Mixing
Number theory for MTRP
$begingroup$
Find all $(k,n)$, for which $(2^k-1)^n + (2^k+1)^n$ is a perfect square.
I did something like this.
$(2^k-1)^n + (2^k+1)^n = d^2$
Since both are odd, $d$ must be even.
I considered this as $mod 3$, $mod 4$, since $x^2$ congruent to $0$ or $1$ $mod(3,4)$. But this too did not help me.
I now have no idea how to proceed further.
I could not solve this problem but i tried very much.
Check problem 25. You may also note down the other problems as well.
number-theory
$endgroup$
|
show 1 more comment
$begingroup$
Find all $(k,n)$, for which $(2^k-1)^n + (2^k+1)^n$ is a perfect square.
I did something like this.
$(2^k-1)^n + (2^k+1)^n = d^2$
Since both are odd, $d$ must be even.
I considered this as $mod 3$, $mod 4$, since $x^2$ congruent to $0$ or $1$ $mod(3,4)$. But this too did not help me.
I now have no idea how to proceed further.
I could not solve this problem but i tried very much.
Check problem 25. You may also note down the other problems as well.
number-theory
$endgroup$
$begingroup$
Please dont downvote me... i dont know how to use latex.. i am new here
$endgroup$
– user636268
Jan 19 at 11:46
$begingroup$
Hi, this website can teach you how to type latex: en.wikibooks.org/wiki/LaTeX/Mathematics
$endgroup$
– kelvin hong 方
Jan 19 at 11:47
$begingroup$
Do i need to install some kind of applications?
$endgroup$
– user636268
Jan 19 at 11:49
$begingroup$
There is no need to install anything for you to use LaTeX in MSE.
$endgroup$
– kelvin hong 方
Jan 19 at 11:52
$begingroup$
Tomorrow is exam.. please somebody answer fast
$endgroup$
– user636268
Jan 19 at 12:00
|
show 1 more comment
$begingroup$
Find all $(k,n)$, for which $(2^k-1)^n + (2^k+1)^n$ is a perfect square.
I did something like this.
$(2^k-1)^n + (2^k+1)^n = d^2$
Since both are odd, $d$ must be even.
I considered this as $mod 3$, $mod 4$, since $x^2$ congruent to $0$ or $1$ $mod(3,4)$. But this too did not help me.
I now have no idea how to proceed further.
I could not solve this problem but i tried very much.
Check problem 25. You may also note down the other problems as well.
number-theory
$endgroup$
Find all $(k,n)$, for which $(2^k-1)^n + (2^k+1)^n$ is a perfect square.
I did something like this.
$(2^k-1)^n + (2^k+1)^n = d^2$
Since both are odd, $d$ must be even.
I considered this as $mod 3$, $mod 4$, since $x^2$ congruent to $0$ or $1$ $mod(3,4)$. But this too did not help me.
I now have no idea how to proceed further.
I could not solve this problem but i tried very much.
Check problem 25. You may also note down the other problems as well.
number-theory
number-theory
edited Jan 19 at 11:55
asked Jan 19 at 11:39
user636268
$begingroup$
Please dont downvote me... i dont know how to use latex.. i am new here
$endgroup$
– user636268
Jan 19 at 11:46
$begingroup$
Hi, this website can teach you how to type latex: en.wikibooks.org/wiki/LaTeX/Mathematics
$endgroup$
– kelvin hong 方
Jan 19 at 11:47
$begingroup$
Do i need to install some kind of applications?
$endgroup$
– user636268
Jan 19 at 11:49
$begingroup$
There is no need to install anything for you to use LaTeX in MSE.
$endgroup$
– kelvin hong 方
Jan 19 at 11:52
$begingroup$
Tomorrow is exam.. please somebody answer fast
$endgroup$
– user636268
Jan 19 at 12:00
|
show 1 more comment
$begingroup$
Please dont downvote me... i dont know how to use latex.. i am new here
$endgroup$
– user636268
Jan 19 at 11:46
$begingroup$
Hi, this website can teach you how to type latex: en.wikibooks.org/wiki/LaTeX/Mathematics
$endgroup$
– kelvin hong 方
Jan 19 at 11:47
$begingroup$
Do i need to install some kind of applications?
$endgroup$
– user636268
Jan 19 at 11:49
$begingroup$
There is no need to install anything for you to use LaTeX in MSE.
$endgroup$
– kelvin hong 方
Jan 19 at 11:52
$begingroup$
Tomorrow is exam.. please somebody answer fast
$endgroup$
– user636268
Jan 19 at 12:00
$begingroup$
Please dont downvote me... i dont know how to use latex.. i am new here
$endgroup$
– user636268
Jan 19 at 11:46
$begingroup$
Please dont downvote me... i dont know how to use latex.. i am new here
$endgroup$
– user636268
Jan 19 at 11:46
$begingroup$
Hi, this website can teach you how to type latex: en.wikibooks.org/wiki/LaTeX/Mathematics
$endgroup$
– kelvin hong 方
Jan 19 at 11:47
$begingroup$
Hi, this website can teach you how to type latex: en.wikibooks.org/wiki/LaTeX/Mathematics
$endgroup$
– kelvin hong 方
Jan 19 at 11:47
$begingroup$
Do i need to install some kind of applications?
$endgroup$
– user636268
Jan 19 at 11:49
$begingroup$
Do i need to install some kind of applications?
$endgroup$
– user636268
Jan 19 at 11:49
$begingroup$
There is no need to install anything for you to use LaTeX in MSE.
$endgroup$
– kelvin hong 方
Jan 19 at 11:52
$begingroup$
There is no need to install anything for you to use LaTeX in MSE.
$endgroup$
– kelvin hong 方
Jan 19 at 11:52
$begingroup$
Tomorrow is exam.. please somebody answer fast
$endgroup$
– user636268
Jan 19 at 12:00
$begingroup$
Tomorrow is exam.. please somebody answer fast
$endgroup$
– user636268
Jan 19 at 12:00
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
My solution is not complete, but I hope someone can fill in the missing things.
My result: $n$ cannot be even, and also it should be an odd number with $nequiv 1pmod 8$.
Let's proceed: Suppose $$d^2=(2^k-1)^n+(2^k+1)^n$$
If $ngeq 2$ is even, then using binomial expansion we have $$d^2=2[(2^k)^n+binom{n}{2}(2^k)^{n-2}+cdots+1]$$
Clearly $d^2$ is divisible by $2$ but not $4$, hence every even $n$ doesn't work for sure.
If $ngeq 1$ is odd, we first take $n=1$, and we see $$d^2=2^{k+1}$$
Hence we have $(n,k)=(1,1),(1,3),(1,5), cdots$.
If $ngeq 3$, then begin{align*}d^2&=2[(2^k)^n+binom n2(2^k)^{n-2}+cdots+binom{n}{n-1}2^k]\
&=2^{k+1}bigg[(2^k)^{n-1}+binom{n}{2}(2^k)^{n-3}+cdots+binom{n}{n-3}(2^k)^2+nbigg]end{align*}
We see that since $n$ is odd, the big square bracket is odd, hence $2^{k+1}$ must itself be a perfect square, hence $k$ is odd. Then the big square bracket should itself be an odd square, and we see any odd square should have the property $1pmod 8$ since for any positive integer $m$ we have $$(2m+1)^2equiv 4m^2+4m+1equiv 4m(m+1)+1equiv 1pmod 8.$$
Hence when $n$ is an odd number with $nequiv 1pmod 8$, $k$ is an odd number, but I cannot found any example when $n> 1$, there should be no solution at this stage but I cannot prove it.
answered Jan 19 at 12:04
kelvin hong 方kelvin hong 方
72218
$endgroup$
$begingroup$
Thanks for your response!
$endgroup$
– user636268
Jan 19 at 12:17
$begingroup$
What if we substitute some numbef with a special property, say a perfect square itself and check?
$endgroup$
– user636268
Jan 19 at 12:23
$begingroup$
You can try it, but as my proof saying, the minimum next integer value of $n$ you can try is $n=9$, which makes calculation becomes huge very fast, if you have a powerful calculator you can try to do that.
$endgroup$
– kelvin hong 方
Jan 19 at 12:36
$begingroup$
If that does not help🤔
$endgroup$
– user636268
Jan 19 at 12:39
2
$begingroup$
Yes! That is why i am using MSE
$endgroup$
– user636268
Jan 19 at 12:42
|
show 1 more comment
Your Answer
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1 Answer
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$begingroup$
My solution is not complete, but I hope someone can fill in the missing things.
My result: $n$ cannot be even, and also it should be an odd number with $nequiv 1pmod 8$.
Let's proceed: Suppose $$d^2=(2^k-1)^n+(2^k+1)^n$$
If $ngeq 2$ is even, then using binomial expansion we have $$d^2=2[(2^k)^n+binom{n}{2}(2^k)^{n-2}+cdots+1]$$
Clearly $d^2$ is divisible by $2$ but not $4$, hence every even $n$ doesn't work for sure.
If $ngeq 1$ is odd, we first take $n=1$, and we see $$d^2=2^{k+1}$$
Hence we have $(n,k)=(1,1),(1,3),(1,5), cdots$.
If $ngeq 3$, then begin{align*}d^2&=2[(2^k)^n+binom n2(2^k)^{n-2}+cdots+binom{n}{n-1}2^k]\
&=2^{k+1}bigg[(2^k)^{n-1}+binom{n}{2}(2^k)^{n-3}+cdots+binom{n}{n-3}(2^k)^2+nbigg]end{align*}
We see that since $n$ is odd, the big square bracket is odd, hence $2^{k+1}$ must itself be a perfect square, hence $k$ is odd. Then the big square bracket should itself be an odd square, and we see any odd square should have the property $1pmod 8$ since for any positive integer $m$ we have $$(2m+1)^2equiv 4m^2+4m+1equiv 4m(m+1)+1equiv 1pmod 8.$$
Hence when $n$ is an odd number with $nequiv 1pmod 8$, $k$ is an odd number, but I cannot found any example when $n> 1$, there should be no solution at this stage but I cannot prove it.
answered Jan 19 at 12:04
kelvin hong 方kelvin hong 方
72218
$endgroup$
$begingroup$
Thanks for your response!
$endgroup$
– user636268
Jan 19 at 12:17
$begingroup$
What if we substitute some numbef with a special property, say a perfect square itself and check?
$endgroup$
– user636268
Jan 19 at 12:23
$begingroup$
You can try it, but as my proof saying, the minimum next integer value of $n$ you can try is $n=9$, which makes calculation becomes huge very fast, if you have a powerful calculator you can try to do that.
$endgroup$
– kelvin hong 方
Jan 19 at 12:36
$begingroup$
If that does not help🤔
$endgroup$
– user636268
Jan 19 at 12:39
2
$begingroup$
Yes! That is why i am using MSE
$endgroup$
– user636268
Jan 19 at 12:42
|
show 1 more comment
$begingroup$
My solution is not complete, but I hope someone can fill in the missing things.
My result: $n$ cannot be even, and also it should be an odd number with $nequiv 1pmod 8$.
Let's proceed: Suppose $$d^2=(2^k-1)^n+(2^k+1)^n$$
If $ngeq 2$ is even, then using binomial expansion we have $$d^2=2[(2^k)^n+binom{n}{2}(2^k)^{n-2}+cdots+1]$$
Clearly $d^2$ is divisible by $2$ but not $4$, hence every even $n$ doesn't work for sure.
If $ngeq 1$ is odd, we first take $n=1$, and we see $$d^2=2^{k+1}$$
Hence we have $(n,k)=(1,1),(1,3),(1,5), cdots$.
If $ngeq 3$, then begin{align*}d^2&=2[(2^k)^n+binom n2(2^k)^{n-2}+cdots+binom{n}{n-1}2^k]\
&=2^{k+1}bigg[(2^k)^{n-1}+binom{n}{2}(2^k)^{n-3}+cdots+binom{n}{n-3}(2^k)^2+nbigg]end{align*}
We see that since $n$ is odd, the big square bracket is odd, hence $2^{k+1}$ must itself be a perfect square, hence $k$ is odd. Then the big square bracket should itself be an odd square, and we see any odd square should have the property $1pmod 8$ since for any positive integer $m$ we have $$(2m+1)^2equiv 4m^2+4m+1equiv 4m(m+1)+1equiv 1pmod 8.$$
Hence when $n$ is an odd number with $nequiv 1pmod 8$, $k$ is an odd number, but I cannot found any example when $n> 1$, there should be no solution at this stage but I cannot prove it.
answered Jan 19 at 12:04
kelvin hong 方kelvin hong 方
72218
$endgroup$
$begingroup$
Thanks for your response!
$endgroup$
– user636268
Jan 19 at 12:17
$begingroup$
What if we substitute some numbef with a special property, say a perfect square itself and check?
$endgroup$
– user636268
Jan 19 at 12:23
$begingroup$
You can try it, but as my proof saying, the minimum next integer value of $n$ you can try is $n=9$, which makes calculation becomes huge very fast, if you have a powerful calculator you can try to do that.
$endgroup$
– kelvin hong 方
Jan 19 at 12:36
$begingroup$
If that does not help🤔
$endgroup$
– user636268
Jan 19 at 12:39
2
$begingroup$
Yes! That is why i am using MSE
$endgroup$
– user636268
Jan 19 at 12:42
|
show 1 more comment
$begingroup$
My solution is not complete, but I hope someone can fill in the missing things.
My result: $n$ cannot be even, and also it should be an odd number with $nequiv 1pmod 8$.
Let's proceed: Suppose $$d^2=(2^k-1)^n+(2^k+1)^n$$
If $ngeq 2$ is even, then using binomial expansion we have $$d^2=2[(2^k)^n+binom{n}{2}(2^k)^{n-2}+cdots+1]$$
Clearly $d^2$ is divisible by $2$ but not $4$, hence every even $n$ doesn't work for sure.
If $ngeq 1$ is odd, we first take $n=1$, and we see $$d^2=2^{k+1}$$
Hence we have $(n,k)=(1,1),(1,3),(1,5), cdots$.
If $ngeq 3$, then begin{align*}d^2&=2[(2^k)^n+binom n2(2^k)^{n-2}+cdots+binom{n}{n-1}2^k]\
&=2^{k+1}bigg[(2^k)^{n-1}+binom{n}{2}(2^k)^{n-3}+cdots+binom{n}{n-3}(2^k)^2+nbigg]end{align*}
We see that since $n$ is odd, the big square bracket is odd, hence $2^{k+1}$ must itself be a perfect square, hence $k$ is odd. Then the big square bracket should itself be an odd square, and we see any odd square should have the property $1pmod 8$ since for any positive integer $m$ we have $$(2m+1)^2equiv 4m^2+4m+1equiv 4m(m+1)+1equiv 1pmod 8.$$
Hence when $n$ is an odd number with $nequiv 1pmod 8$, $k$ is an odd number, but I cannot found any example when $n> 1$, there should be no solution at this stage but I cannot prove it.
answered Jan 19 at 12:04
kelvin hong 方kelvin hong 方
72218
$endgroup$
My solution is not complete, but I hope someone can fill in the missing things.
My result: $n$ cannot be even, and also it should be an odd number with $nequiv 1pmod 8$.
Let's proceed: Suppose $$d^2=(2^k-1)^n+(2^k+1)^n$$
If $ngeq 2$ is even, then using binomial expansion we have $$d^2=2[(2^k)^n+binom{n}{2}(2^k)^{n-2}+cdots+1]$$
Clearly $d^2$ is divisible by $2$ but not $4$, hence every even $n$ doesn't work for sure.
If $ngeq 1$ is odd, we first take $n=1$, and we see $$d^2=2^{k+1}$$
Hence we have $(n,k)=(1,1),(1,3),(1,5), cdots$.
If $ngeq 3$, then begin{align*}d^2&=2[(2^k)^n+binom n2(2^k)^{n-2}+cdots+binom{n}{n-1}2^k]\
&=2^{k+1}bigg[(2^k)^{n-1}+binom{n}{2}(2^k)^{n-3}+cdots+binom{n}{n-3}(2^k)^2+nbigg]end{align*}
We see that since $n$ is odd, the big square bracket is odd, hence $2^{k+1}$ must itself be a perfect square, hence $k$ is odd. Then the big square bracket should itself be an odd square, and we see any odd square should have the property $1pmod 8$ since for any positive integer $m$ we have $$(2m+1)^2equiv 4m^2+4m+1equiv 4m(m+1)+1equiv 1pmod 8.$$
Hence when $n$ is an odd number with $nequiv 1pmod 8$, $k$ is an odd number, but I cannot found any example when $n> 1$, there should be no solution at this stage but I cannot prove it.
answered Jan 19 at 12:04
kelvin hong 方kelvin hong 方
72218
answered Jan 19 at 12:04
kelvin hong 方kelvin hong 方
72218
answered Jan 19 at 12:04
kelvin hong 方kelvin hong 方
72218
answered Jan 19 at 12:04
kelvin hong 方kelvin hong 方
72218
72218
$begingroup$
Thanks for your response!
$endgroup$
– user636268
Jan 19 at 12:17
$begingroup$
What if we substitute some numbef with a special property, say a perfect square itself and check?
$endgroup$
– user636268
Jan 19 at 12:23
$begingroup$
You can try it, but as my proof saying, the minimum next integer value of $n$ you can try is $n=9$, which makes calculation becomes huge very fast, if you have a powerful calculator you can try to do that.
$endgroup$
– kelvin hong 方
Jan 19 at 12:36
$begingroup$
If that does not help🤔
$endgroup$
– user636268
Jan 19 at 12:39
2
$begingroup$
Yes! That is why i am using MSE
$endgroup$
– user636268
Jan 19 at 12:42
|
show 1 more comment
$begingroup$
Thanks for your response!
$endgroup$
– user636268
Jan 19 at 12:17
$begingroup$
What if we substitute some numbef with a special property, say a perfect square itself and check?
$endgroup$
– user636268
Jan 19 at 12:23
$begingroup$
You can try it, but as my proof saying, the minimum next integer value of $n$ you can try is $n=9$, which makes calculation becomes huge very fast, if you have a powerful calculator you can try to do that.
$endgroup$
– kelvin hong 方
Jan 19 at 12:36
$begingroup$
If that does not help🤔
$endgroup$
– user636268
Jan 19 at 12:39
2
$begingroup$
Yes! That is why i am using MSE
$endgroup$
– user636268
Jan 19 at 12:42
$begingroup$
Thanks for your response!
$endgroup$
– user636268
Jan 19 at 12:17
$begingroup$
Thanks for your response!
$endgroup$
– user636268
Jan 19 at 12:17
$begingroup$
What if we substitute some numbef with a special property, say a perfect square itself and check?
$endgroup$
– user636268
Jan 19 at 12:23
$begingroup$
What if we substitute some numbef with a special property, say a perfect square itself and check?
$endgroup$
– user636268
Jan 19 at 12:23
$begingroup$
You can try it, but as my proof saying, the minimum next integer value of $n$ you can try is $n=9$, which makes calculation becomes huge very fast, if you have a powerful calculator you can try to do that.
$endgroup$
– kelvin hong 方
Jan 19 at 12:36
$begingroup$
You can try it, but as my proof saying, the minimum next integer value of $n$ you can try is $n=9$, which makes calculation becomes huge very fast, if you have a powerful calculator you can try to do that.
$endgroup$
– kelvin hong 方
Jan 19 at 12:36
$begingroup$
If that does not help🤔
$endgroup$
– user636268
Jan 19 at 12:39
$begingroup$
If that does not help🤔
$endgroup$
– user636268
Jan 19 at 12:39
2
2
$begingroup$
Yes! That is why i am using MSE
$endgroup$
– user636268
Jan 19 at 12:42
$begingroup$
Yes! That is why i am using MSE
$endgroup$
– user636268
Jan 19 at 12:42
|
show 1 more comment
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$begingroup$
Please dont downvote me... i dont know how to use latex.. i am new here
$endgroup$
– user636268
Jan 19 at 11:46
$begingroup$
Hi, this website can teach you how to type latex: en.wikibooks.org/wiki/LaTeX/Mathematics
$endgroup$
– kelvin hong 方
Jan 19 at 11:47
$begingroup$
Do i need to install some kind of applications?
$endgroup$
– user636268
Jan 19 at 11:49
$begingroup$
There is no need to install anything for you to use LaTeX in MSE.
$endgroup$
– kelvin hong 方
Jan 19 at 11:52
$begingroup$
Tomorrow is exam.. please somebody answer fast
$endgroup$
– user636268
Jan 19 at 12:00