Is it possible to find $x$ in a general form or forms for the equation $2n^2 + 2n + 1$ mod $x = 0$?
$begingroup$
Hello I was wondering is possible to find x in a general form or forms so that there are integer answers for something like the equation below? If so could you point me in the right direction?
$2n^2 + 2n + 1$ mod$ x = 0$
ie
$x = 1,5,13,17$... work and solve the condition.
elementary-number-theory modular-arithmetic
edited Dec 22 '18 at 5:58
AandJ4Ever
114
asked Dec 22 '18 at 5:30
Ryan ToppsRyan Topps
124
$endgroup$
add a comment |
$begingroup$
Hello I was wondering is possible to find x in a general form or forms so that there are integer answers for something like the equation below? If so could you point me in the right direction?
$2n^2 + 2n + 1$ mod$ x = 0$
ie
$x = 1,5,13,17$... work and solve the condition.
elementary-number-theory modular-arithmetic
edited Dec 22 '18 at 5:58
AandJ4Ever
114
asked Dec 22 '18 at 5:30
Ryan ToppsRyan Topps
124
$endgroup$
add a comment |
$begingroup$
Hello I was wondering is possible to find x in a general form or forms so that there are integer answers for something like the equation below? If so could you point me in the right direction?
$2n^2 + 2n + 1$ mod$ x = 0$
ie
$x = 1,5,13,17$... work and solve the condition.
elementary-number-theory modular-arithmetic
edited Dec 22 '18 at 5:58
AandJ4Ever
114
asked Dec 22 '18 at 5:30
Ryan ToppsRyan Topps
124
$endgroup$
Hello I was wondering is possible to find x in a general form or forms so that there are integer answers for something like the equation below? If so could you point me in the right direction?
$2n^2 + 2n + 1$ mod$ x = 0$
ie
$x = 1,5,13,17$... work and solve the condition.
elementary-number-theory modular-arithmetic
elementary-number-theory modular-arithmetic
edited Dec 22 '18 at 5:58
AandJ4Ever
114
asked Dec 22 '18 at 5:30
Ryan ToppsRyan Topps
124
edited Dec 22 '18 at 5:58
AandJ4Ever
114
asked Dec 22 '18 at 5:30
Ryan ToppsRyan Topps
124
edited Dec 22 '18 at 5:58
AandJ4Ever
114
edited Dec 22 '18 at 5:58
AandJ4Ever
114
edited Dec 22 '18 at 5:58
AandJ4Ever
114
114
asked Dec 22 '18 at 5:30
Ryan ToppsRyan Topps
124
asked Dec 22 '18 at 5:30
Ryan ToppsRyan Topps
124
asked Dec 22 '18 at 5:30
Ryan ToppsRyan Topps
124
124
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I think you're asking for what $x$ the equation $2n^2 + 2n + 1$ has a root modulo $x$.
First, since $2n^2 + 2n + 1$ is odd, we can assume that $x$ is odd. Then we can write:
$$2n^2 + 2n + 1 = frac{1}{2}((2n+1)^2 + 1)$$
This allows us to see that the equation has a solution if and only if there is a square root of $-1$ modulo $x$.
By the Chinese Remainder Theorem, this holds exactly when there is a square root of $-1$ modulo $p^k$ for every prime power $p^k$ dividing $x$.
It's a standard result that this happens when $pequiv 1pmod{4}$. This can be seen by looking at the cyclic group of units $pmod{p^k}$ and seeing when it has order divisible by $4$.
In conclusion, your equation has a solution exactly when all of the prime factors of $x$ are $1pmod 4$.
answered Dec 22 '18 at 7:07
SladeSlade
25.2k12665
$endgroup$
$begingroup$
Thank you, you explained that well
$endgroup$
– Ryan Topps
Dec 22 '18 at 7:36
$begingroup$
@RyanTopps I made a mistake; see the updated answer. The prime factors can never be $3pmod{4}$, I was misremembering something else.
$endgroup$
– Slade
Dec 22 '18 at 7:44
add a comment |
$begingroup$
Yes: $f(n)$ mod $x=0$ if and only if $xmid f(n)$.
answered Dec 22 '18 at 5:54
AandJ4EverAandJ4Ever
114
$endgroup$
$begingroup$
17 is not part of f(n) though
$endgroup$
– Ryan Topps
Dec 22 '18 at 5:57
1
$begingroup$
@RyanTopps We have $amid b$ iff there is some integer $k$ s.t. $b=ak$ iff $b=0$ mod $a$.
$endgroup$
– AandJ4Ever
Dec 22 '18 at 6:01
add a comment |
Your Answer
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
I think you're asking for what $x$ the equation $2n^2 + 2n + 1$ has a root modulo $x$.
First, since $2n^2 + 2n + 1$ is odd, we can assume that $x$ is odd. Then we can write:
$$2n^2 + 2n + 1 = frac{1}{2}((2n+1)^2 + 1)$$
This allows us to see that the equation has a solution if and only if there is a square root of $-1$ modulo $x$.
By the Chinese Remainder Theorem, this holds exactly when there is a square root of $-1$ modulo $p^k$ for every prime power $p^k$ dividing $x$.
It's a standard result that this happens when $pequiv 1pmod{4}$. This can be seen by looking at the cyclic group of units $pmod{p^k}$ and seeing when it has order divisible by $4$.
In conclusion, your equation has a solution exactly when all of the prime factors of $x$ are $1pmod 4$.
answered Dec 22 '18 at 7:07
SladeSlade
25.2k12665
$endgroup$
$begingroup$
Thank you, you explained that well
$endgroup$
– Ryan Topps
Dec 22 '18 at 7:36
$begingroup$
@RyanTopps I made a mistake; see the updated answer. The prime factors can never be $3pmod{4}$, I was misremembering something else.
$endgroup$
– Slade
Dec 22 '18 at 7:44
add a comment |
$begingroup$
I think you're asking for what $x$ the equation $2n^2 + 2n + 1$ has a root modulo $x$.
First, since $2n^2 + 2n + 1$ is odd, we can assume that $x$ is odd. Then we can write:
$$2n^2 + 2n + 1 = frac{1}{2}((2n+1)^2 + 1)$$
This allows us to see that the equation has a solution if and only if there is a square root of $-1$ modulo $x$.
By the Chinese Remainder Theorem, this holds exactly when there is a square root of $-1$ modulo $p^k$ for every prime power $p^k$ dividing $x$.
It's a standard result that this happens when $pequiv 1pmod{4}$. This can be seen by looking at the cyclic group of units $pmod{p^k}$ and seeing when it has order divisible by $4$.
In conclusion, your equation has a solution exactly when all of the prime factors of $x$ are $1pmod 4$.
answered Dec 22 '18 at 7:07
SladeSlade
25.2k12665
$endgroup$
$begingroup$
Thank you, you explained that well
$endgroup$
– Ryan Topps
Dec 22 '18 at 7:36
$begingroup$
@RyanTopps I made a mistake; see the updated answer. The prime factors can never be $3pmod{4}$, I was misremembering something else.
$endgroup$
– Slade
Dec 22 '18 at 7:44
add a comment |
$begingroup$
I think you're asking for what $x$ the equation $2n^2 + 2n + 1$ has a root modulo $x$.
First, since $2n^2 + 2n + 1$ is odd, we can assume that $x$ is odd. Then we can write:
$$2n^2 + 2n + 1 = frac{1}{2}((2n+1)^2 + 1)$$
This allows us to see that the equation has a solution if and only if there is a square root of $-1$ modulo $x$.
By the Chinese Remainder Theorem, this holds exactly when there is a square root of $-1$ modulo $p^k$ for every prime power $p^k$ dividing $x$.
It's a standard result that this happens when $pequiv 1pmod{4}$. This can be seen by looking at the cyclic group of units $pmod{p^k}$ and seeing when it has order divisible by $4$.
In conclusion, your equation has a solution exactly when all of the prime factors of $x$ are $1pmod 4$.
answered Dec 22 '18 at 7:07
SladeSlade
25.2k12665
$endgroup$
I think you're asking for what $x$ the equation $2n^2 + 2n + 1$ has a root modulo $x$.
First, since $2n^2 + 2n + 1$ is odd, we can assume that $x$ is odd. Then we can write:
$$2n^2 + 2n + 1 = frac{1}{2}((2n+1)^2 + 1)$$
This allows us to see that the equation has a solution if and only if there is a square root of $-1$ modulo $x$.
By the Chinese Remainder Theorem, this holds exactly when there is a square root of $-1$ modulo $p^k$ for every prime power $p^k$ dividing $x$.
It's a standard result that this happens when $pequiv 1pmod{4}$. This can be seen by looking at the cyclic group of units $pmod{p^k}$ and seeing when it has order divisible by $4$.
In conclusion, your equation has a solution exactly when all of the prime factors of $x$ are $1pmod 4$.
answered Dec 22 '18 at 7:07
SladeSlade
25.2k12665
edited Dec 22 '18 at 7:43
answered Dec 22 '18 at 7:07
SladeSlade
25.2k12665
answered Dec 22 '18 at 7:07
SladeSlade
25.2k12665
answered Dec 22 '18 at 7:07
SladeSlade
25.2k12665
25.2k12665
$begingroup$
Thank you, you explained that well
$endgroup$
– Ryan Topps
Dec 22 '18 at 7:36
$begingroup$
@RyanTopps I made a mistake; see the updated answer. The prime factors can never be $3pmod{4}$, I was misremembering something else.
$endgroup$
– Slade
Dec 22 '18 at 7:44
add a comment |
$begingroup$
Thank you, you explained that well
$endgroup$
– Ryan Topps
Dec 22 '18 at 7:36
$begingroup$
@RyanTopps I made a mistake; see the updated answer. The prime factors can never be $3pmod{4}$, I was misremembering something else.
$endgroup$
– Slade
Dec 22 '18 at 7:44
$begingroup$
Thank you, you explained that well
$endgroup$
– Ryan Topps
Dec 22 '18 at 7:36
$begingroup$
Thank you, you explained that well
$endgroup$
– Ryan Topps
Dec 22 '18 at 7:36
$begingroup$
@RyanTopps I made a mistake; see the updated answer. The prime factors can never be $3pmod{4}$, I was misremembering something else.
$endgroup$
– Slade
Dec 22 '18 at 7:44
$begingroup$
@RyanTopps I made a mistake; see the updated answer. The prime factors can never be $3pmod{4}$, I was misremembering something else.
$endgroup$
– Slade
Dec 22 '18 at 7:44
add a comment |
$begingroup$
Yes: $f(n)$ mod $x=0$ if and only if $xmid f(n)$.
answered Dec 22 '18 at 5:54
AandJ4EverAandJ4Ever
114
$endgroup$
$begingroup$
17 is not part of f(n) though
$endgroup$
– Ryan Topps
Dec 22 '18 at 5:57
1
$begingroup$
@RyanTopps We have $amid b$ iff there is some integer $k$ s.t. $b=ak$ iff $b=0$ mod $a$.
$endgroup$
– AandJ4Ever
Dec 22 '18 at 6:01
add a comment |
$begingroup$
Yes: $f(n)$ mod $x=0$ if and only if $xmid f(n)$.
answered Dec 22 '18 at 5:54
AandJ4EverAandJ4Ever
114
$endgroup$
$begingroup$
17 is not part of f(n) though
$endgroup$
– Ryan Topps
Dec 22 '18 at 5:57
1
$begingroup$
@RyanTopps We have $amid b$ iff there is some integer $k$ s.t. $b=ak$ iff $b=0$ mod $a$.
$endgroup$
– AandJ4Ever
Dec 22 '18 at 6:01
add a comment |
$begingroup$
Yes: $f(n)$ mod $x=0$ if and only if $xmid f(n)$.
answered Dec 22 '18 at 5:54
AandJ4EverAandJ4Ever
114
$endgroup$
Yes: $f(n)$ mod $x=0$ if and only if $xmid f(n)$.
answered Dec 22 '18 at 5:54
AandJ4EverAandJ4Ever
114
answered Dec 22 '18 at 5:54
AandJ4EverAandJ4Ever
114
answered Dec 22 '18 at 5:54
AandJ4EverAandJ4Ever
114
answered Dec 22 '18 at 5:54
AandJ4EverAandJ4Ever
114
114
$begingroup$
17 is not part of f(n) though
$endgroup$
– Ryan Topps
Dec 22 '18 at 5:57
1
$begingroup$
@RyanTopps We have $amid b$ iff there is some integer $k$ s.t. $b=ak$ iff $b=0$ mod $a$.
$endgroup$
– AandJ4Ever
Dec 22 '18 at 6:01
add a comment |
$begingroup$
17 is not part of f(n) though
$endgroup$
– Ryan Topps
Dec 22 '18 at 5:57
1
$begingroup$
@RyanTopps We have $amid b$ iff there is some integer $k$ s.t. $b=ak$ iff $b=0$ mod $a$.
$endgroup$
– AandJ4Ever
Dec 22 '18 at 6:01
$begingroup$
17 is not part of f(n) though
$endgroup$
– Ryan Topps
Dec 22 '18 at 5:57
$begingroup$
17 is not part of f(n) though
$endgroup$
– Ryan Topps
Dec 22 '18 at 5:57
1
1
$begingroup$
@RyanTopps We have $amid b$ iff there is some integer $k$ s.t. $b=ak$ iff $b=0$ mod $a$.
$endgroup$
– AandJ4Ever
Dec 22 '18 at 6:01
$begingroup$
@RyanTopps We have $amid b$ iff there is some integer $k$ s.t. $b=ak$ iff $b=0$ mod $a$.
$endgroup$
– AandJ4Ever
Dec 22 '18 at 6:01
add a comment |
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