Mixing
How to solve for the sides of a rectangle whose sides are natural numbers given its area is a known natural...
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This may not be the best way to formulate the question but I am looking for a method to solve the following equation $d cdot n'=n$ where $d, n', n in mathbb{N}$ and $n neq1$ is known. How should I approach this problem? Are there guaranteed solutions?
natural-numbers
asked Jan 1 at 12:38
ClimbingTheCurveClimbingTheCurve
12
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add a comment |
$begingroup$
This may not be the best way to formulate the question but I am looking for a method to solve the following equation $d cdot n'=n$ where $d, n', n in mathbb{N}$ and $n neq1$ is known. How should I approach this problem? Are there guaranteed solutions?
natural-numbers
asked Jan 1 at 12:38
ClimbingTheCurveClimbingTheCurve
12
$endgroup$
3
$begingroup$
You have basically asked if it is possible to find a factorization of $n$. If you forbid the trivial factorization of $1cdot n = n$, then this problem is at least as hard as determining if $n$ is prime, and probably not harder than breaking the (multiprime) RSA encryption scheme.
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– InequalitiesEverywhere
Jan 1 at 12:48
add a comment |
$begingroup$
This may not be the best way to formulate the question but I am looking for a method to solve the following equation $d cdot n'=n$ where $d, n', n in mathbb{N}$ and $n neq1$ is known. How should I approach this problem? Are there guaranteed solutions?
natural-numbers
asked Jan 1 at 12:38
ClimbingTheCurveClimbingTheCurve
12
$endgroup$
This may not be the best way to formulate the question but I am looking for a method to solve the following equation $d cdot n'=n$ where $d, n', n in mathbb{N}$ and $n neq1$ is known. How should I approach this problem? Are there guaranteed solutions?
natural-numbers
natural-numbers
asked Jan 1 at 12:38
ClimbingTheCurveClimbingTheCurve
12
asked Jan 1 at 12:38
ClimbingTheCurveClimbingTheCurve
12
edited Jan 1 at 13:22
ClimbingTheCurve
asked Jan 1 at 12:38
ClimbingTheCurveClimbingTheCurve
12
asked Jan 1 at 12:38
ClimbingTheCurveClimbingTheCurve
12
asked Jan 1 at 12:38
ClimbingTheCurveClimbingTheCurve
12
12
3
$begingroup$
You have basically asked if it is possible to find a factorization of $n$. If you forbid the trivial factorization of $1cdot n = n$, then this problem is at least as hard as determining if $n$ is prime, and probably not harder than breaking the (multiprime) RSA encryption scheme.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 12:48
add a comment |
3
$begingroup$
You have basically asked if it is possible to find a factorization of $n$. If you forbid the trivial factorization of $1cdot n = n$, then this problem is at least as hard as determining if $n$ is prime, and probably not harder than breaking the (multiprime) RSA encryption scheme.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 12:48
3
3
$begingroup$
You have basically asked if it is possible to find a factorization of $n$. If you forbid the trivial factorization of $1cdot n = n$, then this problem is at least as hard as determining if $n$ is prime, and probably not harder than breaking the (multiprime) RSA encryption scheme.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 12:48
$begingroup$
You have basically asked if it is possible to find a factorization of $n$. If you forbid the trivial factorization of $1cdot n = n$, then this problem is at least as hard as determining if $n$ is prime, and probably not harder than breaking the (multiprime) RSA encryption scheme.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 12:48
add a comment |
1 Answer
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They would be any two factors of $n$.
- If $n$ is prime, then the numbers are $1,n$.
- If $n=1$, $d=n'=1$.
- If $n$ is composite, take any two factors $d, n'$ of $n$ such that $dn' = n$.
Solutions are indeed guaranteed for all $n in mathbb{N}$.
This follows from the facts that by definition the factors of a natural number are in turn also natural numbers, and that all numbers are prime, composite, or $1$.
answered Jan 1 at 12:43
Eevee TrainerEevee Trainer
5,3551836
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$begingroup$
Should you not mention that your final claim is the Fundamental Theorem of Arithmetic? en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic
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– DavidG
Jan 8 at 5:07
add a comment |
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1 Answer
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$begingroup$
They would be any two factors of $n$.
- If $n$ is prime, then the numbers are $1,n$.
- If $n=1$, $d=n'=1$.
- If $n$ is composite, take any two factors $d, n'$ of $n$ such that $dn' = n$.
Solutions are indeed guaranteed for all $n in mathbb{N}$.
This follows from the facts that by definition the factors of a natural number are in turn also natural numbers, and that all numbers are prime, composite, or $1$.
answered Jan 1 at 12:43
Eevee TrainerEevee Trainer
5,3551836
$endgroup$
$begingroup$
Should you not mention that your final claim is the Fundamental Theorem of Arithmetic? en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic
$endgroup$
– DavidG
Jan 8 at 5:07
add a comment |
$begingroup$
They would be any two factors of $n$.
- If $n$ is prime, then the numbers are $1,n$.
- If $n=1$, $d=n'=1$.
- If $n$ is composite, take any two factors $d, n'$ of $n$ such that $dn' = n$.
Solutions are indeed guaranteed for all $n in mathbb{N}$.
This follows from the facts that by definition the factors of a natural number are in turn also natural numbers, and that all numbers are prime, composite, or $1$.
answered Jan 1 at 12:43
Eevee TrainerEevee Trainer
5,3551836
$endgroup$
$begingroup$
Should you not mention that your final claim is the Fundamental Theorem of Arithmetic? en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic
$endgroup$
– DavidG
Jan 8 at 5:07
add a comment |
$begingroup$
They would be any two factors of $n$.
- If $n$ is prime, then the numbers are $1,n$.
- If $n=1$, $d=n'=1$.
- If $n$ is composite, take any two factors $d, n'$ of $n$ such that $dn' = n$.
Solutions are indeed guaranteed for all $n in mathbb{N}$.
This follows from the facts that by definition the factors of a natural number are in turn also natural numbers, and that all numbers are prime, composite, or $1$.
answered Jan 1 at 12:43
Eevee TrainerEevee Trainer
5,3551836
$endgroup$
They would be any two factors of $n$.
- If $n$ is prime, then the numbers are $1,n$.
- If $n=1$, $d=n'=1$.
- If $n$ is composite, take any two factors $d, n'$ of $n$ such that $dn' = n$.
Solutions are indeed guaranteed for all $n in mathbb{N}$.
This follows from the facts that by definition the factors of a natural number are in turn also natural numbers, and that all numbers are prime, composite, or $1$.
answered Jan 1 at 12:43
Eevee TrainerEevee Trainer
5,3551836
answered Jan 1 at 12:43
Eevee TrainerEevee Trainer
5,3551836
answered Jan 1 at 12:43
Eevee TrainerEevee Trainer
5,3551836
answered Jan 1 at 12:43
Eevee TrainerEevee Trainer
5,3551836
5,3551836
$begingroup$
Should you not mention that your final claim is the Fundamental Theorem of Arithmetic? en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic
$endgroup$
– DavidG
Jan 8 at 5:07
add a comment |
$begingroup$
Should you not mention that your final claim is the Fundamental Theorem of Arithmetic? en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic
$endgroup$
– DavidG
Jan 8 at 5:07
$begingroup$
Should you not mention that your final claim is the Fundamental Theorem of Arithmetic? en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic
$endgroup$
– DavidG
Jan 8 at 5:07
$begingroup$
Should you not mention that your final claim is the Fundamental Theorem of Arithmetic? en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic
$endgroup$
– DavidG
Jan 8 at 5:07
add a comment |
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$begingroup$
You have basically asked if it is possible to find a factorization of $n$. If you forbid the trivial factorization of $1cdot n = n$, then this problem is at least as hard as determining if $n$ is prime, and probably not harder than breaking the (multiprime) RSA encryption scheme.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 12:48