Can anything be said about whether an integer has an even or odd number of prime factors?












1












$begingroup$


Because of the odd nature of my question, I am interested in partial solutions if a complete solution is not known. Specifically, the question I am trying to answer is the following:




Is there a method to show that a number does not have precisely two prime factors if we know it has at most three?




More generally, any condition uniquely satisfied by integers with an odd number of prime factors, or with an even number of prime factors, but not numbers with both, will be extremely helpful.



Edit: As per helpful comments, I am assuming the integer is entirely square free; that is, my question refers to distinct prime factors. (If there is an answer or partial answer that does not refer to specifically distinct prime factors, however, it would still be appreciated).










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  • 1




    $begingroup$
    A related problem is to compute the value of the Mobius function. You can see links at here and maybe here if you're interested.
    $endgroup$
    – Ben
    Jan 11 at 23:11






  • 1




    $begingroup$
    It would be interesting to know if you consider $p^2$ to have 1 or 2 prime factors, for this problem.
    $endgroup$
    – Ingix
    Jan 11 at 23:14










  • $begingroup$
    To determine the number of prime factors is not significantly easier than integer factorization (even if we assume we know that the number is squarefree). If you can find a non-trivial factor, you can much more efficient distinguish between two or three prime factors because in this case you only need a primality test.
    $endgroup$
    – Peter
    Jan 12 at 12:45


















1












$begingroup$


Because of the odd nature of my question, I am interested in partial solutions if a complete solution is not known. Specifically, the question I am trying to answer is the following:




Is there a method to show that a number does not have precisely two prime factors if we know it has at most three?




More generally, any condition uniquely satisfied by integers with an odd number of prime factors, or with an even number of prime factors, but not numbers with both, will be extremely helpful.



Edit: As per helpful comments, I am assuming the integer is entirely square free; that is, my question refers to distinct prime factors. (If there is an answer or partial answer that does not refer to specifically distinct prime factors, however, it would still be appreciated).










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    A related problem is to compute the value of the Mobius function. You can see links at here and maybe here if you're interested.
    $endgroup$
    – Ben
    Jan 11 at 23:11






  • 1




    $begingroup$
    It would be interesting to know if you consider $p^2$ to have 1 or 2 prime factors, for this problem.
    $endgroup$
    – Ingix
    Jan 11 at 23:14










  • $begingroup$
    To determine the number of prime factors is not significantly easier than integer factorization (even if we assume we know that the number is squarefree). If you can find a non-trivial factor, you can much more efficient distinguish between two or three prime factors because in this case you only need a primality test.
    $endgroup$
    – Peter
    Jan 12 at 12:45
















1












1








1





$begingroup$


Because of the odd nature of my question, I am interested in partial solutions if a complete solution is not known. Specifically, the question I am trying to answer is the following:




Is there a method to show that a number does not have precisely two prime factors if we know it has at most three?




More generally, any condition uniquely satisfied by integers with an odd number of prime factors, or with an even number of prime factors, but not numbers with both, will be extremely helpful.



Edit: As per helpful comments, I am assuming the integer is entirely square free; that is, my question refers to distinct prime factors. (If there is an answer or partial answer that does not refer to specifically distinct prime factors, however, it would still be appreciated).










share|cite|improve this question











$endgroup$




Because of the odd nature of my question, I am interested in partial solutions if a complete solution is not known. Specifically, the question I am trying to answer is the following:




Is there a method to show that a number does not have precisely two prime factors if we know it has at most three?




More generally, any condition uniquely satisfied by integers with an odd number of prime factors, or with an even number of prime factors, but not numbers with both, will be extremely helpful.



Edit: As per helpful comments, I am assuming the integer is entirely square free; that is, my question refers to distinct prime factors. (If there is an answer or partial answer that does not refer to specifically distinct prime factors, however, it would still be appreciated).







number-theory elementary-number-theory prime-numbers prime-factorization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 23:25







Tejas Rao

















asked Jan 11 at 22:49









Tejas RaoTejas Rao

33711




33711








  • 1




    $begingroup$
    A related problem is to compute the value of the Mobius function. You can see links at here and maybe here if you're interested.
    $endgroup$
    – Ben
    Jan 11 at 23:11






  • 1




    $begingroup$
    It would be interesting to know if you consider $p^2$ to have 1 or 2 prime factors, for this problem.
    $endgroup$
    – Ingix
    Jan 11 at 23:14










  • $begingroup$
    To determine the number of prime factors is not significantly easier than integer factorization (even if we assume we know that the number is squarefree). If you can find a non-trivial factor, you can much more efficient distinguish between two or three prime factors because in this case you only need a primality test.
    $endgroup$
    – Peter
    Jan 12 at 12:45
















  • 1




    $begingroup$
    A related problem is to compute the value of the Mobius function. You can see links at here and maybe here if you're interested.
    $endgroup$
    – Ben
    Jan 11 at 23:11






  • 1




    $begingroup$
    It would be interesting to know if you consider $p^2$ to have 1 or 2 prime factors, for this problem.
    $endgroup$
    – Ingix
    Jan 11 at 23:14










  • $begingroup$
    To determine the number of prime factors is not significantly easier than integer factorization (even if we assume we know that the number is squarefree). If you can find a non-trivial factor, you can much more efficient distinguish between two or three prime factors because in this case you only need a primality test.
    $endgroup$
    – Peter
    Jan 12 at 12:45










1




1




$begingroup$
A related problem is to compute the value of the Mobius function. You can see links at here and maybe here if you're interested.
$endgroup$
– Ben
Jan 11 at 23:11




$begingroup$
A related problem is to compute the value of the Mobius function. You can see links at here and maybe here if you're interested.
$endgroup$
– Ben
Jan 11 at 23:11




1




1




$begingroup$
It would be interesting to know if you consider $p^2$ to have 1 or 2 prime factors, for this problem.
$endgroup$
– Ingix
Jan 11 at 23:14




$begingroup$
It would be interesting to know if you consider $p^2$ to have 1 or 2 prime factors, for this problem.
$endgroup$
– Ingix
Jan 11 at 23:14












$begingroup$
To determine the number of prime factors is not significantly easier than integer factorization (even if we assume we know that the number is squarefree). If you can find a non-trivial factor, you can much more efficient distinguish between two or three prime factors because in this case you only need a primality test.
$endgroup$
– Peter
Jan 12 at 12:45






$begingroup$
To determine the number of prime factors is not significantly easier than integer factorization (even if we assume we know that the number is squarefree). If you can find a non-trivial factor, you can much more efficient distinguish between two or three prime factors because in this case you only need a primality test.
$endgroup$
– Peter
Jan 12 at 12:45












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