Mixing
What is the maximum value of $(a+ b+c)$ if $(a^n + b^n + c^n)$ is divisible by $(a+ b+c)$ where the remainder...
$begingroup$
The ‘energy’ of an ordered triple $(a, b, c)$ formed by three
positive integers $a$, $b$ and $c$ is said to be n if the following $c$
$ge bgeq a$, gcd$(a, b, c) = 1$, and $(a^n + b^n + c^n)$ is
divisible (remainder is 0) by $(a +b+ c) $. There are some
possible ordered triple whose ‘energy’ can be of all values of $n ge$
$1$. In this case, for which ordered triple, the value of $(a+b+c)$ is
maximum?
Second part (of the original problem) Determine all triples $(a,b,c)$ that are simultaneously $2004$-good and $2005$-good, but not $2007$-good.
Source: Bangladesh Math Olympiad 2017 junior category (Originally from Canada, 2005).
I can't understand the first line of this question. Any 3 consecutive numbers have a gcd of 1. Moreover if $n=1$, then $(a^n + b^n + c^n) = (a +b+ c) $.
number-theory elementary-number-theory contest-math greatest-common-divisor
edited Jan 11 at 6:00
Aaron
1,909415
asked Jan 11 at 1:46
ShromiShromi
1219
$endgroup$
add a comment |
$begingroup$
The ‘energy’ of an ordered triple $(a, b, c)$ formed by three
positive integers $a$, $b$ and $c$ is said to be n if the following $c$
$ge bgeq a$, gcd$(a, b, c) = 1$, and $(a^n + b^n + c^n)$ is
divisible (remainder is 0) by $(a +b+ c) $. There are some
possible ordered triple whose ‘energy’ can be of all values of $n ge$
$1$. In this case, for which ordered triple, the value of $(a+b+c)$ is
maximum?
Second part (of the original problem) Determine all triples $(a,b,c)$ that are simultaneously $2004$-good and $2005$-good, but not $2007$-good.
Source: Bangladesh Math Olympiad 2017 junior category (Originally from Canada, 2005).
I can't understand the first line of this question. Any 3 consecutive numbers have a gcd of 1. Moreover if $n=1$, then $(a^n + b^n + c^n) = (a +b+ c) $.
number-theory elementary-number-theory contest-math greatest-common-divisor
edited Jan 11 at 6:00
Aaron
1,909415
asked Jan 11 at 1:46
ShromiShromi
1219
$endgroup$
2
$begingroup$
What I think they are saying is if $a+b+cmid a^n+b^n+c^n$ for every $n$ and $gcd(a,b,c) = 1$ then what is the maximum value of $a+b+c$.
$endgroup$
– kingW3
Jan 11 at 2:04
$begingroup$
Could you please explain if there is any valid solution of this question?
$endgroup$
– Shromi
Jan 11 at 3:01
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@Shromi There is at least one valid "trivial" solution of $left(a,b,cright) = left(1,1,1right)$, which of course gives that $a + b + c = 3$. I'm not sure if there are any other larger solutions, with my best guess being that there isn't, but I'm not sure offhand how to prove it, although I am thinking about it.
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– John Omielan
Jan 11 at 3:10
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@Lee Thanks for the comment, however, for your for $left(a, b, cright) = left(1, 2, 3right)$, then $a + b + c = 6$, but $a^2 + b^2 + c^2 = 1 + 4 + 9 = 13$, which is not a multiple of $6$. Note the values must divide for every power of $n$.
$endgroup$
– John Omielan
Jan 11 at 4:08
$begingroup$
@JohnOmielan, yes, I miss that part
$endgroup$
– Lee
Jan 11 at 4:09
add a comment |
$begingroup$
The ‘energy’ of an ordered triple $(a, b, c)$ formed by three
positive integers $a$, $b$ and $c$ is said to be n if the following $c$
$ge bgeq a$, gcd$(a, b, c) = 1$, and $(a^n + b^n + c^n)$ is
divisible (remainder is 0) by $(a +b+ c) $. There are some
possible ordered triple whose ‘energy’ can be of all values of $n ge$
$1$. In this case, for which ordered triple, the value of $(a+b+c)$ is
maximum?
Second part (of the original problem) Determine all triples $(a,b,c)$ that are simultaneously $2004$-good and $2005$-good, but not $2007$-good.
Source: Bangladesh Math Olympiad 2017 junior category (Originally from Canada, 2005).
I can't understand the first line of this question. Any 3 consecutive numbers have a gcd of 1. Moreover if $n=1$, then $(a^n + b^n + c^n) = (a +b+ c) $.
number-theory elementary-number-theory contest-math greatest-common-divisor
edited Jan 11 at 6:00
Aaron
1,909415
asked Jan 11 at 1:46
ShromiShromi
1219
$endgroup$
The ‘energy’ of an ordered triple $(a, b, c)$ formed by three
positive integers $a$, $b$ and $c$ is said to be n if the following $c$
$ge bgeq a$, gcd$(a, b, c) = 1$, and $(a^n + b^n + c^n)$ is
divisible (remainder is 0) by $(a +b+ c) $. There are some
possible ordered triple whose ‘energy’ can be of all values of $n ge$
$1$. In this case, for which ordered triple, the value of $(a+b+c)$ is
maximum?
Second part (of the original problem) Determine all triples $(a,b,c)$ that are simultaneously $2004$-good and $2005$-good, but not $2007$-good.
Source: Bangladesh Math Olympiad 2017 junior category (Originally from Canada, 2005).
I can't understand the first line of this question. Any 3 consecutive numbers have a gcd of 1. Moreover if $n=1$, then $(a^n + b^n + c^n) = (a +b+ c) $.
number-theory elementary-number-theory contest-math greatest-common-divisor
number-theory elementary-number-theory contest-math greatest-common-divisor
edited Jan 11 at 6:00
Aaron
1,909415
asked Jan 11 at 1:46
ShromiShromi
1219
edited Jan 11 at 6:00
Aaron
1,909415
asked Jan 11 at 1:46
ShromiShromi
1219
edited Jan 11 at 6:00
Aaron
1,909415
edited Jan 11 at 6:00
Aaron
1,909415
edited Jan 11 at 6:00
Aaron
1,909415
1,909415
asked Jan 11 at 1:46
ShromiShromi
1219
asked Jan 11 at 1:46
ShromiShromi
1219
asked Jan 11 at 1:46
ShromiShromi
1219
1219
2
$begingroup$
What I think they are saying is if $a+b+cmid a^n+b^n+c^n$ for every $n$ and $gcd(a,b,c) = 1$ then what is the maximum value of $a+b+c$.
$endgroup$
– kingW3
Jan 11 at 2:04
$begingroup$
Could you please explain if there is any valid solution of this question?
$endgroup$
– Shromi
Jan 11 at 3:01
$begingroup$
@Shromi There is at least one valid "trivial" solution of $left(a,b,cright) = left(1,1,1right)$, which of course gives that $a + b + c = 3$. I'm not sure if there are any other larger solutions, with my best guess being that there isn't, but I'm not sure offhand how to prove it, although I am thinking about it.
$endgroup$
– John Omielan
Jan 11 at 3:10
$begingroup$
@Lee Thanks for the comment, however, for your for $left(a, b, cright) = left(1, 2, 3right)$, then $a + b + c = 6$, but $a^2 + b^2 + c^2 = 1 + 4 + 9 = 13$, which is not a multiple of $6$. Note the values must divide for every power of $n$.
$endgroup$
– John Omielan
Jan 11 at 4:08
$begingroup$
@JohnOmielan, yes, I miss that part
$endgroup$
– Lee
Jan 11 at 4:09
add a comment |
2
$begingroup$
What I think they are saying is if $a+b+cmid a^n+b^n+c^n$ for every $n$ and $gcd(a,b,c) = 1$ then what is the maximum value of $a+b+c$.
$endgroup$
– kingW3
Jan 11 at 2:04
$begingroup$
Could you please explain if there is any valid solution of this question?
$endgroup$
– Shromi
Jan 11 at 3:01
$begingroup$
@Shromi There is at least one valid "trivial" solution of $left(a,b,cright) = left(1,1,1right)$, which of course gives that $a + b + c = 3$. I'm not sure if there are any other larger solutions, with my best guess being that there isn't, but I'm not sure offhand how to prove it, although I am thinking about it.
$endgroup$
– John Omielan
Jan 11 at 3:10
$begingroup$
@Lee Thanks for the comment, however, for your for $left(a, b, cright) = left(1, 2, 3right)$, then $a + b + c = 6$, but $a^2 + b^2 + c^2 = 1 + 4 + 9 = 13$, which is not a multiple of $6$. Note the values must divide for every power of $n$.
$endgroup$
– John Omielan
Jan 11 at 4:08
$begingroup$
@JohnOmielan, yes, I miss that part
$endgroup$
– Lee
Jan 11 at 4:09
2
2
$begingroup$
What I think they are saying is if $a+b+cmid a^n+b^n+c^n$ for every $n$ and $gcd(a,b,c) = 1$ then what is the maximum value of $a+b+c$.
$endgroup$
– kingW3
Jan 11 at 2:04
$begingroup$
What I think they are saying is if $a+b+cmid a^n+b^n+c^n$ for every $n$ and $gcd(a,b,c) = 1$ then what is the maximum value of $a+b+c$.
$endgroup$
– kingW3
Jan 11 at 2:04
$begingroup$
Could you please explain if there is any valid solution of this question?
$endgroup$
– Shromi
Jan 11 at 3:01
$begingroup$
Could you please explain if there is any valid solution of this question?
$endgroup$
– Shromi
Jan 11 at 3:01
$begingroup$
@Shromi There is at least one valid "trivial" solution of $left(a,b,cright) = left(1,1,1right)$, which of course gives that $a + b + c = 3$. I'm not sure if there are any other larger solutions, with my best guess being that there isn't, but I'm not sure offhand how to prove it, although I am thinking about it.
$endgroup$
– John Omielan
Jan 11 at 3:10
$begingroup$
@Shromi There is at least one valid "trivial" solution of $left(a,b,cright) = left(1,1,1right)$, which of course gives that $a + b + c = 3$. I'm not sure if there are any other larger solutions, with my best guess being that there isn't, but I'm not sure offhand how to prove it, although I am thinking about it.
$endgroup$
– John Omielan
Jan 11 at 3:10
$begingroup$
@Lee Thanks for the comment, however, for your for $left(a, b, cright) = left(1, 2, 3right)$, then $a + b + c = 6$, but $a^2 + b^2 + c^2 = 1 + 4 + 9 = 13$, which is not a multiple of $6$. Note the values must divide for every power of $n$.
$endgroup$
– John Omielan
Jan 11 at 4:08
$begingroup$
@Lee Thanks for the comment, however, for your for $left(a, b, cright) = left(1, 2, 3right)$, then $a + b + c = 6$, but $a^2 + b^2 + c^2 = 1 + 4 + 9 = 13$, which is not a multiple of $6$. Note the values must divide for every power of $n$.
$endgroup$
– John Omielan
Jan 11 at 4:08
$begingroup$
@JohnOmielan, yes, I miss that part
$endgroup$
– Lee
Jan 11 at 4:09
$begingroup$
@JohnOmielan, yes, I miss that part
$endgroup$
– Lee
Jan 11 at 4:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The answer is indeed $6$. Here is a complete solution.
First, take a prime $p$ such that, $pmid a+b+c$. Note that, if $p$ divides two of $a$ and $b$, then it must divide the third, contradicting with $(a,b,c)=1$. Thus, $p$ either divides none of them, or only one.
- If $pnmid a,b,c$, then by taking $n=p-1$, we have, by Fermat's theorem that $a^{p-1}equiv b^{p-1}equiv c^{p-1}equiv 1 pmod{p}$. Thus, $pmid 3$, hence $p=3$.
- If $pmid a$, and $pnmid b,c$, then we obtain $pmid 2$, by taking $n=p-1$.
Therefore, only prime divisors of $a+b+c$ are $2$ or $3$. Now, for $n=2$, we get using $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ that $a+b+cmid 2(ab+bc+ca)$. Also, for $n=3$, using $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, we have $a+b+cmid 3abc$. Next, note that, if $9mid a+b+c$, then $3mid abc$. Hence, either $a$ or $b$ or $c$ is divisible by $3$. Now, if $3mid a$, then using $a+b+cmid 2(ab+bc+ca)$, we obtain that $3mid bc$, hence $3mid b$ or $3mid c$. However, this, together with $3mid a+b+c$ contradicts with $(a,b,c)=1$. Hence, $9nmid a+b+c$.
Similarly, if $4mid a+b+c$, then note that among $a,b,c$ exactly one is even, suppose it is $a$, that is, $4mid a$. But this gives, $4mid 2(ab+bc+ca)$, yielding $2mid (ab+bc+ca)$, yielding $2mid bc$. However, since $b$ and $c$ are odd, this is a clear contradiction.
Thus, $a+b+c=3$ or $a+b+c=6$. In former, we get $(1,1,1)$, which is really $n$-good for any $n$. For the latter, we have $(3,2,1)$ or $(4,1,1)$. For the former, $(3,2,1)$ is not $n-$ good for any even $n$, using modulo $3$. For the latter, it is easy to see that it is $n$-good for any $n$.
Hence, we are done.
Note (Alternative) There is an alternative way of proving that $a+b+c$ can only admit $2$ or $3$ as its prime divisors. To see this, suppose $p>3$ divides $a+b+c$. Then, $pmid 3abc$ implies $pmid abc$. Using $(a,b,c)=1$,
we see that exactly one of $a,b,c$ is divisible by $p$. Suppose, it is $a$. Then, $pmid ab+bc+ca$, together with $pmid ab+ac$ implies $pmid bc$, which clearly is a contradiction. From here, one can finish in exact same way as in above proof, i.e., prove $4,9nmid a+b+c$, and finish.
answered Jan 11 at 4:28
AaronAaron
1,909415
$endgroup$
$begingroup$
It's a good answer, except for the minor point that the question says that $c ge b ge a$, so you need to reverse the order of your values, e.g., the largest result which works is $left(a, b, cright) = left(1, 1, 4right)$.
$endgroup$
– John Omielan
Jan 11 at 4:33
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well, that requirement, as you can see from my proof, is completely redundant (and in fact, as far as I remember, does not even exist in the original problem statement). This is an old olympiad problem, circa 2005, from Canadian olympiad, where Bangladeshi's seem to have stolen from.
$endgroup$
– Aaron
Jan 11 at 4:37
$begingroup$
Thanks for the info. I agree the condition is redundant, and I didn't realize it was an old Canadian Olympiad problem.
$endgroup$
– John Omielan
Jan 11 at 4:41
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John, this was one of my favorites back when I was still in high school, around 2010. In fact it has a second part too, if you are interested:
$endgroup$
– Aaron
Jan 11 at 4:42
$begingroup$
Find all triples, $(a,b,c)$ that are $2004$-good and $2005$-good, but not $2007$-good.
$endgroup$
– Aaron
Jan 11 at 4:42
|
show 3 more comments
$begingroup$
The question is to find the largest sum of $a, b, c$, given they're all relatively prime to each other and divide
$$a^n + b^n + c^n text{ } forall text{ } n ge 1 tag{1}label{eq1}$$
Since $a, b, c ge 1$, then $a + b + c ge 3$, so it consists of one or more prime factors. Call one of these prime factors $d$. Thus,
$$a + b + c equiv a^n + b^n + c^n equiv 0 pmod d tag{2}label{eq2}$$
As this must hold for all $n ge 1$, consider $n = 3$ and substitute
$$c equiv -a - b pmod d tag{3}label{eq3}$$
into the middle part of the congruence in eqref{eq2} to get
$$a^3 + b^3 + left(-a - bright)^3 equiv 0 pmod d tag{4}label{eq4}$$
This simplifies to
$$-3ab^2 - 3a^2b = -3ableft(a + bright) equiv 0 pmod d tag{5}label{eq5}$$
Thus, $d$ must divide $3$, $a$, $b$ and/or $a + b equiv -c pmod d$. If $d$ is not $3$, then note that if it must divide just one of $a$, $b$ and $-c$. This is because if it divides any $2$, say $a$ and $b$, it must also divide $c$. As the equations are symmetric in $a, b text{ and } c$, WLOG, assume that $d$ divides $a$, i.e., $a equiv 0 pmod d$. Thus, $b + c equiv 0 pmod d$, i.e., $b equiv -c pmod d$. Now, use $n = 2$ in the congruence in the middle part of eqref{eq2} to get
$$a^2 + b^2 + c^2 equiv 0 + left(-cright) + c^2 equiv 2c^2 equiv 0 pmod d tag{6}label{eq6}$$
As $d$ doesn't divide $c$, this means that $d$ must divide $2$, i.e., it must be $2$. Thus, the only possible prime factors of $a + b + c$ are $3$, as mentioned earlier, and $2$ as shown here.
Based on what I have found here, I believe the maximum value is just the product of those $2$ factors, i.e., $6$, which can be obtained from
$$left(a, b, cright) = left(1, 1, 4right) tag{7}label{eq7}$$
You can easily verify that $a^n + b^n + c^n = 1 + 1 + 4^n equiv 0 pmod 6$ for all $n ge 1$ as $1 + 1 = 2$, so $2$ divides it, and $4^n equiv 1^n equiv 1 pmod 3$, so $3$ divides it as well. I am posting this partial solution for now so you have something to start from, as I believe I'm on the right track. I need to go for a while, but I will keep thinking about this and, if I determine the rest of the solution, I will add it later.
answered Jan 11 at 4:20
John OmielanJohn Omielan
2,386212
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
The answer is indeed $6$. Here is a complete solution.
First, take a prime $p$ such that, $pmid a+b+c$. Note that, if $p$ divides two of $a$ and $b$, then it must divide the third, contradicting with $(a,b,c)=1$. Thus, $p$ either divides none of them, or only one.
- If $pnmid a,b,c$, then by taking $n=p-1$, we have, by Fermat's theorem that $a^{p-1}equiv b^{p-1}equiv c^{p-1}equiv 1 pmod{p}$. Thus, $pmid 3$, hence $p=3$.
- If $pmid a$, and $pnmid b,c$, then we obtain $pmid 2$, by taking $n=p-1$.
Therefore, only prime divisors of $a+b+c$ are $2$ or $3$. Now, for $n=2$, we get using $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ that $a+b+cmid 2(ab+bc+ca)$. Also, for $n=3$, using $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, we have $a+b+cmid 3abc$. Next, note that, if $9mid a+b+c$, then $3mid abc$. Hence, either $a$ or $b$ or $c$ is divisible by $3$. Now, if $3mid a$, then using $a+b+cmid 2(ab+bc+ca)$, we obtain that $3mid bc$, hence $3mid b$ or $3mid c$. However, this, together with $3mid a+b+c$ contradicts with $(a,b,c)=1$. Hence, $9nmid a+b+c$.
Similarly, if $4mid a+b+c$, then note that among $a,b,c$ exactly one is even, suppose it is $a$, that is, $4mid a$. But this gives, $4mid 2(ab+bc+ca)$, yielding $2mid (ab+bc+ca)$, yielding $2mid bc$. However, since $b$ and $c$ are odd, this is a clear contradiction.
Thus, $a+b+c=3$ or $a+b+c=6$. In former, we get $(1,1,1)$, which is really $n$-good for any $n$. For the latter, we have $(3,2,1)$ or $(4,1,1)$. For the former, $(3,2,1)$ is not $n-$ good for any even $n$, using modulo $3$. For the latter, it is easy to see that it is $n$-good for any $n$.
Hence, we are done.
Note (Alternative) There is an alternative way of proving that $a+b+c$ can only admit $2$ or $3$ as its prime divisors. To see this, suppose $p>3$ divides $a+b+c$. Then, $pmid 3abc$ implies $pmid abc$. Using $(a,b,c)=1$,
we see that exactly one of $a,b,c$ is divisible by $p$. Suppose, it is $a$. Then, $pmid ab+bc+ca$, together with $pmid ab+ac$ implies $pmid bc$, which clearly is a contradiction. From here, one can finish in exact same way as in above proof, i.e., prove $4,9nmid a+b+c$, and finish.
answered Jan 11 at 4:28
AaronAaron
1,909415
$endgroup$
$begingroup$
It's a good answer, except for the minor point that the question says that $c ge b ge a$, so you need to reverse the order of your values, e.g., the largest result which works is $left(a, b, cright) = left(1, 1, 4right)$.
$endgroup$
– John Omielan
Jan 11 at 4:33
$begingroup$
well, that requirement, as you can see from my proof, is completely redundant (and in fact, as far as I remember, does not even exist in the original problem statement). This is an old olympiad problem, circa 2005, from Canadian olympiad, where Bangladeshi's seem to have stolen from.
$endgroup$
– Aaron
Jan 11 at 4:37
$begingroup$
Thanks for the info. I agree the condition is redundant, and I didn't realize it was an old Canadian Olympiad problem.
$endgroup$
– John Omielan
Jan 11 at 4:41
$begingroup$
John, this was one of my favorites back when I was still in high school, around 2010. In fact it has a second part too, if you are interested:
$endgroup$
– Aaron
Jan 11 at 4:42
$begingroup$
Find all triples, $(a,b,c)$ that are $2004$-good and $2005$-good, but not $2007$-good.
$endgroup$
– Aaron
Jan 11 at 4:42
|
show 3 more comments
$begingroup$
The answer is indeed $6$. Here is a complete solution.
First, take a prime $p$ such that, $pmid a+b+c$. Note that, if $p$ divides two of $a$ and $b$, then it must divide the third, contradicting with $(a,b,c)=1$. Thus, $p$ either divides none of them, or only one.
- If $pnmid a,b,c$, then by taking $n=p-1$, we have, by Fermat's theorem that $a^{p-1}equiv b^{p-1}equiv c^{p-1}equiv 1 pmod{p}$. Thus, $pmid 3$, hence $p=3$.
- If $pmid a$, and $pnmid b,c$, then we obtain $pmid 2$, by taking $n=p-1$.
Therefore, only prime divisors of $a+b+c$ are $2$ or $3$. Now, for $n=2$, we get using $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ that $a+b+cmid 2(ab+bc+ca)$. Also, for $n=3$, using $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, we have $a+b+cmid 3abc$. Next, note that, if $9mid a+b+c$, then $3mid abc$. Hence, either $a$ or $b$ or $c$ is divisible by $3$. Now, if $3mid a$, then using $a+b+cmid 2(ab+bc+ca)$, we obtain that $3mid bc$, hence $3mid b$ or $3mid c$. However, this, together with $3mid a+b+c$ contradicts with $(a,b,c)=1$. Hence, $9nmid a+b+c$.
Similarly, if $4mid a+b+c$, then note that among $a,b,c$ exactly one is even, suppose it is $a$, that is, $4mid a$. But this gives, $4mid 2(ab+bc+ca)$, yielding $2mid (ab+bc+ca)$, yielding $2mid bc$. However, since $b$ and $c$ are odd, this is a clear contradiction.
Thus, $a+b+c=3$ or $a+b+c=6$. In former, we get $(1,1,1)$, which is really $n$-good for any $n$. For the latter, we have $(3,2,1)$ or $(4,1,1)$. For the former, $(3,2,1)$ is not $n-$ good for any even $n$, using modulo $3$. For the latter, it is easy to see that it is $n$-good for any $n$.
Hence, we are done.
Note (Alternative) There is an alternative way of proving that $a+b+c$ can only admit $2$ or $3$ as its prime divisors. To see this, suppose $p>3$ divides $a+b+c$. Then, $pmid 3abc$ implies $pmid abc$. Using $(a,b,c)=1$,
we see that exactly one of $a,b,c$ is divisible by $p$. Suppose, it is $a$. Then, $pmid ab+bc+ca$, together with $pmid ab+ac$ implies $pmid bc$, which clearly is a contradiction. From here, one can finish in exact same way as in above proof, i.e., prove $4,9nmid a+b+c$, and finish.
answered Jan 11 at 4:28
AaronAaron
1,909415
$endgroup$
$begingroup$
It's a good answer, except for the minor point that the question says that $c ge b ge a$, so you need to reverse the order of your values, e.g., the largest result which works is $left(a, b, cright) = left(1, 1, 4right)$.
$endgroup$
– John Omielan
Jan 11 at 4:33
$begingroup$
well, that requirement, as you can see from my proof, is completely redundant (and in fact, as far as I remember, does not even exist in the original problem statement). This is an old olympiad problem, circa 2005, from Canadian olympiad, where Bangladeshi's seem to have stolen from.
$endgroup$
– Aaron
Jan 11 at 4:37
$begingroup$
Thanks for the info. I agree the condition is redundant, and I didn't realize it was an old Canadian Olympiad problem.
$endgroup$
– John Omielan
Jan 11 at 4:41
$begingroup$
John, this was one of my favorites back when I was still in high school, around 2010. In fact it has a second part too, if you are interested:
$endgroup$
– Aaron
Jan 11 at 4:42
$begingroup$
Find all triples, $(a,b,c)$ that are $2004$-good and $2005$-good, but not $2007$-good.
$endgroup$
– Aaron
Jan 11 at 4:42
|
show 3 more comments
$begingroup$
The answer is indeed $6$. Here is a complete solution.
First, take a prime $p$ such that, $pmid a+b+c$. Note that, if $p$ divides two of $a$ and $b$, then it must divide the third, contradicting with $(a,b,c)=1$. Thus, $p$ either divides none of them, or only one.
- If $pnmid a,b,c$, then by taking $n=p-1$, we have, by Fermat's theorem that $a^{p-1}equiv b^{p-1}equiv c^{p-1}equiv 1 pmod{p}$. Thus, $pmid 3$, hence $p=3$.
- If $pmid a$, and $pnmid b,c$, then we obtain $pmid 2$, by taking $n=p-1$.
Therefore, only prime divisors of $a+b+c$ are $2$ or $3$. Now, for $n=2$, we get using $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ that $a+b+cmid 2(ab+bc+ca)$. Also, for $n=3$, using $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, we have $a+b+cmid 3abc$. Next, note that, if $9mid a+b+c$, then $3mid abc$. Hence, either $a$ or $b$ or $c$ is divisible by $3$. Now, if $3mid a$, then using $a+b+cmid 2(ab+bc+ca)$, we obtain that $3mid bc$, hence $3mid b$ or $3mid c$. However, this, together with $3mid a+b+c$ contradicts with $(a,b,c)=1$. Hence, $9nmid a+b+c$.
Similarly, if $4mid a+b+c$, then note that among $a,b,c$ exactly one is even, suppose it is $a$, that is, $4mid a$. But this gives, $4mid 2(ab+bc+ca)$, yielding $2mid (ab+bc+ca)$, yielding $2mid bc$. However, since $b$ and $c$ are odd, this is a clear contradiction.
Thus, $a+b+c=3$ or $a+b+c=6$. In former, we get $(1,1,1)$, which is really $n$-good for any $n$. For the latter, we have $(3,2,1)$ or $(4,1,1)$. For the former, $(3,2,1)$ is not $n-$ good for any even $n$, using modulo $3$. For the latter, it is easy to see that it is $n$-good for any $n$.
Hence, we are done.
Note (Alternative) There is an alternative way of proving that $a+b+c$ can only admit $2$ or $3$ as its prime divisors. To see this, suppose $p>3$ divides $a+b+c$. Then, $pmid 3abc$ implies $pmid abc$. Using $(a,b,c)=1$,
we see that exactly one of $a,b,c$ is divisible by $p$. Suppose, it is $a$. Then, $pmid ab+bc+ca$, together with $pmid ab+ac$ implies $pmid bc$, which clearly is a contradiction. From here, one can finish in exact same way as in above proof, i.e., prove $4,9nmid a+b+c$, and finish.
answered Jan 11 at 4:28
AaronAaron
1,909415
$endgroup$
The answer is indeed $6$. Here is a complete solution.
First, take a prime $p$ such that, $pmid a+b+c$. Note that, if $p$ divides two of $a$ and $b$, then it must divide the third, contradicting with $(a,b,c)=1$. Thus, $p$ either divides none of them, or only one.
- If $pnmid a,b,c$, then by taking $n=p-1$, we have, by Fermat's theorem that $a^{p-1}equiv b^{p-1}equiv c^{p-1}equiv 1 pmod{p}$. Thus, $pmid 3$, hence $p=3$.
- If $pmid a$, and $pnmid b,c$, then we obtain $pmid 2$, by taking $n=p-1$.
Therefore, only prime divisors of $a+b+c$ are $2$ or $3$. Now, for $n=2$, we get using $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ that $a+b+cmid 2(ab+bc+ca)$. Also, for $n=3$, using $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, we have $a+b+cmid 3abc$. Next, note that, if $9mid a+b+c$, then $3mid abc$. Hence, either $a$ or $b$ or $c$ is divisible by $3$. Now, if $3mid a$, then using $a+b+cmid 2(ab+bc+ca)$, we obtain that $3mid bc$, hence $3mid b$ or $3mid c$. However, this, together with $3mid a+b+c$ contradicts with $(a,b,c)=1$. Hence, $9nmid a+b+c$.
Similarly, if $4mid a+b+c$, then note that among $a,b,c$ exactly one is even, suppose it is $a$, that is, $4mid a$. But this gives, $4mid 2(ab+bc+ca)$, yielding $2mid (ab+bc+ca)$, yielding $2mid bc$. However, since $b$ and $c$ are odd, this is a clear contradiction.
Thus, $a+b+c=3$ or $a+b+c=6$. In former, we get $(1,1,1)$, which is really $n$-good for any $n$. For the latter, we have $(3,2,1)$ or $(4,1,1)$. For the former, $(3,2,1)$ is not $n-$ good for any even $n$, using modulo $3$. For the latter, it is easy to see that it is $n$-good for any $n$.
Hence, we are done.
Note (Alternative) There is an alternative way of proving that $a+b+c$ can only admit $2$ or $3$ as its prime divisors. To see this, suppose $p>3$ divides $a+b+c$. Then, $pmid 3abc$ implies $pmid abc$. Using $(a,b,c)=1$,
we see that exactly one of $a,b,c$ is divisible by $p$. Suppose, it is $a$. Then, $pmid ab+bc+ca$, together with $pmid ab+ac$ implies $pmid bc$, which clearly is a contradiction. From here, one can finish in exact same way as in above proof, i.e., prove $4,9nmid a+b+c$, and finish.
answered Jan 11 at 4:28
AaronAaron
1,909415
answered Jan 11 at 4:28
AaronAaron
1,909415
answered Jan 11 at 4:28
AaronAaron
1,909415
answered Jan 11 at 4:28
AaronAaron
1,909415
1,909415
$begingroup$
It's a good answer, except for the minor point that the question says that $c ge b ge a$, so you need to reverse the order of your values, e.g., the largest result which works is $left(a, b, cright) = left(1, 1, 4right)$.
$endgroup$
– John Omielan
Jan 11 at 4:33
$begingroup$
well, that requirement, as you can see from my proof, is completely redundant (and in fact, as far as I remember, does not even exist in the original problem statement). This is an old olympiad problem, circa 2005, from Canadian olympiad, where Bangladeshi's seem to have stolen from.
$endgroup$
– Aaron
Jan 11 at 4:37
$begingroup$
Thanks for the info. I agree the condition is redundant, and I didn't realize it was an old Canadian Olympiad problem.
$endgroup$
– John Omielan
Jan 11 at 4:41
$begingroup$
John, this was one of my favorites back when I was still in high school, around 2010. In fact it has a second part too, if you are interested:
$endgroup$
– Aaron
Jan 11 at 4:42
$begingroup$
Find all triples, $(a,b,c)$ that are $2004$-good and $2005$-good, but not $2007$-good.
$endgroup$
– Aaron
Jan 11 at 4:42
|
show 3 more comments
$begingroup$
It's a good answer, except for the minor point that the question says that $c ge b ge a$, so you need to reverse the order of your values, e.g., the largest result which works is $left(a, b, cright) = left(1, 1, 4right)$.
$endgroup$
– John Omielan
Jan 11 at 4:33
$begingroup$
well, that requirement, as you can see from my proof, is completely redundant (and in fact, as far as I remember, does not even exist in the original problem statement). This is an old olympiad problem, circa 2005, from Canadian olympiad, where Bangladeshi's seem to have stolen from.
$endgroup$
– Aaron
Jan 11 at 4:37
$begingroup$
Thanks for the info. I agree the condition is redundant, and I didn't realize it was an old Canadian Olympiad problem.
$endgroup$
– John Omielan
Jan 11 at 4:41
$begingroup$
John, this was one of my favorites back when I was still in high school, around 2010. In fact it has a second part too, if you are interested:
$endgroup$
– Aaron
Jan 11 at 4:42
$begingroup$
Find all triples, $(a,b,c)$ that are $2004$-good and $2005$-good, but not $2007$-good.
$endgroup$
– Aaron
Jan 11 at 4:42
$begingroup$
It's a good answer, except for the minor point that the question says that $c ge b ge a$, so you need to reverse the order of your values, e.g., the largest result which works is $left(a, b, cright) = left(1, 1, 4right)$.
$endgroup$
– John Omielan
Jan 11 at 4:33
$begingroup$
It's a good answer, except for the minor point that the question says that $c ge b ge a$, so you need to reverse the order of your values, e.g., the largest result which works is $left(a, b, cright) = left(1, 1, 4right)$.
$endgroup$
– John Omielan
Jan 11 at 4:33
$begingroup$
well, that requirement, as you can see from my proof, is completely redundant (and in fact, as far as I remember, does not even exist in the original problem statement). This is an old olympiad problem, circa 2005, from Canadian olympiad, where Bangladeshi's seem to have stolen from.
$endgroup$
– Aaron
Jan 11 at 4:37
$begingroup$
well, that requirement, as you can see from my proof, is completely redundant (and in fact, as far as I remember, does not even exist in the original problem statement). This is an old olympiad problem, circa 2005, from Canadian olympiad, where Bangladeshi's seem to have stolen from.
$endgroup$
– Aaron
Jan 11 at 4:37
$begingroup$
Thanks for the info. I agree the condition is redundant, and I didn't realize it was an old Canadian Olympiad problem.
$endgroup$
– John Omielan
Jan 11 at 4:41
$begingroup$
Thanks for the info. I agree the condition is redundant, and I didn't realize it was an old Canadian Olympiad problem.
$endgroup$
– John Omielan
Jan 11 at 4:41
$begingroup$
John, this was one of my favorites back when I was still in high school, around 2010. In fact it has a second part too, if you are interested:
$endgroup$
– Aaron
Jan 11 at 4:42
$begingroup$
John, this was one of my favorites back when I was still in high school, around 2010. In fact it has a second part too, if you are interested:
$endgroup$
– Aaron
Jan 11 at 4:42
$begingroup$
Find all triples, $(a,b,c)$ that are $2004$-good and $2005$-good, but not $2007$-good.
$endgroup$
– Aaron
Jan 11 at 4:42
$begingroup$
Find all triples, $(a,b,c)$ that are $2004$-good and $2005$-good, but not $2007$-good.
$endgroup$
– Aaron
Jan 11 at 4:42
|
show 3 more comments
$begingroup$
The question is to find the largest sum of $a, b, c$, given they're all relatively prime to each other and divide
$$a^n + b^n + c^n text{ } forall text{ } n ge 1 tag{1}label{eq1}$$
Since $a, b, c ge 1$, then $a + b + c ge 3$, so it consists of one or more prime factors. Call one of these prime factors $d$. Thus,
$$a + b + c equiv a^n + b^n + c^n equiv 0 pmod d tag{2}label{eq2}$$
As this must hold for all $n ge 1$, consider $n = 3$ and substitute
$$c equiv -a - b pmod d tag{3}label{eq3}$$
into the middle part of the congruence in eqref{eq2} to get
$$a^3 + b^3 + left(-a - bright)^3 equiv 0 pmod d tag{4}label{eq4}$$
This simplifies to
$$-3ab^2 - 3a^2b = -3ableft(a + bright) equiv 0 pmod d tag{5}label{eq5}$$
Thus, $d$ must divide $3$, $a$, $b$ and/or $a + b equiv -c pmod d$. If $d$ is not $3$, then note that if it must divide just one of $a$, $b$ and $-c$. This is because if it divides any $2$, say $a$ and $b$, it must also divide $c$. As the equations are symmetric in $a, b text{ and } c$, WLOG, assume that $d$ divides $a$, i.e., $a equiv 0 pmod d$. Thus, $b + c equiv 0 pmod d$, i.e., $b equiv -c pmod d$. Now, use $n = 2$ in the congruence in the middle part of eqref{eq2} to get
$$a^2 + b^2 + c^2 equiv 0 + left(-cright) + c^2 equiv 2c^2 equiv 0 pmod d tag{6}label{eq6}$$
As $d$ doesn't divide $c$, this means that $d$ must divide $2$, i.e., it must be $2$. Thus, the only possible prime factors of $a + b + c$ are $3$, as mentioned earlier, and $2$ as shown here.
Based on what I have found here, I believe the maximum value is just the product of those $2$ factors, i.e., $6$, which can be obtained from
$$left(a, b, cright) = left(1, 1, 4right) tag{7}label{eq7}$$
You can easily verify that $a^n + b^n + c^n = 1 + 1 + 4^n equiv 0 pmod 6$ for all $n ge 1$ as $1 + 1 = 2$, so $2$ divides it, and $4^n equiv 1^n equiv 1 pmod 3$, so $3$ divides it as well. I am posting this partial solution for now so you have something to start from, as I believe I'm on the right track. I need to go for a while, but I will keep thinking about this and, if I determine the rest of the solution, I will add it later.
answered Jan 11 at 4:20
John OmielanJohn Omielan
2,386212
$endgroup$
add a comment |
$begingroup$
The question is to find the largest sum of $a, b, c$, given they're all relatively prime to each other and divide
$$a^n + b^n + c^n text{ } forall text{ } n ge 1 tag{1}label{eq1}$$
Since $a, b, c ge 1$, then $a + b + c ge 3$, so it consists of one or more prime factors. Call one of these prime factors $d$. Thus,
$$a + b + c equiv a^n + b^n + c^n equiv 0 pmod d tag{2}label{eq2}$$
As this must hold for all $n ge 1$, consider $n = 3$ and substitute
$$c equiv -a - b pmod d tag{3}label{eq3}$$
into the middle part of the congruence in eqref{eq2} to get
$$a^3 + b^3 + left(-a - bright)^3 equiv 0 pmod d tag{4}label{eq4}$$
This simplifies to
$$-3ab^2 - 3a^2b = -3ableft(a + bright) equiv 0 pmod d tag{5}label{eq5}$$
Thus, $d$ must divide $3$, $a$, $b$ and/or $a + b equiv -c pmod d$. If $d$ is not $3$, then note that if it must divide just one of $a$, $b$ and $-c$. This is because if it divides any $2$, say $a$ and $b$, it must also divide $c$. As the equations are symmetric in $a, b text{ and } c$, WLOG, assume that $d$ divides $a$, i.e., $a equiv 0 pmod d$. Thus, $b + c equiv 0 pmod d$, i.e., $b equiv -c pmod d$. Now, use $n = 2$ in the congruence in the middle part of eqref{eq2} to get
$$a^2 + b^2 + c^2 equiv 0 + left(-cright) + c^2 equiv 2c^2 equiv 0 pmod d tag{6}label{eq6}$$
As $d$ doesn't divide $c$, this means that $d$ must divide $2$, i.e., it must be $2$. Thus, the only possible prime factors of $a + b + c$ are $3$, as mentioned earlier, and $2$ as shown here.
Based on what I have found here, I believe the maximum value is just the product of those $2$ factors, i.e., $6$, which can be obtained from
$$left(a, b, cright) = left(1, 1, 4right) tag{7}label{eq7}$$
You can easily verify that $a^n + b^n + c^n = 1 + 1 + 4^n equiv 0 pmod 6$ for all $n ge 1$ as $1 + 1 = 2$, so $2$ divides it, and $4^n equiv 1^n equiv 1 pmod 3$, so $3$ divides it as well. I am posting this partial solution for now so you have something to start from, as I believe I'm on the right track. I need to go for a while, but I will keep thinking about this and, if I determine the rest of the solution, I will add it later.
answered Jan 11 at 4:20
John OmielanJohn Omielan
2,386212
$endgroup$
add a comment |
$begingroup$
The question is to find the largest sum of $a, b, c$, given they're all relatively prime to each other and divide
$$a^n + b^n + c^n text{ } forall text{ } n ge 1 tag{1}label{eq1}$$
Since $a, b, c ge 1$, then $a + b + c ge 3$, so it consists of one or more prime factors. Call one of these prime factors $d$. Thus,
$$a + b + c equiv a^n + b^n + c^n equiv 0 pmod d tag{2}label{eq2}$$
As this must hold for all $n ge 1$, consider $n = 3$ and substitute
$$c equiv -a - b pmod d tag{3}label{eq3}$$
into the middle part of the congruence in eqref{eq2} to get
$$a^3 + b^3 + left(-a - bright)^3 equiv 0 pmod d tag{4}label{eq4}$$
This simplifies to
$$-3ab^2 - 3a^2b = -3ableft(a + bright) equiv 0 pmod d tag{5}label{eq5}$$
Thus, $d$ must divide $3$, $a$, $b$ and/or $a + b equiv -c pmod d$. If $d$ is not $3$, then note that if it must divide just one of $a$, $b$ and $-c$. This is because if it divides any $2$, say $a$ and $b$, it must also divide $c$. As the equations are symmetric in $a, b text{ and } c$, WLOG, assume that $d$ divides $a$, i.e., $a equiv 0 pmod d$. Thus, $b + c equiv 0 pmod d$, i.e., $b equiv -c pmod d$. Now, use $n = 2$ in the congruence in the middle part of eqref{eq2} to get
$$a^2 + b^2 + c^2 equiv 0 + left(-cright) + c^2 equiv 2c^2 equiv 0 pmod d tag{6}label{eq6}$$
As $d$ doesn't divide $c$, this means that $d$ must divide $2$, i.e., it must be $2$. Thus, the only possible prime factors of $a + b + c$ are $3$, as mentioned earlier, and $2$ as shown here.
Based on what I have found here, I believe the maximum value is just the product of those $2$ factors, i.e., $6$, which can be obtained from
$$left(a, b, cright) = left(1, 1, 4right) tag{7}label{eq7}$$
You can easily verify that $a^n + b^n + c^n = 1 + 1 + 4^n equiv 0 pmod 6$ for all $n ge 1$ as $1 + 1 = 2$, so $2$ divides it, and $4^n equiv 1^n equiv 1 pmod 3$, so $3$ divides it as well. I am posting this partial solution for now so you have something to start from, as I believe I'm on the right track. I need to go for a while, but I will keep thinking about this and, if I determine the rest of the solution, I will add it later.
answered Jan 11 at 4:20
John OmielanJohn Omielan
2,386212
$endgroup$
The question is to find the largest sum of $a, b, c$, given they're all relatively prime to each other and divide
$$a^n + b^n + c^n text{ } forall text{ } n ge 1 tag{1}label{eq1}$$
Since $a, b, c ge 1$, then $a + b + c ge 3$, so it consists of one or more prime factors. Call one of these prime factors $d$. Thus,
$$a + b + c equiv a^n + b^n + c^n equiv 0 pmod d tag{2}label{eq2}$$
As this must hold for all $n ge 1$, consider $n = 3$ and substitute
$$c equiv -a - b pmod d tag{3}label{eq3}$$
into the middle part of the congruence in eqref{eq2} to get
$$a^3 + b^3 + left(-a - bright)^3 equiv 0 pmod d tag{4}label{eq4}$$
This simplifies to
$$-3ab^2 - 3a^2b = -3ableft(a + bright) equiv 0 pmod d tag{5}label{eq5}$$
Thus, $d$ must divide $3$, $a$, $b$ and/or $a + b equiv -c pmod d$. If $d$ is not $3$, then note that if it must divide just one of $a$, $b$ and $-c$. This is because if it divides any $2$, say $a$ and $b$, it must also divide $c$. As the equations are symmetric in $a, b text{ and } c$, WLOG, assume that $d$ divides $a$, i.e., $a equiv 0 pmod d$. Thus, $b + c equiv 0 pmod d$, i.e., $b equiv -c pmod d$. Now, use $n = 2$ in the congruence in the middle part of eqref{eq2} to get
$$a^2 + b^2 + c^2 equiv 0 + left(-cright) + c^2 equiv 2c^2 equiv 0 pmod d tag{6}label{eq6}$$
As $d$ doesn't divide $c$, this means that $d$ must divide $2$, i.e., it must be $2$. Thus, the only possible prime factors of $a + b + c$ are $3$, as mentioned earlier, and $2$ as shown here.
Based on what I have found here, I believe the maximum value is just the product of those $2$ factors, i.e., $6$, which can be obtained from
$$left(a, b, cright) = left(1, 1, 4right) tag{7}label{eq7}$$
You can easily verify that $a^n + b^n + c^n = 1 + 1 + 4^n equiv 0 pmod 6$ for all $n ge 1$ as $1 + 1 = 2$, so $2$ divides it, and $4^n equiv 1^n equiv 1 pmod 3$, so $3$ divides it as well. I am posting this partial solution for now so you have something to start from, as I believe I'm on the right track. I need to go for a while, but I will keep thinking about this and, if I determine the rest of the solution, I will add it later.
answered Jan 11 at 4:20
John OmielanJohn Omielan
2,386212
answered Jan 11 at 4:20
John OmielanJohn Omielan
2,386212
answered Jan 11 at 4:20
John OmielanJohn Omielan
2,386212
answered Jan 11 at 4:20
John OmielanJohn Omielan
2,386212
2,386212
add a comment |
add a comment |
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What I think they are saying is if $a+b+cmid a^n+b^n+c^n$ for every $n$ and $gcd(a,b,c) = 1$ then what is the maximum value of $a+b+c$.
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– kingW3
Jan 11 at 2:04
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Could you please explain if there is any valid solution of this question?
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– Shromi
Jan 11 at 3:01
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@Shromi There is at least one valid "trivial" solution of $left(a,b,cright) = left(1,1,1right)$, which of course gives that $a + b + c = 3$. I'm not sure if there are any other larger solutions, with my best guess being that there isn't, but I'm not sure offhand how to prove it, although I am thinking about it.
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– John Omielan
Jan 11 at 3:10
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@Lee Thanks for the comment, however, for your for $left(a, b, cright) = left(1, 2, 3right)$, then $a + b + c = 6$, but $a^2 + b^2 + c^2 = 1 + 4 + 9 = 13$, which is not a multiple of $6$. Note the values must divide for every power of $n$.
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– John Omielan
Jan 11 at 4:08
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@JohnOmielan, yes, I miss that part
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– Lee
Jan 11 at 4:09