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Computax Software Free Download Crack REPACK



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Computax software free download crack untuk PC Windows. Computax software free download 2.0.Q:

Do powers of a specific prime have one or two prime factors?

Say we know that $n^4$ has $2$ distinct prime factors for all primes $n$ and $n^5$ has $3$ distinct prime factors for all primes $n$. Can we conclude anything about the number of prime factors of $n^6$ for all $n$?
If there is no algebraic formula for this then can we prove some asymptotic results on this problem (by maybe proving that the number of prime factors for $n^6$ grows like $p(n)^6$ for some function $p$)?


You can’t prove anything (really!) about the exact number of prime factors of $n^6$. (Why not? Look at the divisibility by $n^6$!) The problem is that there is no deep probabilistic reason to expect the number of prime factors to be a function of $n$, because the number of different primes is infinite, and there are infinitely many values of $n$ for which $n^6$ is prime.
But you can prove something about the number of prime factors for “most” values of $n$. Namely, you can prove that it is $1$ for the asymptotically negligible set of values of $n$ satisfying the Erdős-Rényi-Szekeres conjecture. (If you want to make the statements as precise as possible, then the number of prime factors is always $1$ when $n$ is a perfect square, and is always $2$ when $n$ is the fifth power of an odd prime number.)

If one has a little bit of problem with the technical language above, let me know and I’ll provide a more formal version.


Proving that 2x + 2y + 3z cannot be a prime number?

How can I prove that the equation $2x + 2y + 3z=1$ has no prime solutions?
I tried to prove it by contradiction, by supposing that $p$ is the prime factor of $2x + 2y + 3z=1$. Then I proved that $p|2x+2y+3z$ or $p|


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