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Making the sum of 5th power of integers, a perfect square.
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Yesterday this question was posed in a contest. It contains pretty easy questions like asking range of $ab+bc+ca$ when $a^2+b^2+c^2=1$, etc.
But this question is something else. I haven't been able to solve this after $4$ hours.
Question:
Find, $n$ if
$133^5+27^5+84^5+110^5=n^2$.
I checked answer on calculator, it is $248832$.
I tried all sort of things I can do like trying to factor the expression, converting two of the odd nos. into $(m+n)^5+(m-n)^5$ form. But all in vain.
I hope anyone can help me here.
elementary-number-theory contest-math natural-numbers
asked Jul 30 at 8:29
Love Invariants
77715
 |Â
show 5 more comments
up vote
2
down vote
favorite
Yesterday this question was posed in a contest. It contains pretty easy questions like asking range of $ab+bc+ca$ when $a^2+b^2+c^2=1$, etc.
But this question is something else. I haven't been able to solve this after $4$ hours.
Question:
Find, $n$ if
$133^5+27^5+84^5+110^5=n^2$.
I checked answer on calculator, it is $248832$.
I tried all sort of things I can do like trying to factor the expression, converting two of the odd nos. into $(m+n)^5+(m-n)^5$ form. But all in vain.
I hope anyone can help me here.
elementary-number-theory contest-math natural-numbers
asked Jul 30 at 8:29
Love Invariants
77715
Not a very enlightening method but I guess you could try to take the sum modulo different numbers (taking ones which are easy to compute. For example, modulo 27, you get $(-2)^5+0^5+3^5+2^5 equiv 3^5 equiv 0 (textrmmod 27)$ and use Chinese remainder theorem to get $n equiv a(textrmmod N)$. You can bound $n$ to then retrieve the answer.
– daruma
Jul 30 at 8:45
5
It seems $248832=12^5$
– Henry
Jul 30 at 8:48
@daruma-Isn't that too much to ask when you have 10 minutes at max to solve in the paper?? I also thought of that but would take too much time also won't confirm that the answer is true.
– Love Invariants
Jul 30 at 8:49
6
Astonishing that you were asked to figure this out by yourself. The fact that the sum of these four fifth powers is equal to a fifth power, $n^2 = 144^5$, was discovered in a paper in 1966 as a counterexample to Euler's conjecture regarding sums of powers. They used a computer, not any analytic techniques. Look up Wikipedia for "Euler's conjecture counterexamples"
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 30 at 8:57
1
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó It is indeed true that it was published as a counter-example to a conjecture of Euler, but the difficult part is finding the counter-example, not proving that the sum of the four numbers is a fifth power. Once you are given the numbers, it is trivial to check that the claimed equality is true. The difficult part is finding the numbers in the first place.
– Dylan
Jul 31 at 20:36
 |Â
show 5 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Yesterday this question was posed in a contest. It contains pretty easy questions like asking range of $ab+bc+ca$ when $a^2+b^2+c^2=1$, etc.
But this question is something else. I haven't been able to solve this after $4$ hours.
Question:
Find, $n$ if
$133^5+27^5+84^5+110^5=n^2$.
I checked answer on calculator, it is $248832$.
I tried all sort of things I can do like trying to factor the expression, converting two of the odd nos. into $(m+n)^5+(m-n)^5$ form. But all in vain.
I hope anyone can help me here.
elementary-number-theory contest-math natural-numbers
asked Jul 30 at 8:29
Love Invariants
77715
Yesterday this question was posed in a contest. It contains pretty easy questions like asking range of $ab+bc+ca$ when $a^2+b^2+c^2=1$, etc.
But this question is something else. I haven't been able to solve this after $4$ hours.
Question:
Find, $n$ if
$133^5+27^5+84^5+110^5=n^2$.
I checked answer on calculator, it is $248832$.
I tried all sort of things I can do like trying to factor the expression, converting two of the odd nos. into $(m+n)^5+(m-n)^5$ form. But all in vain.
I hope anyone can help me here.
elementary-number-theory contest-math natural-numbers
asked Jul 30 at 8:29
Love Invariants
77715
edited Jul 30 at 8:35
asked Jul 30 at 8:29
Love Invariants
77715
asked Jul 30 at 8:29
Love Invariants
77715
asked Jul 30 at 8:29
Love Invariants
77715
77715
Not a very enlightening method but I guess you could try to take the sum modulo different numbers (taking ones which are easy to compute. For example, modulo 27, you get $(-2)^5+0^5+3^5+2^5 equiv 3^5 equiv 0 (textrmmod 27)$ and use Chinese remainder theorem to get $n equiv a(textrmmod N)$. You can bound $n$ to then retrieve the answer.
– daruma
Jul 30 at 8:45
5
It seems $248832=12^5$
– Henry
Jul 30 at 8:48
@daruma-Isn't that too much to ask when you have 10 minutes at max to solve in the paper?? I also thought of that but would take too much time also won't confirm that the answer is true.
– Love Invariants
Jul 30 at 8:49
6
Astonishing that you were asked to figure this out by yourself. The fact that the sum of these four fifth powers is equal to a fifth power, $n^2 = 144^5$, was discovered in a paper in 1966 as a counterexample to Euler's conjecture regarding sums of powers. They used a computer, not any analytic techniques. Look up Wikipedia for "Euler's conjecture counterexamples"
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 30 at 8:57
1
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó It is indeed true that it was published as a counter-example to a conjecture of Euler, but the difficult part is finding the counter-example, not proving that the sum of the four numbers is a fifth power. Once you are given the numbers, it is trivial to check that the claimed equality is true. The difficult part is finding the numbers in the first place.
– Dylan
Jul 31 at 20:36
 |Â
show 5 more comments
Not a very enlightening method but I guess you could try to take the sum modulo different numbers (taking ones which are easy to compute. For example, modulo 27, you get $(-2)^5+0^5+3^5+2^5 equiv 3^5 equiv 0 (textrmmod 27)$ and use Chinese remainder theorem to get $n equiv a(textrmmod N)$. You can bound $n$ to then retrieve the answer.
– daruma
Jul 30 at 8:45
5
It seems $248832=12^5$
– Henry
Jul 30 at 8:48
@daruma-Isn't that too much to ask when you have 10 minutes at max to solve in the paper?? I also thought of that but would take too much time also won't confirm that the answer is true.
– Love Invariants
Jul 30 at 8:49
6
Astonishing that you were asked to figure this out by yourself. The fact that the sum of these four fifth powers is equal to a fifth power, $n^2 = 144^5$, was discovered in a paper in 1966 as a counterexample to Euler's conjecture regarding sums of powers. They used a computer, not any analytic techniques. Look up Wikipedia for "Euler's conjecture counterexamples"
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 30 at 8:57
1
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó It is indeed true that it was published as a counter-example to a conjecture of Euler, but the difficult part is finding the counter-example, not proving that the sum of the four numbers is a fifth power. Once you are given the numbers, it is trivial to check that the claimed equality is true. The difficult part is finding the numbers in the first place.
– Dylan
Jul 31 at 20:36
Not a very enlightening method but I guess you could try to take the sum modulo different numbers (taking ones which are easy to compute. For example, modulo 27, you get $(-2)^5+0^5+3^5+2^5 equiv 3^5 equiv 0 (textrmmod 27)$ and use Chinese remainder theorem to get $n equiv a(textrmmod N)$. You can bound $n$ to then retrieve the answer.
– daruma
Jul 30 at 8:45
Not a very enlightening method but I guess you could try to take the sum modulo different numbers (taking ones which are easy to compute. For example, modulo 27, you get $(-2)^5+0^5+3^5+2^5 equiv 3^5 equiv 0 (textrmmod 27)$ and use Chinese remainder theorem to get $n equiv a(textrmmod N)$. You can bound $n$ to then retrieve the answer.
– daruma
Jul 30 at 8:45
5
5
It seems $248832=12^5$
– Henry
Jul 30 at 8:48
It seems $248832=12^5$
– Henry
Jul 30 at 8:48
@daruma-Isn't that too much to ask when you have 10 minutes at max to solve in the paper?? I also thought of that but would take too much time also won't confirm that the answer is true.
– Love Invariants
Jul 30 at 8:49
@daruma-Isn't that too much to ask when you have 10 minutes at max to solve in the paper?? I also thought of that but would take too much time also won't confirm that the answer is true.
– Love Invariants
Jul 30 at 8:49
6
6
Astonishing that you were asked to figure this out by yourself. The fact that the sum of these four fifth powers is equal to a fifth power, $n^2 = 144^5$, was discovered in a paper in 1966 as a counterexample to Euler's conjecture regarding sums of powers. They used a computer, not any analytic techniques. Look up Wikipedia for "Euler's conjecture counterexamples"
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 30 at 8:57
Astonishing that you were asked to figure this out by yourself. The fact that the sum of these four fifth powers is equal to a fifth power, $n^2 = 144^5$, was discovered in a paper in 1966 as a counterexample to Euler's conjecture regarding sums of powers. They used a computer, not any analytic techniques. Look up Wikipedia for "Euler's conjecture counterexamples"
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 30 at 8:57
1
1
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó It is indeed true that it was published as a counter-example to a conjecture of Euler, but the difficult part is finding the counter-example, not proving that the sum of the four numbers is a fifth power. Once you are given the numbers, it is trivial to check that the claimed equality is true. The difficult part is finding the numbers in the first place.
– Dylan
Jul 31 at 20:36
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó It is indeed true that it was published as a counter-example to a conjecture of Euler, but the difficult part is finding the counter-example, not proving that the sum of the four numbers is a fifth power. Once you are given the numbers, it is trivial to check that the claimed equality is true. The difficult part is finding the numbers in the first place.
– Dylan
Jul 31 at 20:36
 |Â
show 5 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
Solution
Let's make a transformation, to find $n in mathbbN_+$ such that $133^5+27^5+84^5+110^5=m^5.$
Step 1
Estimate the bound of $m$.
Notice that $$133^5<10 times 100^5,~~~27^5<1 times 100^5,~~~84^5<1 times 100^5,~~~110^5<10 times 100^5.$$Thus $$133^5<m^5=133^5+27^5+84^5+110^5<22 times 100^5<200^5.$$
Hence, $$133<m<200.$$
Step 2
Find the unit-digit of $m$.
Since $$m^5 equiv133^5+27^5+84^5+110^5equiv3^5+7^5+4^5+0^5equiv3+7+4+0equiv 4(textrmmod 10) ,$$hence the unitdigit of $m$ is $4.$
Step 3
Find $m$.
Since $$m^5 equiv133^5+27^5+84^5+110^5equiv1^5+0^5+0^5+2^5equiv 0(textrmmod 3) ,$$hence $3|m^5$. Further, $3|m$. Therefore, $m=144$ or $174.$ However, since$$m^5 equiv133^5+27^5+84^5+110^5equiv0^5+6^5+0^5+5^5equiv 2(textrmmod 7) ,$$ and $144^5 equiv 4^5 equiv 2 (textrmmod 7)$, $174^5 equiv 6^5 equiv 6 (textrmmod 7)$,$174$ is not the solution. As a result,$$m=144.$$
answered Jul 30 at 11:22
mengdie1982
2,800216
Yes, if you are told that $n$ is a fifth power it becomes fairly easy. But this information was not available to the OP!
– TonyK
Jul 30 at 12:09
@mengdie- Nope bro this solution is wrong. I wasn't given sum of those numbers is 5th power of some no.
– Love Invariants
Jul 30 at 13:35
This method could be generalized to solve OP. Further more, if $n^2 = 144^5 = (12^2)^5 = (12^5)^2$, then $n = 12^5$.
– xbh
2 days ago
add a comment |Â
up vote
1
down vote
This is less than a complete solution but too long for a comment.
Suppose $n^2 = a^5 + b^5 + c^5 + d^5$. Then:
$$n^2 = (a+b)m + c^5 + d^5quadquad(1)$$
where:
$$m = a^4 -a^3b + a^2b^2 -ab^3 + b^4$$
Applying $(1)$ with $(a,b,c,d) = (133,27,110,84)$:
$$n^2 = 160m + 110^5 + 84^5 = 2^5(5m + 55^5 + 42^5)$$
Hence $2^5 | n^2$ and so $2^3 | n$.
Also, applying $(1)$ with $(a,b,c,d) = (133,110,84,27)$:
$$n^2 = 243m + 84^5 + 27^5 = 3^5(m + 28^5 + 9^5)$$
Hence $3^5 | n^2$ and so $3^3 | n$.
Combining the above results:
$$2^33^3 | n$$
answered Aug 1 at 12:12
Adam Bailey
1,8541118
It is helpful but I don't see how I will reach the correct answer from this.
– Love Invariants
Aug 1 at 16:01
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Solution
Let's make a transformation, to find $n in mathbbN_+$ such that $133^5+27^5+84^5+110^5=m^5.$
Step 1
Estimate the bound of $m$.
Notice that $$133^5<10 times 100^5,~~~27^5<1 times 100^5,~~~84^5<1 times 100^5,~~~110^5<10 times 100^5.$$Thus $$133^5<m^5=133^5+27^5+84^5+110^5<22 times 100^5<200^5.$$
Hence, $$133<m<200.$$
Step 2
Find the unit-digit of $m$.
Since $$m^5 equiv133^5+27^5+84^5+110^5equiv3^5+7^5+4^5+0^5equiv3+7+4+0equiv 4(textrmmod 10) ,$$hence the unitdigit of $m$ is $4.$
Step 3
Find $m$.
Since $$m^5 equiv133^5+27^5+84^5+110^5equiv1^5+0^5+0^5+2^5equiv 0(textrmmod 3) ,$$hence $3|m^5$. Further, $3|m$. Therefore, $m=144$ or $174.$ However, since$$m^5 equiv133^5+27^5+84^5+110^5equiv0^5+6^5+0^5+5^5equiv 2(textrmmod 7) ,$$ and $144^5 equiv 4^5 equiv 2 (textrmmod 7)$, $174^5 equiv 6^5 equiv 6 (textrmmod 7)$,$174$ is not the solution. As a result,$$m=144.$$
answered Jul 30 at 11:22
mengdie1982
2,800216
Yes, if you are told that $n$ is a fifth power it becomes fairly easy. But this information was not available to the OP!
– TonyK
Jul 30 at 12:09
@mengdie- Nope bro this solution is wrong. I wasn't given sum of those numbers is 5th power of some no.
– Love Invariants
Jul 30 at 13:35
This method could be generalized to solve OP. Further more, if $n^2 = 144^5 = (12^2)^5 = (12^5)^2$, then $n = 12^5$.
– xbh
2 days ago
add a comment |Â
up vote
1
down vote
Solution
Let's make a transformation, to find $n in mathbbN_+$ such that $133^5+27^5+84^5+110^5=m^5.$
Step 1
Estimate the bound of $m$.
Notice that $$133^5<10 times 100^5,~~~27^5<1 times 100^5,~~~84^5<1 times 100^5,~~~110^5<10 times 100^5.$$Thus $$133^5<m^5=133^5+27^5+84^5+110^5<22 times 100^5<200^5.$$
Hence, $$133<m<200.$$
Step 2
Find the unit-digit of $m$.
Since $$m^5 equiv133^5+27^5+84^5+110^5equiv3^5+7^5+4^5+0^5equiv3+7+4+0equiv 4(textrmmod 10) ,$$hence the unitdigit of $m$ is $4.$
Step 3
Find $m$.
Since $$m^5 equiv133^5+27^5+84^5+110^5equiv1^5+0^5+0^5+2^5equiv 0(textrmmod 3) ,$$hence $3|m^5$. Further, $3|m$. Therefore, $m=144$ or $174.$ However, since$$m^5 equiv133^5+27^5+84^5+110^5equiv0^5+6^5+0^5+5^5equiv 2(textrmmod 7) ,$$ and $144^5 equiv 4^5 equiv 2 (textrmmod 7)$, $174^5 equiv 6^5 equiv 6 (textrmmod 7)$,$174$ is not the solution. As a result,$$m=144.$$
answered Jul 30 at 11:22
mengdie1982
2,800216
Yes, if you are told that $n$ is a fifth power it becomes fairly easy. But this information was not available to the OP!
– TonyK
Jul 30 at 12:09
@mengdie- Nope bro this solution is wrong. I wasn't given sum of those numbers is 5th power of some no.
– Love Invariants
Jul 30 at 13:35
This method could be generalized to solve OP. Further more, if $n^2 = 144^5 = (12^2)^5 = (12^5)^2$, then $n = 12^5$.
– xbh
2 days ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Solution
Let's make a transformation, to find $n in mathbbN_+$ such that $133^5+27^5+84^5+110^5=m^5.$
Step 1
Estimate the bound of $m$.
Notice that $$133^5<10 times 100^5,~~~27^5<1 times 100^5,~~~84^5<1 times 100^5,~~~110^5<10 times 100^5.$$Thus $$133^5<m^5=133^5+27^5+84^5+110^5<22 times 100^5<200^5.$$
Hence, $$133<m<200.$$
Step 2
Find the unit-digit of $m$.
Since $$m^5 equiv133^5+27^5+84^5+110^5equiv3^5+7^5+4^5+0^5equiv3+7+4+0equiv 4(textrmmod 10) ,$$hence the unitdigit of $m$ is $4.$
Step 3
Find $m$.
Since $$m^5 equiv133^5+27^5+84^5+110^5equiv1^5+0^5+0^5+2^5equiv 0(textrmmod 3) ,$$hence $3|m^5$. Further, $3|m$. Therefore, $m=144$ or $174.$ However, since$$m^5 equiv133^5+27^5+84^5+110^5equiv0^5+6^5+0^5+5^5equiv 2(textrmmod 7) ,$$ and $144^5 equiv 4^5 equiv 2 (textrmmod 7)$, $174^5 equiv 6^5 equiv 6 (textrmmod 7)$,$174$ is not the solution. As a result,$$m=144.$$
answered Jul 30 at 11:22
mengdie1982
2,800216
Solution
Let's make a transformation, to find $n in mathbbN_+$ such that $133^5+27^5+84^5+110^5=m^5.$
Step 1
Estimate the bound of $m$.
Notice that $$133^5<10 times 100^5,~~~27^5<1 times 100^5,~~~84^5<1 times 100^5,~~~110^5<10 times 100^5.$$Thus $$133^5<m^5=133^5+27^5+84^5+110^5<22 times 100^5<200^5.$$
Hence, $$133<m<200.$$
Step 2
Find the unit-digit of $m$.
Since $$m^5 equiv133^5+27^5+84^5+110^5equiv3^5+7^5+4^5+0^5equiv3+7+4+0equiv 4(textrmmod 10) ,$$hence the unitdigit of $m$ is $4.$
Step 3
Find $m$.
Since $$m^5 equiv133^5+27^5+84^5+110^5equiv1^5+0^5+0^5+2^5equiv 0(textrmmod 3) ,$$hence $3|m^5$. Further, $3|m$. Therefore, $m=144$ or $174.$ However, since$$m^5 equiv133^5+27^5+84^5+110^5equiv0^5+6^5+0^5+5^5equiv 2(textrmmod 7) ,$$ and $144^5 equiv 4^5 equiv 2 (textrmmod 7)$, $174^5 equiv 6^5 equiv 6 (textrmmod 7)$,$174$ is not the solution. As a result,$$m=144.$$
answered Jul 30 at 11:22
mengdie1982
2,800216
edited Jul 30 at 14:42
answered Jul 30 at 11:22
mengdie1982
2,800216
answered Jul 30 at 11:22
mengdie1982
2,800216
answered Jul 30 at 11:22
mengdie1982
2,800216
2,800216
Yes, if you are told that $n$ is a fifth power it becomes fairly easy. But this information was not available to the OP!
– TonyK
Jul 30 at 12:09
@mengdie- Nope bro this solution is wrong. I wasn't given sum of those numbers is 5th power of some no.
– Love Invariants
Jul 30 at 13:35
This method could be generalized to solve OP. Further more, if $n^2 = 144^5 = (12^2)^5 = (12^5)^2$, then $n = 12^5$.
– xbh
2 days ago
add a comment |Â
Yes, if you are told that $n$ is a fifth power it becomes fairly easy. But this information was not available to the OP!
– TonyK
Jul 30 at 12:09
@mengdie- Nope bro this solution is wrong. I wasn't given sum of those numbers is 5th power of some no.
– Love Invariants
Jul 30 at 13:35
This method could be generalized to solve OP. Further more, if $n^2 = 144^5 = (12^2)^5 = (12^5)^2$, then $n = 12^5$.
– xbh
2 days ago
Yes, if you are told that $n$ is a fifth power it becomes fairly easy. But this information was not available to the OP!
– TonyK
Jul 30 at 12:09
Yes, if you are told that $n$ is a fifth power it becomes fairly easy. But this information was not available to the OP!
– TonyK
Jul 30 at 12:09
@mengdie- Nope bro this solution is wrong. I wasn't given sum of those numbers is 5th power of some no.
– Love Invariants
Jul 30 at 13:35
@mengdie- Nope bro this solution is wrong. I wasn't given sum of those numbers is 5th power of some no.
– Love Invariants
Jul 30 at 13:35
This method could be generalized to solve OP. Further more, if $n^2 = 144^5 = (12^2)^5 = (12^5)^2$, then $n = 12^5$.
– xbh
2 days ago
This method could be generalized to solve OP. Further more, if $n^2 = 144^5 = (12^2)^5 = (12^5)^2$, then $n = 12^5$.
– xbh
2 days ago
add a comment |Â
up vote
1
down vote
This is less than a complete solution but too long for a comment.
Suppose $n^2 = a^5 + b^5 + c^5 + d^5$. Then:
$$n^2 = (a+b)m + c^5 + d^5quadquad(1)$$
where:
$$m = a^4 -a^3b + a^2b^2 -ab^3 + b^4$$
Applying $(1)$ with $(a,b,c,d) = (133,27,110,84)$:
$$n^2 = 160m + 110^5 + 84^5 = 2^5(5m + 55^5 + 42^5)$$
Hence $2^5 | n^2$ and so $2^3 | n$.
Also, applying $(1)$ with $(a,b,c,d) = (133,110,84,27)$:
$$n^2 = 243m + 84^5 + 27^5 = 3^5(m + 28^5 + 9^5)$$
Hence $3^5 | n^2$ and so $3^3 | n$.
Combining the above results:
$$2^33^3 | n$$
answered Aug 1 at 12:12
Adam Bailey
1,8541118
It is helpful but I don't see how I will reach the correct answer from this.
– Love Invariants
Aug 1 at 16:01
add a comment |Â
up vote
1
down vote
This is less than a complete solution but too long for a comment.
Suppose $n^2 = a^5 + b^5 + c^5 + d^5$. Then:
$$n^2 = (a+b)m + c^5 + d^5quadquad(1)$$
where:
$$m = a^4 -a^3b + a^2b^2 -ab^3 + b^4$$
Applying $(1)$ with $(a,b,c,d) = (133,27,110,84)$:
$$n^2 = 160m + 110^5 + 84^5 = 2^5(5m + 55^5 + 42^5)$$
Hence $2^5 | n^2$ and so $2^3 | n$.
Also, applying $(1)$ with $(a,b,c,d) = (133,110,84,27)$:
$$n^2 = 243m + 84^5 + 27^5 = 3^5(m + 28^5 + 9^5)$$
Hence $3^5 | n^2$ and so $3^3 | n$.
Combining the above results:
$$2^33^3 | n$$
answered Aug 1 at 12:12
Adam Bailey
1,8541118
It is helpful but I don't see how I will reach the correct answer from this.
– Love Invariants
Aug 1 at 16:01
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is less than a complete solution but too long for a comment.
Suppose $n^2 = a^5 + b^5 + c^5 + d^5$. Then:
$$n^2 = (a+b)m + c^5 + d^5quadquad(1)$$
where:
$$m = a^4 -a^3b + a^2b^2 -ab^3 + b^4$$
Applying $(1)$ with $(a,b,c,d) = (133,27,110,84)$:
$$n^2 = 160m + 110^5 + 84^5 = 2^5(5m + 55^5 + 42^5)$$
Hence $2^5 | n^2$ and so $2^3 | n$.
Also, applying $(1)$ with $(a,b,c,d) = (133,110,84,27)$:
$$n^2 = 243m + 84^5 + 27^5 = 3^5(m + 28^5 + 9^5)$$
Hence $3^5 | n^2$ and so $3^3 | n$.
Combining the above results:
$$2^33^3 | n$$
answered Aug 1 at 12:12
Adam Bailey
1,8541118
This is less than a complete solution but too long for a comment.
Suppose $n^2 = a^5 + b^5 + c^5 + d^5$. Then:
$$n^2 = (a+b)m + c^5 + d^5quadquad(1)$$
where:
$$m = a^4 -a^3b + a^2b^2 -ab^3 + b^4$$
Applying $(1)$ with $(a,b,c,d) = (133,27,110,84)$:
$$n^2 = 160m + 110^5 + 84^5 = 2^5(5m + 55^5 + 42^5)$$
Hence $2^5 | n^2$ and so $2^3 | n$.
Also, applying $(1)$ with $(a,b,c,d) = (133,110,84,27)$:
$$n^2 = 243m + 84^5 + 27^5 = 3^5(m + 28^5 + 9^5)$$
Hence $3^5 | n^2$ and so $3^3 | n$.
Combining the above results:
$$2^33^3 | n$$
answered Aug 1 at 12:12
Adam Bailey
1,8541118
answered Aug 1 at 12:12
Adam Bailey
1,8541118
answered Aug 1 at 12:12
Adam Bailey
1,8541118
answered Aug 1 at 12:12
Adam Bailey
1,8541118
1,8541118
It is helpful but I don't see how I will reach the correct answer from this.
– Love Invariants
Aug 1 at 16:01
add a comment |Â
It is helpful but I don't see how I will reach the correct answer from this.
– Love Invariants
Aug 1 at 16:01
It is helpful but I don't see how I will reach the correct answer from this.
– Love Invariants
Aug 1 at 16:01
It is helpful but I don't see how I will reach the correct answer from this.
– Love Invariants
Aug 1 at 16:01
add a comment |Â
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Not a very enlightening method but I guess you could try to take the sum modulo different numbers (taking ones which are easy to compute. For example, modulo 27, you get $(-2)^5+0^5+3^5+2^5 equiv 3^5 equiv 0 (textrmmod 27)$ and use Chinese remainder theorem to get $n equiv a(textrmmod N)$. You can bound $n$ to then retrieve the answer.
– daruma
Jul 30 at 8:45
5
It seems $248832=12^5$
– Henry
Jul 30 at 8:48
@daruma-Isn't that too much to ask when you have 10 minutes at max to solve in the paper?? I also thought of that but would take too much time also won't confirm that the answer is true.
– Love Invariants
Jul 30 at 8:49
6
Astonishing that you were asked to figure this out by yourself. The fact that the sum of these four fifth powers is equal to a fifth power, $n^2 = 144^5$, was discovered in a paper in 1966 as a counterexample to Euler's conjecture regarding sums of powers. They used a computer, not any analytic techniques. Look up Wikipedia for "Euler's conjecture counterexamples"
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 30 at 8:57
1
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó It is indeed true that it was published as a counter-example to a conjecture of Euler, but the difficult part is finding the counter-example, not proving that the sum of the four numbers is a fifth power. Once you are given the numbers, it is trivial to check that the claimed equality is true. The difficult part is finding the numbers in the first place.
– Dylan
Jul 31 at 20:36