Mixing
Solve equation $n^4+n^2+1=p$, where p is prime number
If $n in mathbb N$ and $p$ is prime number solve equation $n^4+n^2+1=p$ i can write that like this
$$n^4+2n^2-n^2+1=p$$
$$(n^2+1)^2-n^2=(n^2+1-n)(n^2+1+n)=p=1 cdot p$$
Since $n^2+1+n>1$ then $n^2+1-n=1$ so $n=0$ or $n=1$ if I put $n=0$ then $p=1$ that is not true since one is not prime number, then $n=1$ and $p=3$, is this ok?
elementary-number-theory proof-verification diophantine-equations
edited Nov 21 '18 at 20:07
Jean-Claude Arbaut
14.7k63464
asked Nov 21 '18 at 20:05
Marko ŠkorićMarko Škorić
70310
add a comment |
If $n in mathbb N$ and $p$ is prime number solve equation $n^4+n^2+1=p$ i can write that like this
$$n^4+2n^2-n^2+1=p$$
$$(n^2+1)^2-n^2=(n^2+1-n)(n^2+1+n)=p=1 cdot p$$
Since $n^2+1+n>1$ then $n^2+1-n=1$ so $n=0$ or $n=1$ if I put $n=0$ then $p=1$ that is not true since one is not prime number, then $n=1$ and $p=3$, is this ok?
elementary-number-theory proof-verification diophantine-equations
edited Nov 21 '18 at 20:07
Jean-Claude Arbaut
14.7k63464
asked Nov 21 '18 at 20:05
Marko ŠkorićMarko Škorić
70310
2
Yes. ${}{}{}{}{}$
– Jean-Claude Arbaut
Nov 21 '18 at 20:09
ah it depends on what level of seriousness you have in number theory really, for stack exchange yes by the looks of the popular vote is saying this proof is ok, but this would not qualify by my standards.
– Adam
Nov 25 '18 at 3:27
You have stated that your aim is to solve an equation when one does this, it means they have obtained all possible solutions to that equation given it's stated definitions. If $p$ can be any prime, you have a solution set that is infinite right? You have stated nothing of the nature of the cardinality of your solution set and so I do recommend you looking into learning about this.
– Adam
Nov 25 '18 at 3:30
add a comment |
If $n in mathbb N$ and $p$ is prime number solve equation $n^4+n^2+1=p$ i can write that like this
$$n^4+2n^2-n^2+1=p$$
$$(n^2+1)^2-n^2=(n^2+1-n)(n^2+1+n)=p=1 cdot p$$
Since $n^2+1+n>1$ then $n^2+1-n=1$ so $n=0$ or $n=1$ if I put $n=0$ then $p=1$ that is not true since one is not prime number, then $n=1$ and $p=3$, is this ok?
elementary-number-theory proof-verification diophantine-equations
edited Nov 21 '18 at 20:07
Jean-Claude Arbaut
14.7k63464
asked Nov 21 '18 at 20:05
Marko ŠkorićMarko Škorić
70310
If $n in mathbb N$ and $p$ is prime number solve equation $n^4+n^2+1=p$ i can write that like this
$$n^4+2n^2-n^2+1=p$$
$$(n^2+1)^2-n^2=(n^2+1-n)(n^2+1+n)=p=1 cdot p$$
Since $n^2+1+n>1$ then $n^2+1-n=1$ so $n=0$ or $n=1$ if I put $n=0$ then $p=1$ that is not true since one is not prime number, then $n=1$ and $p=3$, is this ok?
elementary-number-theory proof-verification diophantine-equations
elementary-number-theory proof-verification diophantine-equations
edited Nov 21 '18 at 20:07
Jean-Claude Arbaut
14.7k63464
asked Nov 21 '18 at 20:05
Marko ŠkorićMarko Škorić
70310
edited Nov 21 '18 at 20:07
Jean-Claude Arbaut
14.7k63464
asked Nov 21 '18 at 20:05
Marko ŠkorićMarko Škorić
70310
edited Nov 21 '18 at 20:07
Jean-Claude Arbaut
14.7k63464
edited Nov 21 '18 at 20:07
Jean-Claude Arbaut
14.7k63464
edited Nov 21 '18 at 20:07
Jean-Claude Arbaut
14.7k63464
14.7k63464
asked Nov 21 '18 at 20:05
Marko ŠkorićMarko Škorić
70310
asked Nov 21 '18 at 20:05
Marko ŠkorićMarko Škorić
70310
asked Nov 21 '18 at 20:05
Marko ŠkorićMarko Škorić
70310
70310
2
Yes. ${}{}{}{}{}$
– Jean-Claude Arbaut
Nov 21 '18 at 20:09
ah it depends on what level of seriousness you have in number theory really, for stack exchange yes by the looks of the popular vote is saying this proof is ok, but this would not qualify by my standards.
– Adam
Nov 25 '18 at 3:27
You have stated that your aim is to solve an equation when one does this, it means they have obtained all possible solutions to that equation given it's stated definitions. If $p$ can be any prime, you have a solution set that is infinite right? You have stated nothing of the nature of the cardinality of your solution set and so I do recommend you looking into learning about this.
– Adam
Nov 25 '18 at 3:30
add a comment |
2
Yes. ${}{}{}{}{}$
– Jean-Claude Arbaut
Nov 21 '18 at 20:09
ah it depends on what level of seriousness you have in number theory really, for stack exchange yes by the looks of the popular vote is saying this proof is ok, but this would not qualify by my standards.
– Adam
Nov 25 '18 at 3:27
You have stated that your aim is to solve an equation when one does this, it means they have obtained all possible solutions to that equation given it's stated definitions. If $p$ can be any prime, you have a solution set that is infinite right? You have stated nothing of the nature of the cardinality of your solution set and so I do recommend you looking into learning about this.
– Adam
Nov 25 '18 at 3:30
2
2
Yes. ${}{}{}{}{}$
– Jean-Claude Arbaut
Nov 21 '18 at 20:09
Yes. ${}{}{}{}{}$
– Jean-Claude Arbaut
Nov 21 '18 at 20:09
ah it depends on what level of seriousness you have in number theory really, for stack exchange yes by the looks of the popular vote is saying this proof is ok, but this would not qualify by my standards.
– Adam
Nov 25 '18 at 3:27
ah it depends on what level of seriousness you have in number theory really, for stack exchange yes by the looks of the popular vote is saying this proof is ok, but this would not qualify by my standards.
– Adam
Nov 25 '18 at 3:27
You have stated that your aim is to solve an equation when one does this, it means they have obtained all possible solutions to that equation given it's stated definitions. If $p$ can be any prime, you have a solution set that is infinite right? You have stated nothing of the nature of the cardinality of your solution set and so I do recommend you looking into learning about this.
– Adam
Nov 25 '18 at 3:30
You have stated that your aim is to solve an equation when one does this, it means they have obtained all possible solutions to that equation given it's stated definitions. If $p$ can be any prime, you have a solution set that is infinite right? You have stated nothing of the nature of the cardinality of your solution set and so I do recommend you looking into learning about this.
– Adam
Nov 25 '18 at 3:30
add a comment |
1 Answer
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Besides punctuation, which I'd use more of in your case, the proof is perfect. Perhaps elaborate on some things a little more (like emphasizing you're taking $n^4+2n^2+1$ and turning those terms specifically into $(n^2+1)^2$, or showing that $n^2+1+n>1$) but the level of detail and preciseness always depends on the context.
In short, nicely done. Good work!
answered Nov 21 '18 at 20:22
vrugtehagelvrugtehagel
10.7k1549
add a comment |
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Besides punctuation, which I'd use more of in your case, the proof is perfect. Perhaps elaborate on some things a little more (like emphasizing you're taking $n^4+2n^2+1$ and turning those terms specifically into $(n^2+1)^2$, or showing that $n^2+1+n>1$) but the level of detail and preciseness always depends on the context.
In short, nicely done. Good work!
answered Nov 21 '18 at 20:22
vrugtehagelvrugtehagel
10.7k1549
add a comment |
Besides punctuation, which I'd use more of in your case, the proof is perfect. Perhaps elaborate on some things a little more (like emphasizing you're taking $n^4+2n^2+1$ and turning those terms specifically into $(n^2+1)^2$, or showing that $n^2+1+n>1$) but the level of detail and preciseness always depends on the context.
In short, nicely done. Good work!
answered Nov 21 '18 at 20:22
vrugtehagelvrugtehagel
10.7k1549
add a comment |
Besides punctuation, which I'd use more of in your case, the proof is perfect. Perhaps elaborate on some things a little more (like emphasizing you're taking $n^4+2n^2+1$ and turning those terms specifically into $(n^2+1)^2$, or showing that $n^2+1+n>1$) but the level of detail and preciseness always depends on the context.
In short, nicely done. Good work!
answered Nov 21 '18 at 20:22
vrugtehagelvrugtehagel
10.7k1549
Besides punctuation, which I'd use more of in your case, the proof is perfect. Perhaps elaborate on some things a little more (like emphasizing you're taking $n^4+2n^2+1$ and turning those terms specifically into $(n^2+1)^2$, or showing that $n^2+1+n>1$) but the level of detail and preciseness always depends on the context.
In short, nicely done. Good work!
answered Nov 21 '18 at 20:22
vrugtehagelvrugtehagel
10.7k1549
answered Nov 21 '18 at 20:22
vrugtehagelvrugtehagel
10.7k1549
answered Nov 21 '18 at 20:22
vrugtehagelvrugtehagel
10.7k1549
answered Nov 21 '18 at 20:22
vrugtehagelvrugtehagel
10.7k1549
10.7k1549
add a comment |
add a comment |
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Yes. ${}{}{}{}{}$
– Jean-Claude Arbaut
Nov 21 '18 at 20:09
ah it depends on what level of seriousness you have in number theory really, for stack exchange yes by the looks of the popular vote is saying this proof is ok, but this would not qualify by my standards.
– Adam
Nov 25 '18 at 3:27
You have stated that your aim is to solve an equation when one does this, it means they have obtained all possible solutions to that equation given it's stated definitions. If $p$ can be any prime, you have a solution set that is infinite right? You have stated nothing of the nature of the cardinality of your solution set and so I do recommend you looking into learning about this.
– Adam
Nov 25 '18 at 3:30