Mixing
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Nontrivial integer solutions of $sum_i=1^3 a_i ^3=sum_i=1^3 b_i ^3$ and $sum_i=1^3 a_i =sum_i=1^3 b_i$
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The goal is to understand a set of nontrivial solutions of cubic polynomials
$$
sum_i=1^3 a_i ^3=sum_i=1^3 b_i ^3
Rightarrow a_1^3+a_2^3+a_3^3=b_1^3+b_2^3+b_3^3, tag1
$$
$$
sum_i=1^3 a_i =sum_i=1^3 b_i Rightarrow a_1+a_2+a_3=b_1+b_2+b_3 , tag2
$$
for $a_i,b_iin mathbbZ$
where we demand $(a_1,a_2,a_3)neq (b_1,b_2,b_3)$ or any of its permutation (total $3!=6$ cases).
Question: Are there any nontrivial solutions? (e.g. $a_i,b_iin mathbbZ$
where we demand $(a_1,a_2,a_3)neq (b_1,b_2,b_3)$ and other $3!$ permutations.) How many simple but nontrivial solutions are there in the smaller values of $a,b,c,d$? (The simple form the solutions are the better.) Or can one prove or disprove that there are nontrivial solutions? (Even better, if we can get a quick algorithm to generate the solutions.)A non-trivial example or a proof of non-existence is required to be accepted as a final answer.
Note add:
My trials/attempts: Since it is encouraged to show one's own attempt.
First, it looks like if we have only $i=1,2$ instead of summing over $i=1,2,3$, we have $sum_i=1^2 a_i ^3=sum_i=1^2 b_i ^3 $ and $sum_i=1^3 a_i =sum_i=1^3 b_i$, the nontrivial solutions seem to be impossible (?), but no rigorous proof has be shown yet.
Let me share a few comments on what these two equations boil down to:
If we take (2)$^3$-(1), we get
$$
(a_1+a_2)(a_1+a_3)(a_2+a_3)=(b_1+b_2)(b_1+b_3)(b_2+b_3), tag3
$$
we can further use (2) and (3) to get a lot more interesting constraints. However, I haven't got any other linear relation constraint as useful as (2) yet.
number-theory polynomials prime-numbers arithmetic
asked Jul 17 at 1:56
wonderich
1,66521226
add a comment |Â
up vote
2
down vote
favorite
The goal is to understand a set of nontrivial solutions of cubic polynomials
$$
sum_i=1^3 a_i ^3=sum_i=1^3 b_i ^3
Rightarrow a_1^3+a_2^3+a_3^3=b_1^3+b_2^3+b_3^3, tag1
$$
$$
sum_i=1^3 a_i =sum_i=1^3 b_i Rightarrow a_1+a_2+a_3=b_1+b_2+b_3 , tag2
$$
for $a_i,b_iin mathbbZ$
where we demand $(a_1,a_2,a_3)neq (b_1,b_2,b_3)$ or any of its permutation (total $3!=6$ cases).
Question: Are there any nontrivial solutions? (e.g. $a_i,b_iin mathbbZ$
where we demand $(a_1,a_2,a_3)neq (b_1,b_2,b_3)$ and other $3!$ permutations.) How many simple but nontrivial solutions are there in the smaller values of $a,b,c,d$? (The simple form the solutions are the better.) Or can one prove or disprove that there are nontrivial solutions? (Even better, if we can get a quick algorithm to generate the solutions.)A non-trivial example or a proof of non-existence is required to be accepted as a final answer.
Note add:
My trials/attempts: Since it is encouraged to show one's own attempt.
First, it looks like if we have only $i=1,2$ instead of summing over $i=1,2,3$, we have $sum_i=1^2 a_i ^3=sum_i=1^2 b_i ^3 $ and $sum_i=1^3 a_i =sum_i=1^3 b_i$, the nontrivial solutions seem to be impossible (?), but no rigorous proof has be shown yet.
Let me share a few comments on what these two equations boil down to:
If we take (2)$^3$-(1), we get
$$
(a_1+a_2)(a_1+a_3)(a_2+a_3)=(b_1+b_2)(b_1+b_3)(b_2+b_3), tag3
$$
we can further use (2) and (3) to get a lot more interesting constraints. However, I haven't got any other linear relation constraint as useful as (2) yet.
number-theory polynomials prime-numbers arithmetic
asked Jul 17 at 1:56
wonderich
1,66521226
math.stackexchange.com/questions/2593606/…
– individ
Jul 17 at 5:05
1
See $S 13$ Form: $a^k+b^k+c^k = d^k+e^k+f^k, k = 1,3$ of A Collection of Algeberaic identities collected by T.Piezas III.
– achille hui
Jul 17 at 14:56
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The goal is to understand a set of nontrivial solutions of cubic polynomials
$$
sum_i=1^3 a_i ^3=sum_i=1^3 b_i ^3
Rightarrow a_1^3+a_2^3+a_3^3=b_1^3+b_2^3+b_3^3, tag1
$$
$$
sum_i=1^3 a_i =sum_i=1^3 b_i Rightarrow a_1+a_2+a_3=b_1+b_2+b_3 , tag2
$$
for $a_i,b_iin mathbbZ$
where we demand $(a_1,a_2,a_3)neq (b_1,b_2,b_3)$ or any of its permutation (total $3!=6$ cases).
Question: Are there any nontrivial solutions? (e.g. $a_i,b_iin mathbbZ$
where we demand $(a_1,a_2,a_3)neq (b_1,b_2,b_3)$ and other $3!$ permutations.) How many simple but nontrivial solutions are there in the smaller values of $a,b,c,d$? (The simple form the solutions are the better.) Or can one prove or disprove that there are nontrivial solutions? (Even better, if we can get a quick algorithm to generate the solutions.)A non-trivial example or a proof of non-existence is required to be accepted as a final answer.
Note add:
My trials/attempts: Since it is encouraged to show one's own attempt.
First, it looks like if we have only $i=1,2$ instead of summing over $i=1,2,3$, we have $sum_i=1^2 a_i ^3=sum_i=1^2 b_i ^3 $ and $sum_i=1^3 a_i =sum_i=1^3 b_i$, the nontrivial solutions seem to be impossible (?), but no rigorous proof has be shown yet.
Let me share a few comments on what these two equations boil down to:
If we take (2)$^3$-(1), we get
$$
(a_1+a_2)(a_1+a_3)(a_2+a_3)=(b_1+b_2)(b_1+b_3)(b_2+b_3), tag3
$$
we can further use (2) and (3) to get a lot more interesting constraints. However, I haven't got any other linear relation constraint as useful as (2) yet.
number-theory polynomials prime-numbers arithmetic
asked Jul 17 at 1:56
wonderich
1,66521226
The goal is to understand a set of nontrivial solutions of cubic polynomials
$$
sum_i=1^3 a_i ^3=sum_i=1^3 b_i ^3
Rightarrow a_1^3+a_2^3+a_3^3=b_1^3+b_2^3+b_3^3, tag1
$$
$$
sum_i=1^3 a_i =sum_i=1^3 b_i Rightarrow a_1+a_2+a_3=b_1+b_2+b_3 , tag2
$$
for $a_i,b_iin mathbbZ$
where we demand $(a_1,a_2,a_3)neq (b_1,b_2,b_3)$ or any of its permutation (total $3!=6$ cases).
Question: Are there any nontrivial solutions? (e.g. $a_i,b_iin mathbbZ$
where we demand $(a_1,a_2,a_3)neq (b_1,b_2,b_3)$ and other $3!$ permutations.) How many simple but nontrivial solutions are there in the smaller values of $a,b,c,d$? (The simple form the solutions are the better.) Or can one prove or disprove that there are nontrivial solutions? (Even better, if we can get a quick algorithm to generate the solutions.)A non-trivial example or a proof of non-existence is required to be accepted as a final answer.
Note add:
My trials/attempts: Since it is encouraged to show one's own attempt.
First, it looks like if we have only $i=1,2$ instead of summing over $i=1,2,3$, we have $sum_i=1^2 a_i ^3=sum_i=1^2 b_i ^3 $ and $sum_i=1^3 a_i =sum_i=1^3 b_i$, the nontrivial solutions seem to be impossible (?), but no rigorous proof has be shown yet.
Let me share a few comments on what these two equations boil down to:
If we take (2)$^3$-(1), we get
$$
(a_1+a_2)(a_1+a_3)(a_2+a_3)=(b_1+b_2)(b_1+b_3)(b_2+b_3), tag3
$$
we can further use (2) and (3) to get a lot more interesting constraints. However, I haven't got any other linear relation constraint as useful as (2) yet.
number-theory polynomials prime-numbers arithmetic
asked Jul 17 at 1:56
wonderich
1,66521226
asked Jul 17 at 1:56
wonderich
1,66521226
asked Jul 17 at 1:56
wonderich
1,66521226
asked Jul 17 at 1:56
wonderich
1,66521226
1,66521226
math.stackexchange.com/questions/2593606/…
– individ
Jul 17 at 5:05
1
See $S 13$ Form: $a^k+b^k+c^k = d^k+e^k+f^k, k = 1,3$ of A Collection of Algeberaic identities collected by T.Piezas III.
– achille hui
Jul 17 at 14:56
add a comment |Â
math.stackexchange.com/questions/2593606/…
– individ
Jul 17 at 5:05
1
See $S 13$ Form: $a^k+b^k+c^k = d^k+e^k+f^k, k = 1,3$ of A Collection of Algeberaic identities collected by T.Piezas III.
– achille hui
Jul 17 at 14:56
math.stackexchange.com/questions/2593606/…
– individ
Jul 17 at 5:05
math.stackexchange.com/questions/2593606/…
– individ
Jul 17 at 5:05
1
1
See $S 13$ Form: $a^k+b^k+c^k = d^k+e^k+f^k, k = 1,3$ of A Collection of Algeberaic identities collected by T.Piezas III.
– achille hui
Jul 17 at 14:56
See $S 13$ Form: $a^k+b^k+c^k = d^k+e^k+f^k, k = 1,3$ of A Collection of Algeberaic identities collected by T.Piezas III.
– achille hui
Jul 17 at 14:56
add a comment |Â
4 Answers
4
active
oldest
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up vote
4
down vote
Some solutions:
$$matrixa_1 &= 1,; a_2 &= 5,; a_3 &= 5,; b_1 &= 2,; b_2 &= 3,; b_3 &= 6cr
a_1 &= 1,; a_2 &= 9,; a_3 &= 9,; b_1 &= 4,; b_2 &= 4,; b_3 &= 11cr
a_1 &= 1,; a_2 &= 10,; a_3 &= 14,; b_1 &= 3,; b_2 &= 7,; b_3 &= 15cr
a_1 &= 1,; a_2 &= 10,; a_3 &= 15,; b_1 &= 4,; b_2 &= 6,; b_3 &= 16cr
a_1 &= 2,; a_2 &= 8,; a_3 &= 10,; b_1 &= 4,; b_2 &= 5,; b_3 &= 11cr
a_1 &= 2,; a_2 &= 10,; a_3 &= 12,; b_1 &= 3,; b_2 &= 8,; b_3 &= 13cr
a_1 &= 2,; a_2 &= 10,; a_3 &= 10,; b_1 &= 4,; b_2 &= 6,; b_3 &= 12cr
a_1 &= 5,; a_2 &= 10,; a_3 &= 11,; b_1 &= 6,; b_2 &= 8,; b_3 &= 12cr
$$
EDIT: Here's a nontrivial infinite family of solutions (of course you can also multiply any solution by a constant factor).
$$a_1 = t, ; a_2 = 5 t, ; a_3 = 7 t - 2, ; b_1 = t + 1, ; b_2 = 5 t - 2,; b_3 = 7 t - 1 $$
answered Jul 17 at 2:58
Robert Israel
304k22201442
any generating formula or just brute force?
– Momo
Jul 17 at 2:59
Semi-brute. I looped over $1 le a_1 le a_2 le 10$ and $a_1 < b_1 le b_2 le 10$. Given these, solve the equation in the remaining two variables $a_3, b_3$, and if they are positive integers with $a_3 ge a_2$ and $b_3 ge b_2$ then output the result.
– Robert Israel
Jul 17 at 3:02
add a comment |Â
up vote
1
down vote
Here is one non-trivial solution:
$$ 1 + 5 + 5 = 2 + 3 + 6 $$
$$ 1^3 + 5^3 + 5^3 = 2^3 + 3^3 + 6^3 $$
The first line sums are $11$ and the second line sums are $251$.
answered Jul 17 at 2:32
hardmath
28.2k94592
This is good - how do you obtain it? +1 (hopefully not just brutal force!)
– wonderich
Jul 17 at 2:35
I like this solution - this looks really interesting.
– wonderich
Jul 17 at 2:36
I remembered a friend of mine sharing a problem like this with a small solution, something about rearranging a handful of coins into three piles so that the total number of coins is fixed, and the sums of the cubes of each pile is the same in both arrangements. So in a sense I had to reconstruct it by "brute force", knowing that a small solution would be found.
– hardmath
Jul 17 at 2:39
add a comment |Â
up vote
1
down vote
if you put
$alpha_1=a_2+a_3$
$alpha_2=a_1+a_3$
$alpha_3=a_1+a_2$
$beta_1=b_2+b_3$
$beta_2=b_1+b_3$
$beta_3=b_1+b_2$
Then your (3) becomes $alpha_1alpha_2alpha_3=beta_1beta_2beta_3$ and (2) becomes $alpha_1+alpha_2+alpha_3=beta_1+beta_2+beta_3$
Of course, some of the solutions of the system above give fractional $a_i,b_i$, since the determinant of the transformation is $2$. But if we get a fractional solution, we can multiply it by 2 to get an integer solution.
The link Two sets of 3 positive integers with equal sum and product will give you some examples of non-trivial $alpha_i,beta_i$
For example, the pairs $(8,12,15)$ and $(9,10,16)$ will generate the nontrivial solution of $(2.5,5.5,9.5)$ and $(1.5,7.5,8.5)$, which by doubling gives: $(5,11,19)$ and $(3,15,17)$. It would be interesting though if somebody would come with a closed-form formula to generate them.
answered Jul 17 at 2:27
Momo
11.9k21330
yes, thanks +1, will be nice to have nontrivial solutions and algorithm for them in integers.
– wonderich
Jul 17 at 2:29
thanks for the link
– wonderich
Jul 17 at 2:37
add a comment |Â
up vote
1
down vote
$$left{beginaligned&X_1+X_2+X_3=Y_1+Y_2+Y_3\&X_1^3+X_2^3+X_3^3=Y_1^3+Y_2^3+Y_3^3endalignedright.$$
$$X_1=(b+c-p)(a^2+a(b+c-p)+b(c-p))-cp(c-p)$$
$$X_2=(b+c-p)(a^2+a(b+c-3p)+b(c-p))+p(2p-2b-c)(c-p)$$
$$X_3=(b+c-p)(a^2-a(b+c-p)-b(c-p))+(2ab-2ap+cp)(c-p)$$
$$Y_1=(b+c-p)(a^2+a(c-b-p)+b(c-p))+(2bp-2b^2-cp)(c-p)$$
$$Y_2=(b+c-p)(a^2+a(b-c-p)+b(c-p))+c(p-2b)(c-p)$$
$$Y_3=(b+c-p)(a^2+a(b+c-p)+(b-2p)(c-p))+(2ab+cp-2ap)(c-p)$$
answered Jul 17 at 14:21
individ
3,1941715
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Some solutions:
$$matrixa_1 &= 1,; a_2 &= 5,; a_3 &= 5,; b_1 &= 2,; b_2 &= 3,; b_3 &= 6cr
a_1 &= 1,; a_2 &= 9,; a_3 &= 9,; b_1 &= 4,; b_2 &= 4,; b_3 &= 11cr
a_1 &= 1,; a_2 &= 10,; a_3 &= 14,; b_1 &= 3,; b_2 &= 7,; b_3 &= 15cr
a_1 &= 1,; a_2 &= 10,; a_3 &= 15,; b_1 &= 4,; b_2 &= 6,; b_3 &= 16cr
a_1 &= 2,; a_2 &= 8,; a_3 &= 10,; b_1 &= 4,; b_2 &= 5,; b_3 &= 11cr
a_1 &= 2,; a_2 &= 10,; a_3 &= 12,; b_1 &= 3,; b_2 &= 8,; b_3 &= 13cr
a_1 &= 2,; a_2 &= 10,; a_3 &= 10,; b_1 &= 4,; b_2 &= 6,; b_3 &= 12cr
a_1 &= 5,; a_2 &= 10,; a_3 &= 11,; b_1 &= 6,; b_2 &= 8,; b_3 &= 12cr
$$
EDIT: Here's a nontrivial infinite family of solutions (of course you can also multiply any solution by a constant factor).
$$a_1 = t, ; a_2 = 5 t, ; a_3 = 7 t - 2, ; b_1 = t + 1, ; b_2 = 5 t - 2,; b_3 = 7 t - 1 $$
answered Jul 17 at 2:58
Robert Israel
304k22201442
any generating formula or just brute force?
– Momo
Jul 17 at 2:59
Semi-brute. I looped over $1 le a_1 le a_2 le 10$ and $a_1 < b_1 le b_2 le 10$. Given these, solve the equation in the remaining two variables $a_3, b_3$, and if they are positive integers with $a_3 ge a_2$ and $b_3 ge b_2$ then output the result.
– Robert Israel
Jul 17 at 3:02
add a comment |Â
up vote
4
down vote
Some solutions:
$$matrixa_1 &= 1,; a_2 &= 5,; a_3 &= 5,; b_1 &= 2,; b_2 &= 3,; b_3 &= 6cr
a_1 &= 1,; a_2 &= 9,; a_3 &= 9,; b_1 &= 4,; b_2 &= 4,; b_3 &= 11cr
a_1 &= 1,; a_2 &= 10,; a_3 &= 14,; b_1 &= 3,; b_2 &= 7,; b_3 &= 15cr
a_1 &= 1,; a_2 &= 10,; a_3 &= 15,; b_1 &= 4,; b_2 &= 6,; b_3 &= 16cr
a_1 &= 2,; a_2 &= 8,; a_3 &= 10,; b_1 &= 4,; b_2 &= 5,; b_3 &= 11cr
a_1 &= 2,; a_2 &= 10,; a_3 &= 12,; b_1 &= 3,; b_2 &= 8,; b_3 &= 13cr
a_1 &= 2,; a_2 &= 10,; a_3 &= 10,; b_1 &= 4,; b_2 &= 6,; b_3 &= 12cr
a_1 &= 5,; a_2 &= 10,; a_3 &= 11,; b_1 &= 6,; b_2 &= 8,; b_3 &= 12cr
$$
EDIT: Here's a nontrivial infinite family of solutions (of course you can also multiply any solution by a constant factor).
$$a_1 = t, ; a_2 = 5 t, ; a_3 = 7 t - 2, ; b_1 = t + 1, ; b_2 = 5 t - 2,; b_3 = 7 t - 1 $$
answered Jul 17 at 2:58
Robert Israel
304k22201442
any generating formula or just brute force?
– Momo
Jul 17 at 2:59
Semi-brute. I looped over $1 le a_1 le a_2 le 10$ and $a_1 < b_1 le b_2 le 10$. Given these, solve the equation in the remaining two variables $a_3, b_3$, and if they are positive integers with $a_3 ge a_2$ and $b_3 ge b_2$ then output the result.
– Robert Israel
Jul 17 at 3:02
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Some solutions:
$$matrixa_1 &= 1,; a_2 &= 5,; a_3 &= 5,; b_1 &= 2,; b_2 &= 3,; b_3 &= 6cr
a_1 &= 1,; a_2 &= 9,; a_3 &= 9,; b_1 &= 4,; b_2 &= 4,; b_3 &= 11cr
a_1 &= 1,; a_2 &= 10,; a_3 &= 14,; b_1 &= 3,; b_2 &= 7,; b_3 &= 15cr
a_1 &= 1,; a_2 &= 10,; a_3 &= 15,; b_1 &= 4,; b_2 &= 6,; b_3 &= 16cr
a_1 &= 2,; a_2 &= 8,; a_3 &= 10,; b_1 &= 4,; b_2 &= 5,; b_3 &= 11cr
a_1 &= 2,; a_2 &= 10,; a_3 &= 12,; b_1 &= 3,; b_2 &= 8,; b_3 &= 13cr
a_1 &= 2,; a_2 &= 10,; a_3 &= 10,; b_1 &= 4,; b_2 &= 6,; b_3 &= 12cr
a_1 &= 5,; a_2 &= 10,; a_3 &= 11,; b_1 &= 6,; b_2 &= 8,; b_3 &= 12cr
$$
EDIT: Here's a nontrivial infinite family of solutions (of course you can also multiply any solution by a constant factor).
$$a_1 = t, ; a_2 = 5 t, ; a_3 = 7 t - 2, ; b_1 = t + 1, ; b_2 = 5 t - 2,; b_3 = 7 t - 1 $$
answered Jul 17 at 2:58
Robert Israel
304k22201442
Some solutions:
$$matrixa_1 &= 1,; a_2 &= 5,; a_3 &= 5,; b_1 &= 2,; b_2 &= 3,; b_3 &= 6cr
a_1 &= 1,; a_2 &= 9,; a_3 &= 9,; b_1 &= 4,; b_2 &= 4,; b_3 &= 11cr
a_1 &= 1,; a_2 &= 10,; a_3 &= 14,; b_1 &= 3,; b_2 &= 7,; b_3 &= 15cr
a_1 &= 1,; a_2 &= 10,; a_3 &= 15,; b_1 &= 4,; b_2 &= 6,; b_3 &= 16cr
a_1 &= 2,; a_2 &= 8,; a_3 &= 10,; b_1 &= 4,; b_2 &= 5,; b_3 &= 11cr
a_1 &= 2,; a_2 &= 10,; a_3 &= 12,; b_1 &= 3,; b_2 &= 8,; b_3 &= 13cr
a_1 &= 2,; a_2 &= 10,; a_3 &= 10,; b_1 &= 4,; b_2 &= 6,; b_3 &= 12cr
a_1 &= 5,; a_2 &= 10,; a_3 &= 11,; b_1 &= 6,; b_2 &= 8,; b_3 &= 12cr
$$
EDIT: Here's a nontrivial infinite family of solutions (of course you can also multiply any solution by a constant factor).
$$a_1 = t, ; a_2 = 5 t, ; a_3 = 7 t - 2, ; b_1 = t + 1, ; b_2 = 5 t - 2,; b_3 = 7 t - 1 $$
answered Jul 17 at 2:58
Robert Israel
304k22201442
edited Jul 17 at 7:03
answered Jul 17 at 2:58
Robert Israel
304k22201442
answered Jul 17 at 2:58
Robert Israel
304k22201442
answered Jul 17 at 2:58
Robert Israel
304k22201442
304k22201442
any generating formula or just brute force?
– Momo
Jul 17 at 2:59
Semi-brute. I looped over $1 le a_1 le a_2 le 10$ and $a_1 < b_1 le b_2 le 10$. Given these, solve the equation in the remaining two variables $a_3, b_3$, and if they are positive integers with $a_3 ge a_2$ and $b_3 ge b_2$ then output the result.
– Robert Israel
Jul 17 at 3:02
add a comment |Â
any generating formula or just brute force?
– Momo
Jul 17 at 2:59
Semi-brute. I looped over $1 le a_1 le a_2 le 10$ and $a_1 < b_1 le b_2 le 10$. Given these, solve the equation in the remaining two variables $a_3, b_3$, and if they are positive integers with $a_3 ge a_2$ and $b_3 ge b_2$ then output the result.
– Robert Israel
Jul 17 at 3:02
any generating formula or just brute force?
– Momo
Jul 17 at 2:59
any generating formula or just brute force?
– Momo
Jul 17 at 2:59
Semi-brute. I looped over $1 le a_1 le a_2 le 10$ and $a_1 < b_1 le b_2 le 10$. Given these, solve the equation in the remaining two variables $a_3, b_3$, and if they are positive integers with $a_3 ge a_2$ and $b_3 ge b_2$ then output the result.
– Robert Israel
Jul 17 at 3:02
Semi-brute. I looped over $1 le a_1 le a_2 le 10$ and $a_1 < b_1 le b_2 le 10$. Given these, solve the equation in the remaining two variables $a_3, b_3$, and if they are positive integers with $a_3 ge a_2$ and $b_3 ge b_2$ then output the result.
– Robert Israel
Jul 17 at 3:02
add a comment |Â
up vote
1
down vote
Here is one non-trivial solution:
$$ 1 + 5 + 5 = 2 + 3 + 6 $$
$$ 1^3 + 5^3 + 5^3 = 2^3 + 3^3 + 6^3 $$
The first line sums are $11$ and the second line sums are $251$.
answered Jul 17 at 2:32
hardmath
28.2k94592
This is good - how do you obtain it? +1 (hopefully not just brutal force!)
– wonderich
Jul 17 at 2:35
I like this solution - this looks really interesting.
– wonderich
Jul 17 at 2:36
I remembered a friend of mine sharing a problem like this with a small solution, something about rearranging a handful of coins into three piles so that the total number of coins is fixed, and the sums of the cubes of each pile is the same in both arrangements. So in a sense I had to reconstruct it by "brute force", knowing that a small solution would be found.
– hardmath
Jul 17 at 2:39
add a comment |Â
up vote
1
down vote
Here is one non-trivial solution:
$$ 1 + 5 + 5 = 2 + 3 + 6 $$
$$ 1^3 + 5^3 + 5^3 = 2^3 + 3^3 + 6^3 $$
The first line sums are $11$ and the second line sums are $251$.
answered Jul 17 at 2:32
hardmath
28.2k94592
This is good - how do you obtain it? +1 (hopefully not just brutal force!)
– wonderich
Jul 17 at 2:35
I like this solution - this looks really interesting.
– wonderich
Jul 17 at 2:36
I remembered a friend of mine sharing a problem like this with a small solution, something about rearranging a handful of coins into three piles so that the total number of coins is fixed, and the sums of the cubes of each pile is the same in both arrangements. So in a sense I had to reconstruct it by "brute force", knowing that a small solution would be found.
– hardmath
Jul 17 at 2:39
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here is one non-trivial solution:
$$ 1 + 5 + 5 = 2 + 3 + 6 $$
$$ 1^3 + 5^3 + 5^3 = 2^3 + 3^3 + 6^3 $$
The first line sums are $11$ and the second line sums are $251$.
answered Jul 17 at 2:32
hardmath
28.2k94592
Here is one non-trivial solution:
$$ 1 + 5 + 5 = 2 + 3 + 6 $$
$$ 1^3 + 5^3 + 5^3 = 2^3 + 3^3 + 6^3 $$
The first line sums are $11$ and the second line sums are $251$.
answered Jul 17 at 2:32
hardmath
28.2k94592
answered Jul 17 at 2:32
hardmath
28.2k94592
answered Jul 17 at 2:32
hardmath
28.2k94592
answered Jul 17 at 2:32
hardmath
28.2k94592
28.2k94592
This is good - how do you obtain it? +1 (hopefully not just brutal force!)
– wonderich
Jul 17 at 2:35
I like this solution - this looks really interesting.
– wonderich
Jul 17 at 2:36
I remembered a friend of mine sharing a problem like this with a small solution, something about rearranging a handful of coins into three piles so that the total number of coins is fixed, and the sums of the cubes of each pile is the same in both arrangements. So in a sense I had to reconstruct it by "brute force", knowing that a small solution would be found.
– hardmath
Jul 17 at 2:39
add a comment |Â
This is good - how do you obtain it? +1 (hopefully not just brutal force!)
– wonderich
Jul 17 at 2:35
I like this solution - this looks really interesting.
– wonderich
Jul 17 at 2:36
I remembered a friend of mine sharing a problem like this with a small solution, something about rearranging a handful of coins into three piles so that the total number of coins is fixed, and the sums of the cubes of each pile is the same in both arrangements. So in a sense I had to reconstruct it by "brute force", knowing that a small solution would be found.
– hardmath
Jul 17 at 2:39
This is good - how do you obtain it? +1 (hopefully not just brutal force!)
– wonderich
Jul 17 at 2:35
This is good - how do you obtain it? +1 (hopefully not just brutal force!)
– wonderich
Jul 17 at 2:35
I like this solution - this looks really interesting.
– wonderich
Jul 17 at 2:36
I like this solution - this looks really interesting.
– wonderich
Jul 17 at 2:36
I remembered a friend of mine sharing a problem like this with a small solution, something about rearranging a handful of coins into three piles so that the total number of coins is fixed, and the sums of the cubes of each pile is the same in both arrangements. So in a sense I had to reconstruct it by "brute force", knowing that a small solution would be found.
– hardmath
Jul 17 at 2:39
I remembered a friend of mine sharing a problem like this with a small solution, something about rearranging a handful of coins into three piles so that the total number of coins is fixed, and the sums of the cubes of each pile is the same in both arrangements. So in a sense I had to reconstruct it by "brute force", knowing that a small solution would be found.
– hardmath
Jul 17 at 2:39
add a comment |Â
up vote
1
down vote
if you put
$alpha_1=a_2+a_3$
$alpha_2=a_1+a_3$
$alpha_3=a_1+a_2$
$beta_1=b_2+b_3$
$beta_2=b_1+b_3$
$beta_3=b_1+b_2$
Then your (3) becomes $alpha_1alpha_2alpha_3=beta_1beta_2beta_3$ and (2) becomes $alpha_1+alpha_2+alpha_3=beta_1+beta_2+beta_3$
Of course, some of the solutions of the system above give fractional $a_i,b_i$, since the determinant of the transformation is $2$. But if we get a fractional solution, we can multiply it by 2 to get an integer solution.
The link Two sets of 3 positive integers with equal sum and product will give you some examples of non-trivial $alpha_i,beta_i$
For example, the pairs $(8,12,15)$ and $(9,10,16)$ will generate the nontrivial solution of $(2.5,5.5,9.5)$ and $(1.5,7.5,8.5)$, which by doubling gives: $(5,11,19)$ and $(3,15,17)$. It would be interesting though if somebody would come with a closed-form formula to generate them.
answered Jul 17 at 2:27
Momo
11.9k21330
yes, thanks +1, will be nice to have nontrivial solutions and algorithm for them in integers.
– wonderich
Jul 17 at 2:29
thanks for the link
– wonderich
Jul 17 at 2:37
add a comment |Â
up vote
1
down vote
if you put
$alpha_1=a_2+a_3$
$alpha_2=a_1+a_3$
$alpha_3=a_1+a_2$
$beta_1=b_2+b_3$
$beta_2=b_1+b_3$
$beta_3=b_1+b_2$
Then your (3) becomes $alpha_1alpha_2alpha_3=beta_1beta_2beta_3$ and (2) becomes $alpha_1+alpha_2+alpha_3=beta_1+beta_2+beta_3$
Of course, some of the solutions of the system above give fractional $a_i,b_i$, since the determinant of the transformation is $2$. But if we get a fractional solution, we can multiply it by 2 to get an integer solution.
The link Two sets of 3 positive integers with equal sum and product will give you some examples of non-trivial $alpha_i,beta_i$
For example, the pairs $(8,12,15)$ and $(9,10,16)$ will generate the nontrivial solution of $(2.5,5.5,9.5)$ and $(1.5,7.5,8.5)$, which by doubling gives: $(5,11,19)$ and $(3,15,17)$. It would be interesting though if somebody would come with a closed-form formula to generate them.
answered Jul 17 at 2:27
Momo
11.9k21330
yes, thanks +1, will be nice to have nontrivial solutions and algorithm for them in integers.
– wonderich
Jul 17 at 2:29
thanks for the link
– wonderich
Jul 17 at 2:37
add a comment |Â
up vote
1
down vote
up vote
1
down vote
if you put
$alpha_1=a_2+a_3$
$alpha_2=a_1+a_3$
$alpha_3=a_1+a_2$
$beta_1=b_2+b_3$
$beta_2=b_1+b_3$
$beta_3=b_1+b_2$
Then your (3) becomes $alpha_1alpha_2alpha_3=beta_1beta_2beta_3$ and (2) becomes $alpha_1+alpha_2+alpha_3=beta_1+beta_2+beta_3$
Of course, some of the solutions of the system above give fractional $a_i,b_i$, since the determinant of the transformation is $2$. But if we get a fractional solution, we can multiply it by 2 to get an integer solution.
The link Two sets of 3 positive integers with equal sum and product will give you some examples of non-trivial $alpha_i,beta_i$
For example, the pairs $(8,12,15)$ and $(9,10,16)$ will generate the nontrivial solution of $(2.5,5.5,9.5)$ and $(1.5,7.5,8.5)$, which by doubling gives: $(5,11,19)$ and $(3,15,17)$. It would be interesting though if somebody would come with a closed-form formula to generate them.
answered Jul 17 at 2:27
Momo
11.9k21330
if you put
$alpha_1=a_2+a_3$
$alpha_2=a_1+a_3$
$alpha_3=a_1+a_2$
$beta_1=b_2+b_3$
$beta_2=b_1+b_3$
$beta_3=b_1+b_2$
Then your (3) becomes $alpha_1alpha_2alpha_3=beta_1beta_2beta_3$ and (2) becomes $alpha_1+alpha_2+alpha_3=beta_1+beta_2+beta_3$
Of course, some of the solutions of the system above give fractional $a_i,b_i$, since the determinant of the transformation is $2$. But if we get a fractional solution, we can multiply it by 2 to get an integer solution.
The link Two sets of 3 positive integers with equal sum and product will give you some examples of non-trivial $alpha_i,beta_i$
For example, the pairs $(8,12,15)$ and $(9,10,16)$ will generate the nontrivial solution of $(2.5,5.5,9.5)$ and $(1.5,7.5,8.5)$, which by doubling gives: $(5,11,19)$ and $(3,15,17)$. It would be interesting though if somebody would come with a closed-form formula to generate them.
answered Jul 17 at 2:27
Momo
11.9k21330
edited Jul 17 at 2:52
answered Jul 17 at 2:27
Momo
11.9k21330
answered Jul 17 at 2:27
Momo
11.9k21330
answered Jul 17 at 2:27
Momo
11.9k21330
11.9k21330
yes, thanks +1, will be nice to have nontrivial solutions and algorithm for them in integers.
– wonderich
Jul 17 at 2:29
thanks for the link
– wonderich
Jul 17 at 2:37
add a comment |Â
yes, thanks +1, will be nice to have nontrivial solutions and algorithm for them in integers.
– wonderich
Jul 17 at 2:29
thanks for the link
– wonderich
Jul 17 at 2:37
yes, thanks +1, will be nice to have nontrivial solutions and algorithm for them in integers.
– wonderich
Jul 17 at 2:29
yes, thanks +1, will be nice to have nontrivial solutions and algorithm for them in integers.
– wonderich
Jul 17 at 2:29
thanks for the link
– wonderich
Jul 17 at 2:37
thanks for the link
– wonderich
Jul 17 at 2:37
add a comment |Â
up vote
1
down vote
$$left{beginaligned&X_1+X_2+X_3=Y_1+Y_2+Y_3\&X_1^3+X_2^3+X_3^3=Y_1^3+Y_2^3+Y_3^3endalignedright.$$
$$X_1=(b+c-p)(a^2+a(b+c-p)+b(c-p))-cp(c-p)$$
$$X_2=(b+c-p)(a^2+a(b+c-3p)+b(c-p))+p(2p-2b-c)(c-p)$$
$$X_3=(b+c-p)(a^2-a(b+c-p)-b(c-p))+(2ab-2ap+cp)(c-p)$$
$$Y_1=(b+c-p)(a^2+a(c-b-p)+b(c-p))+(2bp-2b^2-cp)(c-p)$$
$$Y_2=(b+c-p)(a^2+a(b-c-p)+b(c-p))+c(p-2b)(c-p)$$
$$Y_3=(b+c-p)(a^2+a(b+c-p)+(b-2p)(c-p))+(2ab+cp-2ap)(c-p)$$
answered Jul 17 at 14:21
individ
3,1941715
add a comment |Â
up vote
1
down vote
$$left{beginaligned&X_1+X_2+X_3=Y_1+Y_2+Y_3\&X_1^3+X_2^3+X_3^3=Y_1^3+Y_2^3+Y_3^3endalignedright.$$
$$X_1=(b+c-p)(a^2+a(b+c-p)+b(c-p))-cp(c-p)$$
$$X_2=(b+c-p)(a^2+a(b+c-3p)+b(c-p))+p(2p-2b-c)(c-p)$$
$$X_3=(b+c-p)(a^2-a(b+c-p)-b(c-p))+(2ab-2ap+cp)(c-p)$$
$$Y_1=(b+c-p)(a^2+a(c-b-p)+b(c-p))+(2bp-2b^2-cp)(c-p)$$
$$Y_2=(b+c-p)(a^2+a(b-c-p)+b(c-p))+c(p-2b)(c-p)$$
$$Y_3=(b+c-p)(a^2+a(b+c-p)+(b-2p)(c-p))+(2ab+cp-2ap)(c-p)$$
answered Jul 17 at 14:21
individ
3,1941715
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$left{beginaligned&X_1+X_2+X_3=Y_1+Y_2+Y_3\&X_1^3+X_2^3+X_3^3=Y_1^3+Y_2^3+Y_3^3endalignedright.$$
$$X_1=(b+c-p)(a^2+a(b+c-p)+b(c-p))-cp(c-p)$$
$$X_2=(b+c-p)(a^2+a(b+c-3p)+b(c-p))+p(2p-2b-c)(c-p)$$
$$X_3=(b+c-p)(a^2-a(b+c-p)-b(c-p))+(2ab-2ap+cp)(c-p)$$
$$Y_1=(b+c-p)(a^2+a(c-b-p)+b(c-p))+(2bp-2b^2-cp)(c-p)$$
$$Y_2=(b+c-p)(a^2+a(b-c-p)+b(c-p))+c(p-2b)(c-p)$$
$$Y_3=(b+c-p)(a^2+a(b+c-p)+(b-2p)(c-p))+(2ab+cp-2ap)(c-p)$$
answered Jul 17 at 14:21
individ
3,1941715
$$left{beginaligned&X_1+X_2+X_3=Y_1+Y_2+Y_3\&X_1^3+X_2^3+X_3^3=Y_1^3+Y_2^3+Y_3^3endalignedright.$$
$$X_1=(b+c-p)(a^2+a(b+c-p)+b(c-p))-cp(c-p)$$
$$X_2=(b+c-p)(a^2+a(b+c-3p)+b(c-p))+p(2p-2b-c)(c-p)$$
$$X_3=(b+c-p)(a^2-a(b+c-p)-b(c-p))+(2ab-2ap+cp)(c-p)$$
$$Y_1=(b+c-p)(a^2+a(c-b-p)+b(c-p))+(2bp-2b^2-cp)(c-p)$$
$$Y_2=(b+c-p)(a^2+a(b-c-p)+b(c-p))+c(p-2b)(c-p)$$
$$Y_3=(b+c-p)(a^2+a(b+c-p)+(b-2p)(c-p))+(2ab+cp-2ap)(c-p)$$
answered Jul 17 at 14:21
individ
3,1941715
answered Jul 17 at 14:21
individ
3,1941715
answered Jul 17 at 14:21
individ
3,1941715
answered Jul 17 at 14:21
individ
3,1941715
3,1941715
add a comment |Â
add a comment |Â
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math.stackexchange.com/questions/2593606/…
– individ
Jul 17 at 5:05
1
See $S 13$ Form: $a^k+b^k+c^k = d^k+e^k+f^k, k = 1,3$ of A Collection of Algeberaic identities collected by T.Piezas III.
– achille hui
Jul 17 at 14:56