Prove that for all positive integers $n>1, (n^2+1)-n$ is not a perfect square.
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In this hard question, I tried to use mathematical induction but couldn't as it is a hard problem. I have been trying for a week now so I came to stack exchange with this question. Any help would be nice.
Also the problems goes on and asks to prove for $2(n^2+1)-n$
elementary-number-theory
asked yesterday
user587054
47211
add a comment |
In this hard question, I tried to use mathematical induction but couldn't as it is a hard problem. I have been trying for a week now so I came to stack exchange with this question. Any help would be nice.
Also the problems goes on and asks to prove for $2(n^2+1)-n$
elementary-number-theory
asked yesterday
user587054
47211
Did you really mean to write $1(n^2+1)-n$ or is there a typo?
– bof
yesterday
I think it is supposed to be all positive integers greater than $1$, as when $n=1$ it is a square.
– Erik Parkinson
yesterday
I forgot to add the second part
– user587054
yesterday
add a comment |
In this hard question, I tried to use mathematical induction but couldn't as it is a hard problem. I have been trying for a week now so I came to stack exchange with this question. Any help would be nice.
Also the problems goes on and asks to prove for $2(n^2+1)-n$
elementary-number-theory
asked yesterday
user587054
47211
In this hard question, I tried to use mathematical induction but couldn't as it is a hard problem. I have been trying for a week now so I came to stack exchange with this question. Any help would be nice.
Also the problems goes on and asks to prove for $2(n^2+1)-n$
elementary-number-theory
elementary-number-theory
asked yesterday
user587054
47211
asked yesterday
user587054
47211
edited yesterday
asked yesterday
user587054
47211
asked yesterday
user587054
47211
asked yesterday
user587054
47211
47211
Did you really mean to write $1(n^2+1)-n$ or is there a typo?
– bof
yesterday
I think it is supposed to be all positive integers greater than $1$, as when $n=1$ it is a square.
– Erik Parkinson
yesterday
I forgot to add the second part
– user587054
yesterday
add a comment |
Did you really mean to write $1(n^2+1)-n$ or is there a typo?
– bof
yesterday
I think it is supposed to be all positive integers greater than $1$, as when $n=1$ it is a square.
– Erik Parkinson
yesterday
I forgot to add the second part
– user587054
yesterday
Did you really mean to write $1(n^2+1)-n$ or is there a typo?
– bof
yesterday
Did you really mean to write $1(n^2+1)-n$ or is there a typo?
– bof
yesterday
I think it is supposed to be all positive integers greater than $1$, as when $n=1$ it is a square.
– Erik Parkinson
yesterday
I think it is supposed to be all positive integers greater than $1$, as when $n=1$ it is a square.
– Erik Parkinson
yesterday
I forgot to add the second part
– user587054
yesterday
I forgot to add the second part
– user587054
yesterday
add a comment |
3 Answers
3
active
oldest
votes
The trick is that for all $n ge 2$,
$$(n-1)^2 = n^2-2n+1 < n^2-n+1 < n^2$$
So $n^2-n+1$ will lie between all consecutive square numbers and thus can't be square.
answered yesterday
Erik Parkinson
9059
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Note that for $n^2-n+1$ and $n >1$, the following inequality holds true:
$$n^2-2n+1 < n^2-n+1 < n^2$$
$$(n-1)^2 < n^2-n+1 < n^2$$
where $(n-1)^2$ and $n^2$ are two consecutive squares. What can be concluded from this?
answered yesterday
KM101
5,5361423
Thank you but can you please help me with the second part I updated on the post. Thank you anyways
– user587054
yesterday
add a comment |
A hint for the second part. Consider the possibilities for $n pmod3$, that is, $n$ may be of the form $3k$, $3k+1$ or $3k+2$. Do any of these $n$ yield expressions for $2(n^2+1)-n$ that are possible squares $pmod3$? And if any is, is it also a possible square $pmod9$?
answered yesterday
Adam Bailey
1,9221318
add a comment |
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
The trick is that for all $n ge 2$,
$$(n-1)^2 = n^2-2n+1 < n^2-n+1 < n^2$$
So $n^2-n+1$ will lie between all consecutive square numbers and thus can't be square.
answered yesterday
Erik Parkinson
9059
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
The trick is that for all $n ge 2$,
$$(n-1)^2 = n^2-2n+1 < n^2-n+1 < n^2$$
So $n^2-n+1$ will lie between all consecutive square numbers and thus can't be square.
answered yesterday
Erik Parkinson
9059
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
The trick is that for all $n ge 2$,
$$(n-1)^2 = n^2-2n+1 < n^2-n+1 < n^2$$
So $n^2-n+1$ will lie between all consecutive square numbers and thus can't be square.
answered yesterday
Erik Parkinson
9059
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The trick is that for all $n ge 2$,
$$(n-1)^2 = n^2-2n+1 < n^2-n+1 < n^2$$
So $n^2-n+1$ will lie between all consecutive square numbers and thus can't be square.
answered yesterday
Erik Parkinson
9059
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Erik Parkinson
9059
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Erik Parkinson
9059
answered yesterday
Erik Parkinson
9059
9059
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
Note that for $n^2-n+1$ and $n >1$, the following inequality holds true:
$$n^2-2n+1 < n^2-n+1 < n^2$$
$$(n-1)^2 < n^2-n+1 < n^2$$
where $(n-1)^2$ and $n^2$ are two consecutive squares. What can be concluded from this?
answered yesterday
KM101
5,5361423
Thank you but can you please help me with the second part I updated on the post. Thank you anyways
– user587054
yesterday
add a comment |
Note that for $n^2-n+1$ and $n >1$, the following inequality holds true:
$$n^2-2n+1 < n^2-n+1 < n^2$$
$$(n-1)^2 < n^2-n+1 < n^2$$
where $(n-1)^2$ and $n^2$ are two consecutive squares. What can be concluded from this?
answered yesterday
KM101
5,5361423
Thank you but can you please help me with the second part I updated on the post. Thank you anyways
– user587054
yesterday
add a comment |
Note that for $n^2-n+1$ and $n >1$, the following inequality holds true:
$$n^2-2n+1 < n^2-n+1 < n^2$$
$$(n-1)^2 < n^2-n+1 < n^2$$
where $(n-1)^2$ and $n^2$ are two consecutive squares. What can be concluded from this?
answered yesterday
KM101
5,5361423
Note that for $n^2-n+1$ and $n >1$, the following inequality holds true:
$$n^2-2n+1 < n^2-n+1 < n^2$$
$$(n-1)^2 < n^2-n+1 < n^2$$
where $(n-1)^2$ and $n^2$ are two consecutive squares. What can be concluded from this?
answered yesterday
KM101
5,5361423
edited yesterday
answered yesterday
KM101
5,5361423
answered yesterday
KM101
5,5361423
answered yesterday
KM101
5,5361423
5,5361423
Thank you but can you please help me with the second part I updated on the post. Thank you anyways
– user587054
yesterday
add a comment |
Thank you but can you please help me with the second part I updated on the post. Thank you anyways
– user587054
yesterday
Thank you but can you please help me with the second part I updated on the post. Thank you anyways
– user587054
yesterday
Thank you but can you please help me with the second part I updated on the post. Thank you anyways
– user587054
yesterday
add a comment |
A hint for the second part. Consider the possibilities for $n pmod3$, that is, $n$ may be of the form $3k$, $3k+1$ or $3k+2$. Do any of these $n$ yield expressions for $2(n^2+1)-n$ that are possible squares $pmod3$? And if any is, is it also a possible square $pmod9$?
answered yesterday
Adam Bailey
1,9221318
add a comment |
A hint for the second part. Consider the possibilities for $n pmod3$, that is, $n$ may be of the form $3k$, $3k+1$ or $3k+2$. Do any of these $n$ yield expressions for $2(n^2+1)-n$ that are possible squares $pmod3$? And if any is, is it also a possible square $pmod9$?
answered yesterday
Adam Bailey
1,9221318
add a comment |
A hint for the second part. Consider the possibilities for $n pmod3$, that is, $n$ may be of the form $3k$, $3k+1$ or $3k+2$. Do any of these $n$ yield expressions for $2(n^2+1)-n$ that are possible squares $pmod3$? And if any is, is it also a possible square $pmod9$?
answered yesterday
Adam Bailey
1,9221318
A hint for the second part. Consider the possibilities for $n pmod3$, that is, $n$ may be of the form $3k$, $3k+1$ or $3k+2$. Do any of these $n$ yield expressions for $2(n^2+1)-n$ that are possible squares $pmod3$? And if any is, is it also a possible square $pmod9$?
answered yesterday
Adam Bailey
1,9221318
answered yesterday
Adam Bailey
1,9221318
answered yesterday
Adam Bailey
1,9221318
answered yesterday
Adam Bailey
1,9221318
1,9221318
add a comment |
add a comment |
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Did you really mean to write $1(n^2+1)-n$ or is there a typo?
– bof
yesterday
I think it is supposed to be all positive integers greater than $1$, as when $n=1$ it is a square.
– Erik Parkinson
yesterday
I forgot to add the second part
– user587054
yesterday