Integer roots of natural number
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Is it guaranteed that for all $n in N$, there are two positive integers $a, b$ such that $a^{b}=n$ and $a neq n$?
I don't actually need to prove this, it just needs to be true for a certain algorithm i made to be correct.
Thanks in advance.
elementary-number-theory
asked Dec 5 at 18:21
Mateus Buarque
1589
add a comment |
up vote
-3
down vote
favorite
Is it guaranteed that for all $n in N$, there are two positive integers $a, b$ such that $a^{b}=n$ and $a neq n$?
I don't actually need to prove this, it just needs to be true for a certain algorithm i made to be correct.
Thanks in advance.
elementary-number-theory
asked Dec 5 at 18:21
Mateus Buarque
1589
add a comment |
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
Is it guaranteed that for all $n in N$, there are two positive integers $a, b$ such that $a^{b}=n$ and $a neq n$?
I don't actually need to prove this, it just needs to be true for a certain algorithm i made to be correct.
Thanks in advance.
elementary-number-theory
asked Dec 5 at 18:21
Mateus Buarque
1589
Is it guaranteed that for all $n in N$, there are two positive integers $a, b$ such that $a^{b}=n$ and $a neq n$?
I don't actually need to prove this, it just needs to be true for a certain algorithm i made to be correct.
Thanks in advance.
elementary-number-theory
elementary-number-theory
asked Dec 5 at 18:21
Mateus Buarque
1589
asked Dec 5 at 18:21
Mateus Buarque
1589
asked Dec 5 at 18:21
Mateus Buarque
1589
asked Dec 5 at 18:21
Mateus Buarque
1589
asked Dec 5 at 18:21
Mateus Buarque
1589
1589
add a comment |
add a comment |
3 Answers
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3
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This is clearly false. Take for example $n=2$ or any other prime.
answered Dec 5 at 18:23
ajotatxe
53k23890
Any positive integer if the exponents of its factorization have gcd=1 is a counterexample.
– ajotatxe
Dec 5 at 18:30
add a comment |
up vote
1
down vote
No. This is true only for numbers $n$ whose prime factorisation $n = prodlimits_{i=1}^kp_i^{e_i}$ with the $p_i$ distinct primes has the property that there is some integer $b geq 2$ such that $k|e_i$ for all $i$. (This is a very long way from being all integers).
Proof:
For an $n$ that does have this property, take $$a = prodlimits_{i=1}^kp_i^{frac{e_i}{b}},$$
which is an integer since $b|e_i$ for all $i$, and clearly $a^b = n$.
For the converse, for any $n$ which does not have this property, if there is $(a,b)$ such that $a^b = n$, then $a$ has some prime factorisation $a = prodlimits_{i=1}^jp_i^{f_i}$, so $a^b = prodlimits_{i=1}^jp_i^{f_ib}$, so by the uniqueness of prime factorisations, $e_i = f_ib$ for all $i$ (allowing for some $e_i, f_i$ to be $0$ means that we can avoid any problems with the primes differing).
answered Dec 5 at 18:35
user3482749
2,166414
add a comment |
up vote
0
down vote
The list of counterexamples is https://oeis.org/A007916 "Numbers that are not perfect powers": 2, 3, 5, 6, 7, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, ...
answered Dec 5 at 18:28
Chris Culter
19.7k43381
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
This is clearly false. Take for example $n=2$ or any other prime.
answered Dec 5 at 18:23
ajotatxe
53k23890
Any positive integer if the exponents of its factorization have gcd=1 is a counterexample.
– ajotatxe
Dec 5 at 18:30
add a comment |
up vote
3
down vote
This is clearly false. Take for example $n=2$ or any other prime.
answered Dec 5 at 18:23
ajotatxe
53k23890
Any positive integer if the exponents of its factorization have gcd=1 is a counterexample.
– ajotatxe
Dec 5 at 18:30
add a comment |
up vote
3
down vote
up vote
3
down vote
This is clearly false. Take for example $n=2$ or any other prime.
answered Dec 5 at 18:23
ajotatxe
53k23890
This is clearly false. Take for example $n=2$ or any other prime.
answered Dec 5 at 18:23
ajotatxe
53k23890
answered Dec 5 at 18:23
ajotatxe
53k23890
answered Dec 5 at 18:23
ajotatxe
53k23890
answered Dec 5 at 18:23
ajotatxe
53k23890
53k23890
Any positive integer if the exponents of its factorization have gcd=1 is a counterexample.
– ajotatxe
Dec 5 at 18:30
add a comment |
Any positive integer if the exponents of its factorization have gcd=1 is a counterexample.
– ajotatxe
Dec 5 at 18:30
Any positive integer if the exponents of its factorization have gcd=1 is a counterexample.
– ajotatxe
Dec 5 at 18:30
Any positive integer if the exponents of its factorization have gcd=1 is a counterexample.
– ajotatxe
Dec 5 at 18:30
add a comment |
up vote
1
down vote
No. This is true only for numbers $n$ whose prime factorisation $n = prodlimits_{i=1}^kp_i^{e_i}$ with the $p_i$ distinct primes has the property that there is some integer $b geq 2$ such that $k|e_i$ for all $i$. (This is a very long way from being all integers).
Proof:
For an $n$ that does have this property, take $$a = prodlimits_{i=1}^kp_i^{frac{e_i}{b}},$$
which is an integer since $b|e_i$ for all $i$, and clearly $a^b = n$.
For the converse, for any $n$ which does not have this property, if there is $(a,b)$ such that $a^b = n$, then $a$ has some prime factorisation $a = prodlimits_{i=1}^jp_i^{f_i}$, so $a^b = prodlimits_{i=1}^jp_i^{f_ib}$, so by the uniqueness of prime factorisations, $e_i = f_ib$ for all $i$ (allowing for some $e_i, f_i$ to be $0$ means that we can avoid any problems with the primes differing).
answered Dec 5 at 18:35
user3482749
2,166414
add a comment |
up vote
1
down vote
No. This is true only for numbers $n$ whose prime factorisation $n = prodlimits_{i=1}^kp_i^{e_i}$ with the $p_i$ distinct primes has the property that there is some integer $b geq 2$ such that $k|e_i$ for all $i$. (This is a very long way from being all integers).
Proof:
For an $n$ that does have this property, take $$a = prodlimits_{i=1}^kp_i^{frac{e_i}{b}},$$
which is an integer since $b|e_i$ for all $i$, and clearly $a^b = n$.
For the converse, for any $n$ which does not have this property, if there is $(a,b)$ such that $a^b = n$, then $a$ has some prime factorisation $a = prodlimits_{i=1}^jp_i^{f_i}$, so $a^b = prodlimits_{i=1}^jp_i^{f_ib}$, so by the uniqueness of prime factorisations, $e_i = f_ib$ for all $i$ (allowing for some $e_i, f_i$ to be $0$ means that we can avoid any problems with the primes differing).
answered Dec 5 at 18:35
user3482749
2,166414
add a comment |
up vote
1
down vote
up vote
1
down vote
No. This is true only for numbers $n$ whose prime factorisation $n = prodlimits_{i=1}^kp_i^{e_i}$ with the $p_i$ distinct primes has the property that there is some integer $b geq 2$ such that $k|e_i$ for all $i$. (This is a very long way from being all integers).
Proof:
For an $n$ that does have this property, take $$a = prodlimits_{i=1}^kp_i^{frac{e_i}{b}},$$
which is an integer since $b|e_i$ for all $i$, and clearly $a^b = n$.
For the converse, for any $n$ which does not have this property, if there is $(a,b)$ such that $a^b = n$, then $a$ has some prime factorisation $a = prodlimits_{i=1}^jp_i^{f_i}$, so $a^b = prodlimits_{i=1}^jp_i^{f_ib}$, so by the uniqueness of prime factorisations, $e_i = f_ib$ for all $i$ (allowing for some $e_i, f_i$ to be $0$ means that we can avoid any problems with the primes differing).
answered Dec 5 at 18:35
user3482749
2,166414
No. This is true only for numbers $n$ whose prime factorisation $n = prodlimits_{i=1}^kp_i^{e_i}$ with the $p_i$ distinct primes has the property that there is some integer $b geq 2$ such that $k|e_i$ for all $i$. (This is a very long way from being all integers).
Proof:
For an $n$ that does have this property, take $$a = prodlimits_{i=1}^kp_i^{frac{e_i}{b}},$$
which is an integer since $b|e_i$ for all $i$, and clearly $a^b = n$.
For the converse, for any $n$ which does not have this property, if there is $(a,b)$ such that $a^b = n$, then $a$ has some prime factorisation $a = prodlimits_{i=1}^jp_i^{f_i}$, so $a^b = prodlimits_{i=1}^jp_i^{f_ib}$, so by the uniqueness of prime factorisations, $e_i = f_ib$ for all $i$ (allowing for some $e_i, f_i$ to be $0$ means that we can avoid any problems with the primes differing).
answered Dec 5 at 18:35
user3482749
2,166414
answered Dec 5 at 18:35
user3482749
2,166414
answered Dec 5 at 18:35
user3482749
2,166414
answered Dec 5 at 18:35
user3482749
2,166414
2,166414
add a comment |
add a comment |
up vote
0
down vote
The list of counterexamples is https://oeis.org/A007916 "Numbers that are not perfect powers": 2, 3, 5, 6, 7, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, ...
answered Dec 5 at 18:28
Chris Culter
19.7k43381
add a comment |
up vote
0
down vote
The list of counterexamples is https://oeis.org/A007916 "Numbers that are not perfect powers": 2, 3, 5, 6, 7, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, ...
answered Dec 5 at 18:28
Chris Culter
19.7k43381
add a comment |
up vote
0
down vote
up vote
0
down vote
The list of counterexamples is https://oeis.org/A007916 "Numbers that are not perfect powers": 2, 3, 5, 6, 7, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, ...
answered Dec 5 at 18:28
Chris Culter
19.7k43381
The list of counterexamples is https://oeis.org/A007916 "Numbers that are not perfect powers": 2, 3, 5, 6, 7, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, ...
answered Dec 5 at 18:28
Chris Culter
19.7k43381
answered Dec 5 at 18:28
Chris Culter
19.7k43381
answered Dec 5 at 18:28
Chris Culter
19.7k43381
answered Dec 5 at 18:28
Chris Culter
19.7k43381
19.7k43381
add a comment |
add a comment |
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