Mixing
Solve $a^3 + b^3 + c^3 = 6abc$
$begingroup$
Find solutions for $a^3 + b^3 + c^3 = 6abc$ in $mathbb{N}$, such that $gcd(a,b,c) = 1$, except for $(1,2,3)$ and its permutations.
Using trial and error I found out that if $a,b,c$ are solution of the equation, then they are in arithmetic progression. I've managed to prove that conjecture, assuming that $c>b>a$ and let $k$ be their common difference in the arithmetic progression. Then WLOG we have:
$$b = c-k quad quad a = c-2k$$
Now the equation looks like:
$$(c-2k)^3 + (c-k)^3 + c^3 = 6(c-2)(c-1)c$$
After expanding we have:
$$c^3 - 6kc^2 + 12ck^2 - 8k^3 + c^3 - 3kc^2 + 3ck^2 -k^3 + c^3 = 6c^3 - 18kc^2 + 12ck^2$$
$$3c^3 - 9kc^2 + 15ck^2 - 9k^3 = 6c^3 - 18kc^2 + 12ck^2$$
$$c^3 - 3kc^2 + 5ck^2 - 3k^3 = 2c^3 - 6kc^2 + 4ck^2$$
$$-c^3 + 3kc^2 + ck^2 - 3k^3 = 0$$
Now it's easy to see that if $k=c$, then the LHS will be zero, so one of the zeroes of the polynomial is $c_1 = k$, now factorizing we have:
$$(c-k)(3a^2 + 2ax - x^2) = 0$$
$$(c-k)(c+k)(c-3k) = 0$$
Now we have three distinct cases:
Case 1: $c = k$
This implies that $b = 0$ and $a = -k$. But because $k in mathbb{N}$, both $a,b notin mathbb{N}$, violating the initial conditions.
Case 2: $c = -k$
Obviously the initial condition is already violated, becasue $k in mathbb{N}$, so from the relation $c notin mathbb{N}$
Case 2: $c = 3k$
This implies that $b = 2k$ and $a = k$. Now we have one 3-tuple $(3k,2k,k)$ and it's permutation as solution, where $k in mathbb{N}$. But it's easy to note that $k$ is a common factor for $a,b,c$ so we have:
$$gcd(a,b,c) = k$$
But because we want $gcd(a,b,c) = 1$, this implies that $k=1$, which means we have only one solution for $a^3 + b^3 + c^3 = 6abc$ in $mathbb{N}$, such that $gcd(a,b,c) = 1$ and it $(1,2,3)$, solution that is already given.
Now my question is what I'm missing. Is there really no other solutions such that $gcd(a,b,c) = 1$? Or maybe there is a different way to obtain solution except for my method using arithmetic progression?
sequences-and-series polynomials factoring divisibility
asked Sep 25 '13 at 0:21
Stefan4024Stefan4024
30.6k63579
$endgroup$
|
show 2 more comments
$begingroup$
Find solutions for $a^3 + b^3 + c^3 = 6abc$ in $mathbb{N}$, such that $gcd(a,b,c) = 1$, except for $(1,2,3)$ and its permutations.
Using trial and error I found out that if $a,b,c$ are solution of the equation, then they are in arithmetic progression. I've managed to prove that conjecture, assuming that $c>b>a$ and let $k$ be their common difference in the arithmetic progression. Then WLOG we have:
$$b = c-k quad quad a = c-2k$$
Now the equation looks like:
$$(c-2k)^3 + (c-k)^3 + c^3 = 6(c-2)(c-1)c$$
After expanding we have:
$$c^3 - 6kc^2 + 12ck^2 - 8k^3 + c^3 - 3kc^2 + 3ck^2 -k^3 + c^3 = 6c^3 - 18kc^2 + 12ck^2$$
$$3c^3 - 9kc^2 + 15ck^2 - 9k^3 = 6c^3 - 18kc^2 + 12ck^2$$
$$c^3 - 3kc^2 + 5ck^2 - 3k^3 = 2c^3 - 6kc^2 + 4ck^2$$
$$-c^3 + 3kc^2 + ck^2 - 3k^3 = 0$$
Now it's easy to see that if $k=c$, then the LHS will be zero, so one of the zeroes of the polynomial is $c_1 = k$, now factorizing we have:
$$(c-k)(3a^2 + 2ax - x^2) = 0$$
$$(c-k)(c+k)(c-3k) = 0$$
Now we have three distinct cases:
Case 1: $c = k$
This implies that $b = 0$ and $a = -k$. But because $k in mathbb{N}$, both $a,b notin mathbb{N}$, violating the initial conditions.
Case 2: $c = -k$
Obviously the initial condition is already violated, becasue $k in mathbb{N}$, so from the relation $c notin mathbb{N}$
Case 2: $c = 3k$
This implies that $b = 2k$ and $a = k$. Now we have one 3-tuple $(3k,2k,k)$ and it's permutation as solution, where $k in mathbb{N}$. But it's easy to note that $k$ is a common factor for $a,b,c$ so we have:
$$gcd(a,b,c) = k$$
But because we want $gcd(a,b,c) = 1$, this implies that $k=1$, which means we have only one solution for $a^3 + b^3 + c^3 = 6abc$ in $mathbb{N}$, such that $gcd(a,b,c) = 1$ and it $(1,2,3)$, solution that is already given.
Now my question is what I'm missing. Is there really no other solutions such that $gcd(a,b,c) = 1$? Or maybe there is a different way to obtain solution except for my method using arithmetic progression?
sequences-and-series polynomials factoring divisibility
asked Sep 25 '13 at 0:21
Stefan4024Stefan4024
30.6k63579
$endgroup$
$begingroup$
saying that $a,b,c$ is an arithmetic progression satisfies the conditions is not correct, since $1,4,7$ doesn't satisfy the equation. What you mean is that ${ a, b, c } = {k, 2k, 3k } $, which agrees with your cases.
$endgroup$
– Calvin Lin
Sep 25 '13 at 0:50
$begingroup$
My bad. I didn't mean that. I meaned if $a,b,c$ are solution then it implies $a,b,c$ are in arithmetic progression, but it doesn't mean that every arithmetic progression will provide a solution. And what are you thoughts is there another solution satisfying the initial condition. Is there any other way to obtain it? Or there is just one?
$endgroup$
– Stefan4024
Sep 25 '13 at 0:59
$begingroup$
If the arithmetic progression condition is correct, then $(1,2,3)$ is the only primitive solution.
$endgroup$
– Daniel Fischer
Sep 25 '13 at 1:34
$begingroup$
I know that, but is there another way to generate (a,b,c) as integer solution except for arithemtic progression.
$endgroup$
– Stefan4024
Sep 25 '13 at 1:35
$begingroup$
@DanielFischer Is there any way to prove that at least one $a,b,c$ must be divisible by 3? If that's proven then it's easy to prove that (1,2,3) is the only primitive solution.
$endgroup$
– Stefan4024
Sep 25 '13 at 2:13
|
show 2 more comments
$begingroup$
Find solutions for $a^3 + b^3 + c^3 = 6abc$ in $mathbb{N}$, such that $gcd(a,b,c) = 1$, except for $(1,2,3)$ and its permutations.
Using trial and error I found out that if $a,b,c$ are solution of the equation, then they are in arithmetic progression. I've managed to prove that conjecture, assuming that $c>b>a$ and let $k$ be their common difference in the arithmetic progression. Then WLOG we have:
$$b = c-k quad quad a = c-2k$$
Now the equation looks like:
$$(c-2k)^3 + (c-k)^3 + c^3 = 6(c-2)(c-1)c$$
After expanding we have:
$$c^3 - 6kc^2 + 12ck^2 - 8k^3 + c^3 - 3kc^2 + 3ck^2 -k^3 + c^3 = 6c^3 - 18kc^2 + 12ck^2$$
$$3c^3 - 9kc^2 + 15ck^2 - 9k^3 = 6c^3 - 18kc^2 + 12ck^2$$
$$c^3 - 3kc^2 + 5ck^2 - 3k^3 = 2c^3 - 6kc^2 + 4ck^2$$
$$-c^3 + 3kc^2 + ck^2 - 3k^3 = 0$$
Now it's easy to see that if $k=c$, then the LHS will be zero, so one of the zeroes of the polynomial is $c_1 = k$, now factorizing we have:
$$(c-k)(3a^2 + 2ax - x^2) = 0$$
$$(c-k)(c+k)(c-3k) = 0$$
Now we have three distinct cases:
Case 1: $c = k$
This implies that $b = 0$ and $a = -k$. But because $k in mathbb{N}$, both $a,b notin mathbb{N}$, violating the initial conditions.
Case 2: $c = -k$
Obviously the initial condition is already violated, becasue $k in mathbb{N}$, so from the relation $c notin mathbb{N}$
Case 2: $c = 3k$
This implies that $b = 2k$ and $a = k$. Now we have one 3-tuple $(3k,2k,k)$ and it's permutation as solution, where $k in mathbb{N}$. But it's easy to note that $k$ is a common factor for $a,b,c$ so we have:
$$gcd(a,b,c) = k$$
But because we want $gcd(a,b,c) = 1$, this implies that $k=1$, which means we have only one solution for $a^3 + b^3 + c^3 = 6abc$ in $mathbb{N}$, such that $gcd(a,b,c) = 1$ and it $(1,2,3)$, solution that is already given.
Now my question is what I'm missing. Is there really no other solutions such that $gcd(a,b,c) = 1$? Or maybe there is a different way to obtain solution except for my method using arithmetic progression?
sequences-and-series polynomials factoring divisibility
asked Sep 25 '13 at 0:21
Stefan4024Stefan4024
30.6k63579
$endgroup$
Find solutions for $a^3 + b^3 + c^3 = 6abc$ in $mathbb{N}$, such that $gcd(a,b,c) = 1$, except for $(1,2,3)$ and its permutations.
Using trial and error I found out that if $a,b,c$ are solution of the equation, then they are in arithmetic progression. I've managed to prove that conjecture, assuming that $c>b>a$ and let $k$ be their common difference in the arithmetic progression. Then WLOG we have:
$$b = c-k quad quad a = c-2k$$
Now the equation looks like:
$$(c-2k)^3 + (c-k)^3 + c^3 = 6(c-2)(c-1)c$$
After expanding we have:
$$c^3 - 6kc^2 + 12ck^2 - 8k^3 + c^3 - 3kc^2 + 3ck^2 -k^3 + c^3 = 6c^3 - 18kc^2 + 12ck^2$$
$$3c^3 - 9kc^2 + 15ck^2 - 9k^3 = 6c^3 - 18kc^2 + 12ck^2$$
$$c^3 - 3kc^2 + 5ck^2 - 3k^3 = 2c^3 - 6kc^2 + 4ck^2$$
$$-c^3 + 3kc^2 + ck^2 - 3k^3 = 0$$
Now it's easy to see that if $k=c$, then the LHS will be zero, so one of the zeroes of the polynomial is $c_1 = k$, now factorizing we have:
$$(c-k)(3a^2 + 2ax - x^2) = 0$$
$$(c-k)(c+k)(c-3k) = 0$$
Now we have three distinct cases:
Case 1: $c = k$
This implies that $b = 0$ and $a = -k$. But because $k in mathbb{N}$, both $a,b notin mathbb{N}$, violating the initial conditions.
Case 2: $c = -k$
Obviously the initial condition is already violated, becasue $k in mathbb{N}$, so from the relation $c notin mathbb{N}$
Case 2: $c = 3k$
This implies that $b = 2k$ and $a = k$. Now we have one 3-tuple $(3k,2k,k)$ and it's permutation as solution, where $k in mathbb{N}$. But it's easy to note that $k$ is a common factor for $a,b,c$ so we have:
$$gcd(a,b,c) = k$$
But because we want $gcd(a,b,c) = 1$, this implies that $k=1$, which means we have only one solution for $a^3 + b^3 + c^3 = 6abc$ in $mathbb{N}$, such that $gcd(a,b,c) = 1$ and it $(1,2,3)$, solution that is already given.
Now my question is what I'm missing. Is there really no other solutions such that $gcd(a,b,c) = 1$? Or maybe there is a different way to obtain solution except for my method using arithmetic progression?
sequences-and-series polynomials factoring divisibility
sequences-and-series polynomials factoring divisibility
asked Sep 25 '13 at 0:21
Stefan4024Stefan4024
30.6k63579
asked Sep 25 '13 at 0:21
Stefan4024Stefan4024
30.6k63579
edited Sep 25 '13 at 1:00
Stefan4024
asked Sep 25 '13 at 0:21
Stefan4024Stefan4024
30.6k63579
asked Sep 25 '13 at 0:21
Stefan4024Stefan4024
30.6k63579
asked Sep 25 '13 at 0:21
Stefan4024Stefan4024
30.6k63579
30.6k63579
$begingroup$
saying that $a,b,c$ is an arithmetic progression satisfies the conditions is not correct, since $1,4,7$ doesn't satisfy the equation. What you mean is that ${ a, b, c } = {k, 2k, 3k } $, which agrees with your cases.
$endgroup$
– Calvin Lin
Sep 25 '13 at 0:50
$begingroup$
My bad. I didn't mean that. I meaned if $a,b,c$ are solution then it implies $a,b,c$ are in arithmetic progression, but it doesn't mean that every arithmetic progression will provide a solution. And what are you thoughts is there another solution satisfying the initial condition. Is there any other way to obtain it? Or there is just one?
$endgroup$
– Stefan4024
Sep 25 '13 at 0:59
$begingroup$
If the arithmetic progression condition is correct, then $(1,2,3)$ is the only primitive solution.
$endgroup$
– Daniel Fischer
Sep 25 '13 at 1:34
$begingroup$
I know that, but is there another way to generate (a,b,c) as integer solution except for arithemtic progression.
$endgroup$
– Stefan4024
Sep 25 '13 at 1:35
$begingroup$
@DanielFischer Is there any way to prove that at least one $a,b,c$ must be divisible by 3? If that's proven then it's easy to prove that (1,2,3) is the only primitive solution.
$endgroup$
– Stefan4024
Sep 25 '13 at 2:13
|
show 2 more comments
$begingroup$
saying that $a,b,c$ is an arithmetic progression satisfies the conditions is not correct, since $1,4,7$ doesn't satisfy the equation. What you mean is that ${ a, b, c } = {k, 2k, 3k } $, which agrees with your cases.
$endgroup$
– Calvin Lin
Sep 25 '13 at 0:50
$begingroup$
My bad. I didn't mean that. I meaned if $a,b,c$ are solution then it implies $a,b,c$ are in arithmetic progression, but it doesn't mean that every arithmetic progression will provide a solution. And what are you thoughts is there another solution satisfying the initial condition. Is there any other way to obtain it? Or there is just one?
$endgroup$
– Stefan4024
Sep 25 '13 at 0:59
$begingroup$
If the arithmetic progression condition is correct, then $(1,2,3)$ is the only primitive solution.
$endgroup$
– Daniel Fischer
Sep 25 '13 at 1:34
$begingroup$
I know that, but is there another way to generate (a,b,c) as integer solution except for arithemtic progression.
$endgroup$
– Stefan4024
Sep 25 '13 at 1:35
$begingroup$
@DanielFischer Is there any way to prove that at least one $a,b,c$ must be divisible by 3? If that's proven then it's easy to prove that (1,2,3) is the only primitive solution.
$endgroup$
– Stefan4024
Sep 25 '13 at 2:13
$begingroup$
saying that $a,b,c$ is an arithmetic progression satisfies the conditions is not correct, since $1,4,7$ doesn't satisfy the equation. What you mean is that ${ a, b, c } = {k, 2k, 3k } $, which agrees with your cases.
$endgroup$
– Calvin Lin
Sep 25 '13 at 0:50
$begingroup$
saying that $a,b,c$ is an arithmetic progression satisfies the conditions is not correct, since $1,4,7$ doesn't satisfy the equation. What you mean is that ${ a, b, c } = {k, 2k, 3k } $, which agrees with your cases.
$endgroup$
– Calvin Lin
Sep 25 '13 at 0:50
$begingroup$
My bad. I didn't mean that. I meaned if $a,b,c$ are solution then it implies $a,b,c$ are in arithmetic progression, but it doesn't mean that every arithmetic progression will provide a solution. And what are you thoughts is there another solution satisfying the initial condition. Is there any other way to obtain it? Or there is just one?
$endgroup$
– Stefan4024
Sep 25 '13 at 0:59
$begingroup$
My bad. I didn't mean that. I meaned if $a,b,c$ are solution then it implies $a,b,c$ are in arithmetic progression, but it doesn't mean that every arithmetic progression will provide a solution. And what are you thoughts is there another solution satisfying the initial condition. Is there any other way to obtain it? Or there is just one?
$endgroup$
– Stefan4024
Sep 25 '13 at 0:59
$begingroup$
If the arithmetic progression condition is correct, then $(1,2,3)$ is the only primitive solution.
$endgroup$
– Daniel Fischer
Sep 25 '13 at 1:34
$begingroup$
If the arithmetic progression condition is correct, then $(1,2,3)$ is the only primitive solution.
$endgroup$
– Daniel Fischer
Sep 25 '13 at 1:34
$begingroup$
I know that, but is there another way to generate (a,b,c) as integer solution except for arithemtic progression.
$endgroup$
– Stefan4024
Sep 25 '13 at 1:35
$begingroup$
I know that, but is there another way to generate (a,b,c) as integer solution except for arithemtic progression.
$endgroup$
– Stefan4024
Sep 25 '13 at 1:35
$begingroup$
@DanielFischer Is there any way to prove that at least one $a,b,c$ must be divisible by 3? If that's proven then it's easy to prove that (1,2,3) is the only primitive solution.
$endgroup$
– Stefan4024
Sep 25 '13 at 2:13
$begingroup$
@DanielFischer Is there any way to prove that at least one $a,b,c$ must be divisible by 3? If that's proven then it's easy to prove that (1,2,3) is the only primitive solution.
$endgroup$
– Stefan4024
Sep 25 '13 at 2:13
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
It's easy to prove that at least one of the variables needs to be an even number. We know that:
$$6|(n-1)n(n+1)$$
Because in three consecutive numbers, one is divisible with three and at least one is divisible with 2. So we have:
$$6|n^3 - n$$
$$n^3 = n pmod 6$$
Now we have:
$$a^3 + b^3 + c^3 equiv a + b + c equiv 0 pmod 6$$
Beacuse the modulo is an even number that means that the sum $a+b+c$ is an even number also. We know that the sum of 3 odd numbers will be odd number, so it's impossible $a,b,c$ to be odd number, because there won't be solution. So it means that at least one of the variables is an even number.
WLOG we can set $b=2k$. Now we can continue:
$$b-2k = 0$$
Now we can multiply both sides with $b(b+2k)$. Note that won't give another solution, because it'll imply that b is $0$ or a negative number, which violate the condition. Now we have:
$$b(b+2k)(b-2k) = 0$$
$$b(b^2 - 4k^2) = 0$$
$$b^3 - 4bk^2 = 0$$
$$3b^3 - 12bk^2 = 0$$
$$6b^3 - 6bk^2 = 3b^3 + 6bk^2 = 0$$
$$6b(b^2 - k^2) = (b^2 - 3kb^2 + 3bk^2 - k^3) + b^3 + (b^3 + 3kb^2 + 3bk^2 + k^3) = 0$$
$$6b(b-k)(b+k) = (b-k)^3 + b^3 + (b+k)^3$$
Now if we substitute WLOG:
$$b+k=c quad quad b-k=a$$
$$6abc = a^3 + b^3 + c^3$$
Because $b,k in mathbb{N}$ it means that also $a,c in mathbb{N}$. So this proves that for any $b=2k$, there are integer solutions, such $a=k$ and $c=3k$.
But because $k$ is a factor of all of them it's easy to see that:
$$gcd(a,b,c) = gcd(k,2k,3k) = k$$
Because we want $gcd(a,b,c) = 1$, that implies that $k=1$ and that the only primitive solution of this equation is $(1,2,3)$ and its permutation.
answered Sep 25 '13 at 8:38
Stefan4024Stefan4024
30.6k63579
$endgroup$
1
$begingroup$
This doesn't explain how to find all triples of solutions. Note that right now, you have may merely used algebraic manipulations, without any serious number theory. There is no reason why the answers must appear in an arithmetic progression. For the initial argument, you can shorten it to the cube of an odd number is odd, and the sum of 3 odd numbers is odd.
$endgroup$
– Calvin Lin
Sep 25 '13 at 13:49
$begingroup$
Actually that's my problem, I know that any arithmetic progression of the type {k,2k,3k} would provide a solution, but I can't prove that that's the only way to obtain solutions for this problem.
$endgroup$
– Stefan4024
Sep 25 '13 at 17:29
add a comment |
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1 Answer
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$begingroup$
It's easy to prove that at least one of the variables needs to be an even number. We know that:
$$6|(n-1)n(n+1)$$
Because in three consecutive numbers, one is divisible with three and at least one is divisible with 2. So we have:
$$6|n^3 - n$$
$$n^3 = n pmod 6$$
Now we have:
$$a^3 + b^3 + c^3 equiv a + b + c equiv 0 pmod 6$$
Beacuse the modulo is an even number that means that the sum $a+b+c$ is an even number also. We know that the sum of 3 odd numbers will be odd number, so it's impossible $a,b,c$ to be odd number, because there won't be solution. So it means that at least one of the variables is an even number.
WLOG we can set $b=2k$. Now we can continue:
$$b-2k = 0$$
Now we can multiply both sides with $b(b+2k)$. Note that won't give another solution, because it'll imply that b is $0$ or a negative number, which violate the condition. Now we have:
$$b(b+2k)(b-2k) = 0$$
$$b(b^2 - 4k^2) = 0$$
$$b^3 - 4bk^2 = 0$$
$$3b^3 - 12bk^2 = 0$$
$$6b^3 - 6bk^2 = 3b^3 + 6bk^2 = 0$$
$$6b(b^2 - k^2) = (b^2 - 3kb^2 + 3bk^2 - k^3) + b^3 + (b^3 + 3kb^2 + 3bk^2 + k^3) = 0$$
$$6b(b-k)(b+k) = (b-k)^3 + b^3 + (b+k)^3$$
Now if we substitute WLOG:
$$b+k=c quad quad b-k=a$$
$$6abc = a^3 + b^3 + c^3$$
Because $b,k in mathbb{N}$ it means that also $a,c in mathbb{N}$. So this proves that for any $b=2k$, there are integer solutions, such $a=k$ and $c=3k$.
But because $k$ is a factor of all of them it's easy to see that:
$$gcd(a,b,c) = gcd(k,2k,3k) = k$$
Because we want $gcd(a,b,c) = 1$, that implies that $k=1$ and that the only primitive solution of this equation is $(1,2,3)$ and its permutation.
answered Sep 25 '13 at 8:38
Stefan4024Stefan4024
30.6k63579
$endgroup$
1
$begingroup$
This doesn't explain how to find all triples of solutions. Note that right now, you have may merely used algebraic manipulations, without any serious number theory. There is no reason why the answers must appear in an arithmetic progression. For the initial argument, you can shorten it to the cube of an odd number is odd, and the sum of 3 odd numbers is odd.
$endgroup$
– Calvin Lin
Sep 25 '13 at 13:49
$begingroup$
Actually that's my problem, I know that any arithmetic progression of the type {k,2k,3k} would provide a solution, but I can't prove that that's the only way to obtain solutions for this problem.
$endgroup$
– Stefan4024
Sep 25 '13 at 17:29
add a comment |
$begingroup$
It's easy to prove that at least one of the variables needs to be an even number. We know that:
$$6|(n-1)n(n+1)$$
Because in three consecutive numbers, one is divisible with three and at least one is divisible with 2. So we have:
$$6|n^3 - n$$
$$n^3 = n pmod 6$$
Now we have:
$$a^3 + b^3 + c^3 equiv a + b + c equiv 0 pmod 6$$
Beacuse the modulo is an even number that means that the sum $a+b+c$ is an even number also. We know that the sum of 3 odd numbers will be odd number, so it's impossible $a,b,c$ to be odd number, because there won't be solution. So it means that at least one of the variables is an even number.
WLOG we can set $b=2k$. Now we can continue:
$$b-2k = 0$$
Now we can multiply both sides with $b(b+2k)$. Note that won't give another solution, because it'll imply that b is $0$ or a negative number, which violate the condition. Now we have:
$$b(b+2k)(b-2k) = 0$$
$$b(b^2 - 4k^2) = 0$$
$$b^3 - 4bk^2 = 0$$
$$3b^3 - 12bk^2 = 0$$
$$6b^3 - 6bk^2 = 3b^3 + 6bk^2 = 0$$
$$6b(b^2 - k^2) = (b^2 - 3kb^2 + 3bk^2 - k^3) + b^3 + (b^3 + 3kb^2 + 3bk^2 + k^3) = 0$$
$$6b(b-k)(b+k) = (b-k)^3 + b^3 + (b+k)^3$$
Now if we substitute WLOG:
$$b+k=c quad quad b-k=a$$
$$6abc = a^3 + b^3 + c^3$$
Because $b,k in mathbb{N}$ it means that also $a,c in mathbb{N}$. So this proves that for any $b=2k$, there are integer solutions, such $a=k$ and $c=3k$.
But because $k$ is a factor of all of them it's easy to see that:
$$gcd(a,b,c) = gcd(k,2k,3k) = k$$
Because we want $gcd(a,b,c) = 1$, that implies that $k=1$ and that the only primitive solution of this equation is $(1,2,3)$ and its permutation.
answered Sep 25 '13 at 8:38
Stefan4024Stefan4024
30.6k63579
$endgroup$
1
$begingroup$
This doesn't explain how to find all triples of solutions. Note that right now, you have may merely used algebraic manipulations, without any serious number theory. There is no reason why the answers must appear in an arithmetic progression. For the initial argument, you can shorten it to the cube of an odd number is odd, and the sum of 3 odd numbers is odd.
$endgroup$
– Calvin Lin
Sep 25 '13 at 13:49
$begingroup$
Actually that's my problem, I know that any arithmetic progression of the type {k,2k,3k} would provide a solution, but I can't prove that that's the only way to obtain solutions for this problem.
$endgroup$
– Stefan4024
Sep 25 '13 at 17:29
add a comment |
$begingroup$
It's easy to prove that at least one of the variables needs to be an even number. We know that:
$$6|(n-1)n(n+1)$$
Because in three consecutive numbers, one is divisible with three and at least one is divisible with 2. So we have:
$$6|n^3 - n$$
$$n^3 = n pmod 6$$
Now we have:
$$a^3 + b^3 + c^3 equiv a + b + c equiv 0 pmod 6$$
Beacuse the modulo is an even number that means that the sum $a+b+c$ is an even number also. We know that the sum of 3 odd numbers will be odd number, so it's impossible $a,b,c$ to be odd number, because there won't be solution. So it means that at least one of the variables is an even number.
WLOG we can set $b=2k$. Now we can continue:
$$b-2k = 0$$
Now we can multiply both sides with $b(b+2k)$. Note that won't give another solution, because it'll imply that b is $0$ or a negative number, which violate the condition. Now we have:
$$b(b+2k)(b-2k) = 0$$
$$b(b^2 - 4k^2) = 0$$
$$b^3 - 4bk^2 = 0$$
$$3b^3 - 12bk^2 = 0$$
$$6b^3 - 6bk^2 = 3b^3 + 6bk^2 = 0$$
$$6b(b^2 - k^2) = (b^2 - 3kb^2 + 3bk^2 - k^3) + b^3 + (b^3 + 3kb^2 + 3bk^2 + k^3) = 0$$
$$6b(b-k)(b+k) = (b-k)^3 + b^3 + (b+k)^3$$
Now if we substitute WLOG:
$$b+k=c quad quad b-k=a$$
$$6abc = a^3 + b^3 + c^3$$
Because $b,k in mathbb{N}$ it means that also $a,c in mathbb{N}$. So this proves that for any $b=2k$, there are integer solutions, such $a=k$ and $c=3k$.
But because $k$ is a factor of all of them it's easy to see that:
$$gcd(a,b,c) = gcd(k,2k,3k) = k$$
Because we want $gcd(a,b,c) = 1$, that implies that $k=1$ and that the only primitive solution of this equation is $(1,2,3)$ and its permutation.
answered Sep 25 '13 at 8:38
Stefan4024Stefan4024
30.6k63579
$endgroup$
It's easy to prove that at least one of the variables needs to be an even number. We know that:
$$6|(n-1)n(n+1)$$
Because in three consecutive numbers, one is divisible with three and at least one is divisible with 2. So we have:
$$6|n^3 - n$$
$$n^3 = n pmod 6$$
Now we have:
$$a^3 + b^3 + c^3 equiv a + b + c equiv 0 pmod 6$$
Beacuse the modulo is an even number that means that the sum $a+b+c$ is an even number also. We know that the sum of 3 odd numbers will be odd number, so it's impossible $a,b,c$ to be odd number, because there won't be solution. So it means that at least one of the variables is an even number.
WLOG we can set $b=2k$. Now we can continue:
$$b-2k = 0$$
Now we can multiply both sides with $b(b+2k)$. Note that won't give another solution, because it'll imply that b is $0$ or a negative number, which violate the condition. Now we have:
$$b(b+2k)(b-2k) = 0$$
$$b(b^2 - 4k^2) = 0$$
$$b^3 - 4bk^2 = 0$$
$$3b^3 - 12bk^2 = 0$$
$$6b^3 - 6bk^2 = 3b^3 + 6bk^2 = 0$$
$$6b(b^2 - k^2) = (b^2 - 3kb^2 + 3bk^2 - k^3) + b^3 + (b^3 + 3kb^2 + 3bk^2 + k^3) = 0$$
$$6b(b-k)(b+k) = (b-k)^3 + b^3 + (b+k)^3$$
Now if we substitute WLOG:
$$b+k=c quad quad b-k=a$$
$$6abc = a^3 + b^3 + c^3$$
Because $b,k in mathbb{N}$ it means that also $a,c in mathbb{N}$. So this proves that for any $b=2k$, there are integer solutions, such $a=k$ and $c=3k$.
But because $k$ is a factor of all of them it's easy to see that:
$$gcd(a,b,c) = gcd(k,2k,3k) = k$$
Because we want $gcd(a,b,c) = 1$, that implies that $k=1$ and that the only primitive solution of this equation is $(1,2,3)$ and its permutation.
answered Sep 25 '13 at 8:38
Stefan4024Stefan4024
30.6k63579
answered Sep 25 '13 at 8:38
Stefan4024Stefan4024
30.6k63579
answered Sep 25 '13 at 8:38
Stefan4024Stefan4024
30.6k63579
answered Sep 25 '13 at 8:38
Stefan4024Stefan4024
30.6k63579
30.6k63579
1
$begingroup$
This doesn't explain how to find all triples of solutions. Note that right now, you have may merely used algebraic manipulations, without any serious number theory. There is no reason why the answers must appear in an arithmetic progression. For the initial argument, you can shorten it to the cube of an odd number is odd, and the sum of 3 odd numbers is odd.
$endgroup$
– Calvin Lin
Sep 25 '13 at 13:49
$begingroup$
Actually that's my problem, I know that any arithmetic progression of the type {k,2k,3k} would provide a solution, but I can't prove that that's the only way to obtain solutions for this problem.
$endgroup$
– Stefan4024
Sep 25 '13 at 17:29
add a comment |
1
$begingroup$
This doesn't explain how to find all triples of solutions. Note that right now, you have may merely used algebraic manipulations, without any serious number theory. There is no reason why the answers must appear in an arithmetic progression. For the initial argument, you can shorten it to the cube of an odd number is odd, and the sum of 3 odd numbers is odd.
$endgroup$
– Calvin Lin
Sep 25 '13 at 13:49
$begingroup$
Actually that's my problem, I know that any arithmetic progression of the type {k,2k,3k} would provide a solution, but I can't prove that that's the only way to obtain solutions for this problem.
$endgroup$
– Stefan4024
Sep 25 '13 at 17:29
1
1
$begingroup$
This doesn't explain how to find all triples of solutions. Note that right now, you have may merely used algebraic manipulations, without any serious number theory. There is no reason why the answers must appear in an arithmetic progression. For the initial argument, you can shorten it to the cube of an odd number is odd, and the sum of 3 odd numbers is odd.
$endgroup$
– Calvin Lin
Sep 25 '13 at 13:49
$begingroup$
This doesn't explain how to find all triples of solutions. Note that right now, you have may merely used algebraic manipulations, without any serious number theory. There is no reason why the answers must appear in an arithmetic progression. For the initial argument, you can shorten it to the cube of an odd number is odd, and the sum of 3 odd numbers is odd.
$endgroup$
– Calvin Lin
Sep 25 '13 at 13:49
$begingroup$
Actually that's my problem, I know that any arithmetic progression of the type {k,2k,3k} would provide a solution, but I can't prove that that's the only way to obtain solutions for this problem.
$endgroup$
– Stefan4024
Sep 25 '13 at 17:29
$begingroup$
Actually that's my problem, I know that any arithmetic progression of the type {k,2k,3k} would provide a solution, but I can't prove that that's the only way to obtain solutions for this problem.
$endgroup$
– Stefan4024
Sep 25 '13 at 17:29
add a comment |
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saying that $a,b,c$ is an arithmetic progression satisfies the conditions is not correct, since $1,4,7$ doesn't satisfy the equation. What you mean is that ${ a, b, c } = {k, 2k, 3k } $, which agrees with your cases.
$endgroup$
– Calvin Lin
Sep 25 '13 at 0:50
$begingroup$
My bad. I didn't mean that. I meaned if $a,b,c$ are solution then it implies $a,b,c$ are in arithmetic progression, but it doesn't mean that every arithmetic progression will provide a solution. And what are you thoughts is there another solution satisfying the initial condition. Is there any other way to obtain it? Or there is just one?
$endgroup$
– Stefan4024
Sep 25 '13 at 0:59
$begingroup$
If the arithmetic progression condition is correct, then $(1,2,3)$ is the only primitive solution.
$endgroup$
– Daniel Fischer
Sep 25 '13 at 1:34
$begingroup$
I know that, but is there another way to generate (a,b,c) as integer solution except for arithemtic progression.
$endgroup$
– Stefan4024
Sep 25 '13 at 1:35
$begingroup$
@DanielFischer Is there any way to prove that at least one $a,b,c$ must be divisible by 3? If that's proven then it's easy to prove that (1,2,3) is the only primitive solution.
$endgroup$
– Stefan4024
Sep 25 '13 at 2:13