18 at 22:01
I edited my first answer. What I get now is a necessary condition, which can be checked numerically in a limited range of absolute values of the variables.
– nguyen quang do
Jul 20 at 12:52
I made a mistake again ! $a+bomega$ and $c+domega$ could have a common factor. This seems to be a dead end.
– nguyen quang do
Jul 21 at 8:06
thanks for your nice efforts - +1.
– wonderich
Jul 18 at 17:58
thanks for your nice efforts - +1.
– wonderich
Jul 18 at 17:58
Unfortunately my conclusion was too hasty. There is an obvious contradiction with the Plato number. I think I made an error in the derivation of the equality $a+b=pm(c+d)$. I'll think about it more thoroughly and edit it if possible.
– nguyen quang do
Jul 18 at 19:39
Unfortunately my conclusion was too hasty. There is an obvious contradiction with the Plato number. I think I made an error in the derivation of the equality $a+b=pm(c+d)$. I'll think about it more thoroughly and edit it if possible.
– nguyen quang do
Jul 18 at 19:39
thanks for letting us know -
– wonderich
Jul 18 at 22:01
thanks for letting us know -
– wonderich
Jul 18 at 22:01
I edited my first answer. What I get now is a necessary condition, which can be checked numerically in a limited range of absolute values of the variables.
– nguyen quang do
Jul 20 at 12:52
I edited my first answer. What I get now is a necessary condition, which can be checked numerically in a limited range of absolute values of the variables.
– nguyen quang do
Jul 20 at 12:52
I made a mistake again ! $a+bomega$ and $c+domega$ could have a common factor. This seems to be a dead end.
– nguyen quang do
Jul 21 at 8:06
I made a mistake again ! $a+bomega$ and $c+domega$ could have a common factor. This seems to be a dead end.
– nguyen quang do
Jul 21 at 8:06
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2852906%2fnontrivial-integer-solutions-of-a3b3-c3d3%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
3
Here's one such set: $(1,12,9,10),(2,16,9,15)$
– Henning Makholm
Jul 15 at 22:44
1
Have a look at the paper Characterizing the Sum of Two Cubes.
– Théophile
Jul 15 at 22:45
Could you include your own efforts in trying to answer these two questions? As it is, it looks like an exercise or assigned problem and that you're expecting us to answer it entirely for you.
– amWhy
Jul 15 at 22:45
2
See Euler's general solution here for example.
– dxiv
Jul 15 at 22:50
1
One thing I knew is that Plato's number actually works if you want to see my efforts: $$ 3^3 + 4^3 = (-5)^3 + 6^3=91, $$ also formulating a question may be an effort if you contemplate something bigger than the question seems to be. Thanks for all other's useful answers! These are very beneficial (much more than simply complaining or voting down).
– wonderich
Jul 16 at 2:46