Can I show that a positive power of $2$ plus a perfect square is not divisible by $7$? [closed]
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I've gotten another proof down to just having to prove that a positive power of $2$ plus a perfect square is not divisible by $7$:
$$7nmid(2^k + n^2), ; k,n in mathbbN$$
How can I do this?
discrete-mathematics divisibility
edited Sep 1 at 15:29
Michael Hardy
206k23187466
asked Sep 1 at 4:08
Matt Hough
605
closed as off-topic by TheSimpliFire, Holo, user21820, amWhy, Did Sep 1 at 14:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, Holo, user21820, amWhy, Did
add a comment |Â
up vote
3
down vote
favorite
I've gotten another proof down to just having to prove that a positive power of $2$ plus a perfect square is not divisible by $7$:
$$7nmid(2^k + n^2), ; k,n in mathbbN$$
How can I do this?
discrete-mathematics divisibility
edited Sep 1 at 15:29
Michael Hardy
206k23187466
asked Sep 1 at 4:08
Matt Hough
605
closed as off-topic by TheSimpliFire, Holo, user21820, amWhy, Did Sep 1 at 14:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, Holo, user21820, amWhy, Did
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I've gotten another proof down to just having to prove that a positive power of $2$ plus a perfect square is not divisible by $7$:
$$7nmid(2^k + n^2), ; k,n in mathbbN$$
How can I do this?
discrete-mathematics divisibility
edited Sep 1 at 15:29
Michael Hardy
206k23187466
asked Sep 1 at 4:08
Matt Hough
605
I've gotten another proof down to just having to prove that a positive power of $2$ plus a perfect square is not divisible by $7$:
$$7nmid(2^k + n^2), ; k,n in mathbbN$$
How can I do this?
discrete-mathematics divisibility
discrete-mathematics divisibility
edited Sep 1 at 15:29
Michael Hardy
206k23187466
asked Sep 1 at 4:08
Matt Hough
605
edited Sep 1 at 15:29
Michael Hardy
206k23187466
asked Sep 1 at 4:08
Matt Hough
605
edited Sep 1 at 15:29
Michael Hardy
206k23187466
edited Sep 1 at 15:29
Michael Hardy
206k23187466
edited Sep 1 at 15:29
Michael Hardy
206k23187466
206k23187466
asked Sep 1 at 4:08
Matt Hough
605
asked Sep 1 at 4:08
Matt Hough
605
asked Sep 1 at 4:08
Matt Hough
605
605
closed as off-topic by TheSimpliFire, Holo, user21820, amWhy, Did Sep 1 at 14:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, Holo, user21820, amWhy, Did
closed as off-topic by TheSimpliFire, Holo, user21820, amWhy, Did Sep 1 at 14:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, Holo, user21820, amWhy, Did
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
13
down vote
accepted
Edit:
Note that
$$2^k pmod7 in 1,2,4$$
$$n^2 pmod7 in 0,1,2, 4 $$
We have listed the values that $2^k pmod7$
can take, we just have to make sure that the values that $−2^k pmod7$
can take is not in those values that $n^2 pmod7 $
can take. For example, $1$
is a value that $2^k pmod7$ can attain, $−1 equiv 6 pmod7$
is not a value that $n^2 pmod7$ can take.
Remark:
The original question was $7 notmid 2k+n^2$.
You can't since it is not true.
$$1^2+2(3)=7$$
A general strategy to get more counterexamples is choose any $n$, note the parity of $n$. find $l$ such that $7l > n^2$ where $l$ and $n$ share the same parity. $7l-n^2$ is a positive even number.
answered Sep 1 at 4:11
Siong Thye Goh
81.8k1456104
Oh wow, I didn't even realize. Thanks for that :)
– Matt Hough
Sep 1 at 4:19
Ahh, I just realized where I went wrong. 2k in my proof should be 2^m. I have edited the question.
– Matt Hough
Sep 1 at 4:27
But wouldn't n^2 (mod 7) be congruent to 1,2,4 as well since n is a natural number. I'd gotten that far, I just don't know how to turn that into a proof.
– Matt Hough
Sep 1 at 4:42
We have listed the values that $2^k pmod7$ can take, we just have to make sure that the values that $-2^k pmod7$ can take is not in those values that $n^2 pmod7$ can take. For example, $1$ is a value that $2^k pmod7$ can attain, $-1 equiv 6 pmod7$ is not a value that $n^2 pmod7$ can take.
– Siong Thye Goh
Sep 1 at 4:45
1
This seems to be answering an earlier version of the question, which is confusing... sure, questions shouldn't generally be edited to change their meaning substantially, but now that it's happened, could something be done to bring the question and answer in sync? Whether that should be editing the answer, deleting it, rolling back the question, or something else, I don't know.
– David Z
Sep 1 at 11:58
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
13
down vote
accepted
Edit:
Note that
$$2^k pmod7 in 1,2,4$$
$$n^2 pmod7 in 0,1,2, 4 $$
We have listed the values that $2^k pmod7$
can take, we just have to make sure that the values that $−2^k pmod7$
can take is not in those values that $n^2 pmod7 $
can take. For example, $1$
is a value that $2^k pmod7$ can attain, $−1 equiv 6 pmod7$
is not a value that $n^2 pmod7$ can take.
Remark:
The original question was $7 notmid 2k+n^2$.
You can't since it is not true.
$$1^2+2(3)=7$$
A general strategy to get more counterexamples is choose any $n$, note the parity of $n$. find $l$ such that $7l > n^2$ where $l$ and $n$ share the same parity. $7l-n^2$ is a positive even number.
answered Sep 1 at 4:11
Siong Thye Goh
81.8k1456104
Oh wow, I didn't even realize. Thanks for that :)
– Matt Hough
Sep 1 at 4:19
Ahh, I just realized where I went wrong. 2k in my proof should be 2^m. I have edited the question.
– Matt Hough
Sep 1 at 4:27
But wouldn't n^2 (mod 7) be congruent to 1,2,4 as well since n is a natural number. I'd gotten that far, I just don't know how to turn that into a proof.
– Matt Hough
Sep 1 at 4:42
We have listed the values that $2^k pmod7$ can take, we just have to make sure that the values that $-2^k pmod7$ can take is not in those values that $n^2 pmod7$ can take. For example, $1$ is a value that $2^k pmod7$ can attain, $-1 equiv 6 pmod7$ is not a value that $n^2 pmod7$ can take.
– Siong Thye Goh
Sep 1 at 4:45
1
This seems to be answering an earlier version of the question, which is confusing... sure, questions shouldn't generally be edited to change their meaning substantially, but now that it's happened, could something be done to bring the question and answer in sync? Whether that should be editing the answer, deleting it, rolling back the question, or something else, I don't know.
– David Z
Sep 1 at 11:58
 |Â
show 1 more comment
up vote
13
down vote
accepted
Edit:
Note that
$$2^k pmod7 in 1,2,4$$
$$n^2 pmod7 in 0,1,2, 4 $$
We have listed the values that $2^k pmod7$
can take, we just have to make sure that the values that $−2^k pmod7$
can take is not in those values that $n^2 pmod7 $
can take. For example, $1$
is a value that $2^k pmod7$ can attain, $−1 equiv 6 pmod7$
is not a value that $n^2 pmod7$ can take.
Remark:
The original question was $7 notmid 2k+n^2$.
You can't since it is not true.
$$1^2+2(3)=7$$
A general strategy to get more counterexamples is choose any $n$, note the parity of $n$. find $l$ such that $7l > n^2$ where $l$ and $n$ share the same parity. $7l-n^2$ is a positive even number.
answered Sep 1 at 4:11
Siong Thye Goh
81.8k1456104
Oh wow, I didn't even realize. Thanks for that :)
– Matt Hough
Sep 1 at 4:19
Ahh, I just realized where I went wrong. 2k in my proof should be 2^m. I have edited the question.
– Matt Hough
Sep 1 at 4:27
But wouldn't n^2 (mod 7) be congruent to 1,2,4 as well since n is a natural number. I'd gotten that far, I just don't know how to turn that into a proof.
– Matt Hough
Sep 1 at 4:42
We have listed the values that $2^k pmod7$ can take, we just have to make sure that the values that $-2^k pmod7$ can take is not in those values that $n^2 pmod7$ can take. For example, $1$ is a value that $2^k pmod7$ can attain, $-1 equiv 6 pmod7$ is not a value that $n^2 pmod7$ can take.
– Siong Thye Goh
Sep 1 at 4:45
1
This seems to be answering an earlier version of the question, which is confusing... sure, questions shouldn't generally be edited to change their meaning substantially, but now that it's happened, could something be done to bring the question and answer in sync? Whether that should be editing the answer, deleting it, rolling back the question, or something else, I don't know.
– David Z
Sep 1 at 11:58
 |Â
show 1 more comment
up vote
13
down vote
accepted
up vote
13
down vote
accepted
Edit:
Note that
$$2^k pmod7 in 1,2,4$$
$$n^2 pmod7 in 0,1,2, 4 $$
We have listed the values that $2^k pmod7$
can take, we just have to make sure that the values that $−2^k pmod7$
can take is not in those values that $n^2 pmod7 $
can take. For example, $1$
is a value that $2^k pmod7$ can attain, $−1 equiv 6 pmod7$
is not a value that $n^2 pmod7$ can take.
Remark:
The original question was $7 notmid 2k+n^2$.
You can't since it is not true.
$$1^2+2(3)=7$$
A general strategy to get more counterexamples is choose any $n$, note the parity of $n$. find $l$ such that $7l > n^2$ where $l$ and $n$ share the same parity. $7l-n^2$ is a positive even number.
answered Sep 1 at 4:11
Siong Thye Goh
81.8k1456104
Edit:
Note that
$$2^k pmod7 in 1,2,4$$
$$n^2 pmod7 in 0,1,2, 4 $$
We have listed the values that $2^k pmod7$
can take, we just have to make sure that the values that $−2^k pmod7$
can take is not in those values that $n^2 pmod7 $
can take. For example, $1$
is a value that $2^k pmod7$ can attain, $−1 equiv 6 pmod7$
is not a value that $n^2 pmod7$ can take.
Remark:
The original question was $7 notmid 2k+n^2$.
You can't since it is not true.
$$1^2+2(3)=7$$
A general strategy to get more counterexamples is choose any $n$, note the parity of $n$. find $l$ such that $7l > n^2$ where $l$ and $n$ share the same parity. $7l-n^2$ is a positive even number.
answered Sep 1 at 4:11
Siong Thye Goh
81.8k1456104
edited Sep 1 at 12:41
answered Sep 1 at 4:11
Siong Thye Goh
81.8k1456104
answered Sep 1 at 4:11
Siong Thye Goh
81.8k1456104
answered Sep 1 at 4:11
Siong Thye Goh
81.8k1456104
81.8k1456104
Oh wow, I didn't even realize. Thanks for that :)
– Matt Hough
Sep 1 at 4:19
Ahh, I just realized where I went wrong. 2k in my proof should be 2^m. I have edited the question.
– Matt Hough
Sep 1 at 4:27
But wouldn't n^2 (mod 7) be congruent to 1,2,4 as well since n is a natural number. I'd gotten that far, I just don't know how to turn that into a proof.
– Matt Hough
Sep 1 at 4:42
We have listed the values that $2^k pmod7$ can take, we just have to make sure that the values that $-2^k pmod7$ can take is not in those values that $n^2 pmod7$ can take. For example, $1$ is a value that $2^k pmod7$ can attain, $-1 equiv 6 pmod7$ is not a value that $n^2 pmod7$ can take.
– Siong Thye Goh
Sep 1 at 4:45
1
This seems to be answering an earlier version of the question, which is confusing... sure, questions shouldn't generally be edited to change their meaning substantially, but now that it's happened, could something be done to bring the question and answer in sync? Whether that should be editing the answer, deleting it, rolling back the question, or something else, I don't know.
– David Z
Sep 1 at 11:58
 |Â
show 1 more comment
Oh wow, I didn't even realize. Thanks for that :)
– Matt Hough
Sep 1 at 4:19
Ahh, I just realized where I went wrong. 2k in my proof should be 2^m. I have edited the question.
– Matt Hough
Sep 1 at 4:27
But wouldn't n^2 (mod 7) be congruent to 1,2,4 as well since n is a natural number. I'd gotten that far, I just don't know how to turn that into a proof.
– Matt Hough
Sep 1 at 4:42
We have listed the values that $2^k pmod7$ can take, we just have to make sure that the values that $-2^k pmod7$ can take is not in those values that $n^2 pmod7$ can take. For example, $1$ is a value that $2^k pmod7$ can attain, $-1 equiv 6 pmod7$ is not a value that $n^2 pmod7$ can take.
– Siong Thye Goh
Sep 1 at 4:45
1
This seems to be answering an earlier version of the question, which is confusing... sure, questions shouldn't generally be edited to change their meaning substantially, but now that it's happened, could something be done to bring the question and answer in sync? Whether that should be editing the answer, deleting it, rolling back the question, or something else, I don't know.
– David Z
Sep 1 at 11:58
Oh wow, I didn't even realize. Thanks for that :)
– Matt Hough
Sep 1 at 4:19
Oh wow, I didn't even realize. Thanks for that :)
– Matt Hough
Sep 1 at 4:19
Ahh, I just realized where I went wrong. 2k in my proof should be 2^m. I have edited the question.
– Matt Hough
Sep 1 at 4:27
Ahh, I just realized where I went wrong. 2k in my proof should be 2^m. I have edited the question.
– Matt Hough
Sep 1 at 4:27
But wouldn't n^2 (mod 7) be congruent to 1,2,4 as well since n is a natural number. I'd gotten that far, I just don't know how to turn that into a proof.
– Matt Hough
Sep 1 at 4:42
But wouldn't n^2 (mod 7) be congruent to 1,2,4 as well since n is a natural number. I'd gotten that far, I just don't know how to turn that into a proof.
– Matt Hough
Sep 1 at 4:42
We have listed the values that $2^k pmod7$ can take, we just have to make sure that the values that $-2^k pmod7$ can take is not in those values that $n^2 pmod7$ can take. For example, $1$ is a value that $2^k pmod7$ can attain, $-1 equiv 6 pmod7$ is not a value that $n^2 pmod7$ can take.
– Siong Thye Goh
Sep 1 at 4:45
We have listed the values that $2^k pmod7$ can take, we just have to make sure that the values that $-2^k pmod7$ can take is not in those values that $n^2 pmod7$ can take. For example, $1$ is a value that $2^k pmod7$ can attain, $-1 equiv 6 pmod7$ is not a value that $n^2 pmod7$ can take.
– Siong Thye Goh
Sep 1 at 4:45
1
1
This seems to be answering an earlier version of the question, which is confusing... sure, questions shouldn't generally be edited to change their meaning substantially, but now that it's happened, could something be done to bring the question and answer in sync? Whether that should be editing the answer, deleting it, rolling back the question, or something else, I don't know.
– David Z
Sep 1 at 11:58
This seems to be answering an earlier version of the question, which is confusing... sure, questions shouldn't generally be edited to change their meaning substantially, but now that it's happened, could something be done to bring the question and answer in sync? Whether that should be editing the answer, deleting it, rolling back the question, or something else, I don't know.
– David Z
Sep 1 at 11:58
 |Â
show 1 more comment
