A proof that there is no prime number in the form $4k-1$ that is congruent to 3 modulo 4. [on hold]Do there exist two primes $p<q$ such that $p^n-1mid q^n-1$ for infinitely many $n$?Proving that there exist infinitely many primes of the form $mn+1$.How do I show that a prime that is less than $n$, is not a prime factor of $n$?Find all positive integers $n$ such that $frac2^n-1+1n$ is integer. Where I'm wrong?Dirichlet theorem on primes premiseProve any odd square cannot be of the form $4n+3$Is there $n geq 2$ such that $1^1 + 2^2 + dots + n^n$ is a perfect square?Is there/can there be a model-theoretic proof of this theorem of arithmetic ?Existence of a non-square integer $L$ such that $L$ is a quadratic residue modulo $p^n$ for all $n$Prove a case of Dirichlet's Theorem: that there are infinity many primes of the form $8k+1$
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A proof that there is no prime number in the form $4k-1$ that is congruent to 3 modulo 4. [on hold]
Do there exist two primes $p<q$ such that $p^n-1mid q^n-1$ for infinitely many $n$?Proving that there exist infinitely many primes of the form $mn+1$.How do I show that a prime that is less than $n$, is not a prime factor of $n$?Find all positive integers $n$ such that $frac2^n-1+1n$ is integer. Where I'm wrong?Dirichlet theorem on primes premiseProve any odd square cannot be of the form $4n+3$Is there $n geq 2$ such that $1^1 + 2^2 + dots + n^n$ is a perfect square?Is there/can there be a model-theoretic proof of this theorem of arithmetic ?Existence of a non-square integer $L$ such that $L$ is a quadratic residue modulo $p^n$ for all $n$Prove a case of Dirichlet's Theorem: that there are infinity many primes of the form $8k+1$
$begingroup$
This is my first proof ever. I realize this might be mistaken, which is why I need your help to perfect it in order for me to learn and become better at proving mathematical statements.
A proof that there is no prime number in the form $4k-1$ that is congruent to 3 modulo 4.
The statement $4k-1 equiv 3 mod4$ can be rewriteen as $4k-4 = 4m$ where $m$ is any integer $in mathbbZ$.
Then :
$$4k-1-3 = 4m$$ $$4k - 4 = 4m$$ $$(k-1)=m$$ $$k = m+1$$
Plugging $k$ back, we obtain
$$ 4k-1 equiv 3 mod4$$ $$4m-3 equiv 3 mod4$$ $$4m-6 = 4k$$ ($kinmathbbZ$) $$2n-3 = 2k$$
We arrive at a contradiction.
- I am unsure if this is correct for only primes or for all integer k.
number-theory proof-verification
edited Mar 28 at 15:24
hardmath
29.3k953101
asked Mar 28 at 15:17
user69264user69264
1
New contributor
user69264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as unclear what you're asking by Dietrich Burde, Lord Shark the Unknown, Eevee Trainer, Cesareo, Leucippus Mar 29 at 5:45
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 3 more comments
$begingroup$
This is my first proof ever. I realize this might be mistaken, which is why I need your help to perfect it in order for me to learn and become better at proving mathematical statements.
A proof that there is no prime number in the form $4k-1$ that is congruent to 3 modulo 4.
The statement $4k-1 equiv 3 mod4$ can be rewriteen as $4k-4 = 4m$ where $m$ is any integer $in mathbbZ$.
Then :
$$4k-1-3 = 4m$$ $$4k - 4 = 4m$$ $$(k-1)=m$$ $$k = m+1$$
Plugging $k$ back, we obtain
$$ 4k-1 equiv 3 mod4$$ $$4m-3 equiv 3 mod4$$ $$4m-6 = 4k$$ ($kinmathbbZ$) $$2n-3 = 2k$$
We arrive at a contradiction.
- I am unsure if this is correct for only primes or for all integer k.
number-theory proof-verification
edited Mar 28 at 15:24
hardmath
29.3k953101
asked Mar 28 at 15:17
user69264user69264
1
New contributor
user69264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as unclear what you're asking by Dietrich Burde, Lord Shark the Unknown, Eevee Trainer, Cesareo, Leucippus Mar 29 at 5:45
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
6
$begingroup$
Already the title claim is false. Take $k=1$ and $p=3$.
$endgroup$
– Dietrich Burde
Mar 28 at 15:19
1
$begingroup$
@DietrichBurde Too damn fast, I'm always halfway done typing when you have your comment out. P.S. to OP: Any number of the form $4k-1$ is congruent to $3pmod 4$.
$endgroup$
– Don Thousand
Mar 28 at 15:20
$begingroup$
Note also that any integer that can be expressed as $n=4k -1$ has $n equiv -1 equiv 3 pmod 4$.; the conditions are redundant.
$endgroup$
– Brian
Mar 28 at 15:22
$begingroup$
Every number of the form $4k-1$ is congruent to $3 mod 4$: $4k-1equiv -1 mod 4; textand -1equiv 3 mod 4$
$endgroup$
– Keith Backman
Mar 28 at 15:22
1
$begingroup$
@Yanior Weg: Your tag edits are at best in tension with the OP's intentions, esp. the "fake proofs" tag. Please review the Question and input from the OP carefully before making further edits.
$endgroup$
– hardmath
Mar 28 at 15:26
|
show 3 more comments
$begingroup$
This is my first proof ever. I realize this might be mistaken, which is why I need your help to perfect it in order for me to learn and become better at proving mathematical statements.
A proof that there is no prime number in the form $4k-1$ that is congruent to 3 modulo 4.
The statement $4k-1 equiv 3 mod4$ can be rewriteen as $4k-4 = 4m$ where $m$ is any integer $in mathbbZ$.
Then :
$$4k-1-3 = 4m$$ $$4k - 4 = 4m$$ $$(k-1)=m$$ $$k = m+1$$
Plugging $k$ back, we obtain
$$ 4k-1 equiv 3 mod4$$ $$4m-3 equiv 3 mod4$$ $$4m-6 = 4k$$ ($kinmathbbZ$) $$2n-3 = 2k$$
We arrive at a contradiction.
- I am unsure if this is correct for only primes or for all integer k.
number-theory proof-verification
edited Mar 28 at 15:24
hardmath
29.3k953101
asked Mar 28 at 15:17
user69264user69264
1
New contributor
user69264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
This is my first proof ever. I realize this might be mistaken, which is why I need your help to perfect it in order for me to learn and become better at proving mathematical statements.
A proof that there is no prime number in the form $4k-1$ that is congruent to 3 modulo 4.
The statement $4k-1 equiv 3 mod4$ can be rewriteen as $4k-4 = 4m$ where $m$ is any integer $in mathbbZ$.
Then :
$$4k-1-3 = 4m$$ $$4k - 4 = 4m$$ $$(k-1)=m$$ $$k = m+1$$
Plugging $k$ back, we obtain
$$ 4k-1 equiv 3 mod4$$ $$4m-3 equiv 3 mod4$$ $$4m-6 = 4k$$ ($kinmathbbZ$) $$2n-3 = 2k$$
We arrive at a contradiction.
- I am unsure if this is correct for only primes or for all integer k.
number-theory proof-verification
number-theory proof-verification
edited Mar 28 at 15:24
hardmath
29.3k953101
asked Mar 28 at 15:17
user69264user69264
1
New contributor
user69264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 15:24
hardmath
29.3k953101
asked Mar 28 at 15:17
user69264user69264
1
New contributor
user69264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 15:24
hardmath
29.3k953101
edited Mar 28 at 15:24
hardmath
29.3k953101
edited Mar 28 at 15:24
hardmath
29.3k953101
29.3k953101
asked Mar 28 at 15:17
user69264user69264
1
New contributor
user69264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 15:17
user69264user69264
1
asked Mar 28 at 15:17
user69264user69264
1
1
New contributor
user69264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user69264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user69264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as unclear what you're asking by Dietrich Burde, Lord Shark the Unknown, Eevee Trainer, Cesareo, Leucippus Mar 29 at 5:45
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by Dietrich Burde, Lord Shark the Unknown, Eevee Trainer, Cesareo, Leucippus Mar 29 at 5:45
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
6
$begingroup$
Already the title claim is false. Take $k=1$ and $p=3$.
$endgroup$
– Dietrich Burde
Mar 28 at 15:19
1
$begingroup$
@DietrichBurde Too damn fast, I'm always halfway done typing when you have your comment out. P.S. to OP: Any number of the form $4k-1$ is congruent to $3pmod 4$.
$endgroup$
– Don Thousand
Mar 28 at 15:20
$begingroup$
Note also that any integer that can be expressed as $n=4k -1$ has $n equiv -1 equiv 3 pmod 4$.; the conditions are redundant.
$endgroup$
– Brian
Mar 28 at 15:22
$begingroup$
Every number of the form $4k-1$ is congruent to $3 mod 4$: $4k-1equiv -1 mod 4; textand -1equiv 3 mod 4$
$endgroup$
– Keith Backman
Mar 28 at 15:22
1
$begingroup$
@Yanior Weg: Your tag edits are at best in tension with the OP's intentions, esp. the "fake proofs" tag. Please review the Question and input from the OP carefully before making further edits.
$endgroup$
– hardmath
Mar 28 at 15:26
|
show 3 more comments
6
$begingroup$
Already the title claim is false. Take $k=1$ and $p=3$.
$endgroup$
– Dietrich Burde
Mar 28 at 15:19
1
$begingroup$
@DietrichBurde Too damn fast, I'm always halfway done typing when you have your comment out. P.S. to OP: Any number of the form $4k-1$ is congruent to $3pmod 4$.
$endgroup$
– Don Thousand
Mar 28 at 15:20
$begingroup$
Note also that any integer that can be expressed as $n=4k -1$ has $n equiv -1 equiv 3 pmod 4$.; the conditions are redundant.
$endgroup$
– Brian
Mar 28 at 15:22
$begingroup$
Every number of the form $4k-1$ is congruent to $3 mod 4$: $4k-1equiv -1 mod 4; textand -1equiv 3 mod 4$
$endgroup$
– Keith Backman
Mar 28 at 15:22
1
$begingroup$
@Yanior Weg: Your tag edits are at best in tension with the OP's intentions, esp. the "fake proofs" tag. Please review the Question and input from the OP carefully before making further edits.
$endgroup$
– hardmath
Mar 28 at 15:26
6
6
$begingroup$
Already the title claim is false. Take $k=1$ and $p=3$.
$endgroup$
– Dietrich Burde
Mar 28 at 15:19
$begingroup$
Already the title claim is false. Take $k=1$ and $p=3$.
$endgroup$
– Dietrich Burde
Mar 28 at 15:19
1
1
$begingroup$
@DietrichBurde Too damn fast, I'm always halfway done typing when you have your comment out. P.S. to OP: Any number of the form $4k-1$ is congruent to $3pmod 4$.
$endgroup$
– Don Thousand
Mar 28 at 15:20
$begingroup$
@DietrichBurde Too damn fast, I'm always halfway done typing when you have your comment out. P.S. to OP: Any number of the form $4k-1$ is congruent to $3pmod 4$.
$endgroup$
– Don Thousand
Mar 28 at 15:20
$begingroup$
Note also that any integer that can be expressed as $n=4k -1$ has $n equiv -1 equiv 3 pmod 4$.; the conditions are redundant.
$endgroup$
– Brian
Mar 28 at 15:22
$begingroup$
Note also that any integer that can be expressed as $n=4k -1$ has $n equiv -1 equiv 3 pmod 4$.; the conditions are redundant.
$endgroup$
– Brian
Mar 28 at 15:22
$begingroup$
Every number of the form $4k-1$ is congruent to $3 mod 4$: $4k-1equiv -1 mod 4; textand -1equiv 3 mod 4$
$endgroup$
– Keith Backman
Mar 28 at 15:22
$begingroup$
Every number of the form $4k-1$ is congruent to $3 mod 4$: $4k-1equiv -1 mod 4; textand -1equiv 3 mod 4$
$endgroup$
– Keith Backman
Mar 28 at 15:22
1
1
$begingroup$
@Yanior Weg: Your tag edits are at best in tension with the OP's intentions, esp. the "fake proofs" tag. Please review the Question and input from the OP carefully before making further edits.
$endgroup$
– hardmath
Mar 28 at 15:26
$begingroup$
@Yanior Weg: Your tag edits are at best in tension with the OP's intentions, esp. the "fake proofs" tag. Please review the Question and input from the OP carefully before making further edits.
$endgroup$
– hardmath
Mar 28 at 15:26
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
You have stated the question incorrectly. As stated, what you want proven is just false. For example, when $k=2$, there is a prime number $7=4cdot 2-1$ and $7$ is congruent to 3 mod 4.
answered Mar 28 at 15:19
Mark FischlerMark Fischler
33.9k12552
$endgroup$
add a comment |
$begingroup$
There are infinitely many prime numbers of the form $4k-1.$ Your question doesn't make any sense.
answered Mar 28 at 15:21
Dbchatto67Dbchatto67
2,445522
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have stated the question incorrectly. As stated, what you want proven is just false. For example, when $k=2$, there is a prime number $7=4cdot 2-1$ and $7$ is congruent to 3 mod 4.
answered Mar 28 at 15:19
Mark FischlerMark Fischler
33.9k12552
$endgroup$
add a comment |
$begingroup$
You have stated the question incorrectly. As stated, what you want proven is just false. For example, when $k=2$, there is a prime number $7=4cdot 2-1$ and $7$ is congruent to 3 mod 4.
answered Mar 28 at 15:19
Mark FischlerMark Fischler
33.9k12552
$endgroup$
add a comment |
$begingroup$
You have stated the question incorrectly. As stated, what you want proven is just false. For example, when $k=2$, there is a prime number $7=4cdot 2-1$ and $7$ is congruent to 3 mod 4.
answered Mar 28 at 15:19
Mark FischlerMark Fischler
33.9k12552
$endgroup$
You have stated the question incorrectly. As stated, what you want proven is just false. For example, when $k=2$, there is a prime number $7=4cdot 2-1$ and $7$ is congruent to 3 mod 4.
answered Mar 28 at 15:19
Mark FischlerMark Fischler
33.9k12552
answered Mar 28 at 15:19
Mark FischlerMark Fischler
33.9k12552
answered Mar 28 at 15:19
Mark FischlerMark Fischler
33.9k12552
answered Mar 28 at 15:19
Mark FischlerMark Fischler
33.9k12552
33.9k12552
add a comment |
add a comment |
$begingroup$
There are infinitely many prime numbers of the form $4k-1.$ Your question doesn't make any sense.
answered Mar 28 at 15:21
Dbchatto67Dbchatto67
2,445522
$endgroup$
add a comment |
$begingroup$
There are infinitely many prime numbers of the form $4k-1.$ Your question doesn't make any sense.
answered Mar 28 at 15:21
Dbchatto67Dbchatto67
2,445522
$endgroup$
add a comment |
$begingroup$
There are infinitely many prime numbers of the form $4k-1.$ Your question doesn't make any sense.
answered Mar 28 at 15:21
Dbchatto67Dbchatto67
2,445522
$endgroup$
There are infinitely many prime numbers of the form $4k-1.$ Your question doesn't make any sense.
answered Mar 28 at 15:21
Dbchatto67Dbchatto67
2,445522
answered Mar 28 at 15:21
Dbchatto67Dbchatto67
2,445522
answered Mar 28 at 15:21
Dbchatto67Dbchatto67
2,445522
answered Mar 28 at 15:21
Dbchatto67Dbchatto67
2,445522
2,445522
add a comment |
add a comment |
6
$begingroup$
Already the title claim is false. Take $k=1$ and $p=3$.
$endgroup$
– Dietrich Burde
Mar 28 at 15:19
1
$begingroup$
@DietrichBurde Too damn fast, I'm always halfway done typing when you have your comment out. P.S. to OP: Any number of the form $4k-1$ is congruent to $3pmod 4$.
$endgroup$
– Don Thousand
Mar 28 at 15:20
$begingroup$
Note also that any integer that can be expressed as $n=4k -1$ has $n equiv -1 equiv 3 pmod 4$.; the conditions are redundant.
$endgroup$
– Brian
Mar 28 at 15:22
$begingroup$
Every number of the form $4k-1$ is congruent to $3 mod 4$: $4k-1equiv -1 mod 4; textand -1equiv 3 mod 4$
$endgroup$
– Keith Backman
Mar 28 at 15:22
1
$begingroup$
@Yanior Weg: Your tag edits are at best in tension with the OP's intentions, esp. the "fake proofs" tag. Please review the Question and input from the OP carefully before making further edits.
$endgroup$
– hardmath
Mar 28 at 15:26