Hello and good evening.
Well, the factors of N=39 are 3 and 13. When i calculate k1 and k2: \begin{equation} k1 =\log{(3)}/ \log(39) = 0.2998754... \\ k2 =\log{(13)}/ \log(39) = 0.7001246... \\ \end{equation} i see that k1 + k2 equals to 1.
that means: \begin{equation} 1 = k1 + k2 \\ 3 = 39^{k1} \\ 13 = 39^{k2} \\ \end{equation} When i take the square root of 39, i get: \begin{equation} \sqrt{39} = 6.24 \\ 39^{0.5} = 6.24 \end{equation} The main idea is to calculate the .24 decimal off and variate the 0.5 exponent to the values above like k1 and k2 so that i can get the factors 3 and 13.
Is there any way to achieve this ?
Thanks for reading.
Edit:
The issue is the following one:
When i get the square root of 39, then i will have a value of N. I just wanted to calculate the k1 and k2 values without brute-forcing by a computer code. When i increment 0.5 to something 0.51, it will decrement the other 0.5 to 0.49 till the integer prime values have been found of N like 3 and 13.
\begin{equation} 39^{0.51} = 6.47 \\ 39^{0.49} = 6.020 \\ ... till \\ 39^{0.2998754} = 3 \\ 39^{0.7001246} = 13 \end{equation}