Similar to Lehmar's totient problem
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Find all positive integer such that
$$phi(n)^2mid n^2-1$$
My progress so far:
I proved that $n$ has to be a odd square-free integer. So $n=p_1p_2ldots p_k$. So the question can be restated as following:
$$prod (p_i-1)^2mid (p_1p_2ldots p_k)^2-1$$
From here I farther transformed the question as follows (Not sure if this simplifies the problem but it is definitely beautiful):
If $P(x)$ a polynomial such that all it's roots are distinct odd primes, prove that
$$P(1)^2nmid P(0)^2-1$$
And now I am stuck. Any help is appreciated.
number-theory divisibility totient-function
asked Sep 8 at 10:27
Basudeb Bhowmick
112
add a comment |Â
up vote
2
down vote
favorite
Find all positive integer such that
$$phi(n)^2mid n^2-1$$
My progress so far:
I proved that $n$ has to be a odd square-free integer. So $n=p_1p_2ldots p_k$. So the question can be restated as following:
$$prod (p_i-1)^2mid (p_1p_2ldots p_k)^2-1$$
From here I farther transformed the question as follows (Not sure if this simplifies the problem but it is definitely beautiful):
If $P(x)$ a polynomial such that all it's roots are distinct odd primes, prove that
$$P(1)^2nmid P(0)^2-1$$
And now I am stuck. Any help is appreciated.
number-theory divisibility totient-function
asked Sep 8 at 10:27
Basudeb Bhowmick
112
$n=1,2,3$ are solutions and there is no additional solution upto $n=10^8$
– Peter
Sep 9 at 21:57
Still no additional solution upto $n=10^9$
– Peter
Sep 9 at 22:44
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Find all positive integer such that
$$phi(n)^2mid n^2-1$$
My progress so far:
I proved that $n$ has to be a odd square-free integer. So $n=p_1p_2ldots p_k$. So the question can be restated as following:
$$prod (p_i-1)^2mid (p_1p_2ldots p_k)^2-1$$
From here I farther transformed the question as follows (Not sure if this simplifies the problem but it is definitely beautiful):
If $P(x)$ a polynomial such that all it's roots are distinct odd primes, prove that
$$P(1)^2nmid P(0)^2-1$$
And now I am stuck. Any help is appreciated.
number-theory divisibility totient-function
asked Sep 8 at 10:27
Basudeb Bhowmick
112
Find all positive integer such that
$$phi(n)^2mid n^2-1$$
My progress so far:
I proved that $n$ has to be a odd square-free integer. So $n=p_1p_2ldots p_k$. So the question can be restated as following:
$$prod (p_i-1)^2mid (p_1p_2ldots p_k)^2-1$$
From here I farther transformed the question as follows (Not sure if this simplifies the problem but it is definitely beautiful):
If $P(x)$ a polynomial such that all it's roots are distinct odd primes, prove that
$$P(1)^2nmid P(0)^2-1$$
And now I am stuck. Any help is appreciated.
number-theory divisibility totient-function
number-theory divisibility totient-function
asked Sep 8 at 10:27
Basudeb Bhowmick
112
asked Sep 8 at 10:27
Basudeb Bhowmick
112
asked Sep 8 at 10:27
Basudeb Bhowmick
112
asked Sep 8 at 10:27
Basudeb Bhowmick
112
asked Sep 8 at 10:27
Basudeb Bhowmick
112
112
$n=1,2,3$ are solutions and there is no additional solution upto $n=10^8$
– Peter
Sep 9 at 21:57
Still no additional solution upto $n=10^9$
– Peter
Sep 9 at 22:44
add a comment |Â
$n=1,2,3$ are solutions and there is no additional solution upto $n=10^8$
– Peter
Sep 9 at 21:57
Still no additional solution upto $n=10^9$
– Peter
Sep 9 at 22:44
$n=1,2,3$ are solutions and there is no additional solution upto $n=10^8$
– Peter
Sep 9 at 21:57
$n=1,2,3$ are solutions and there is no additional solution upto $n=10^8$
– Peter
Sep 9 at 21:57
Still no additional solution upto $n=10^9$
– Peter
Sep 9 at 22:44
Still no additional solution upto $n=10^9$
– Peter
Sep 9 at 22:44
add a comment |Â
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$n=1,2,3$ are solutions and there is no additional solution upto $n=10^8$
– Peter
Sep 9 at 21:57
Still no additional solution upto $n=10^9$
– Peter
Sep 9 at 22:44