I am trying to disprove a conjecture, and I have gotten it such that the conjecture is only true if $$\prod_{i=1}^{g}{(\frac{j_i^{L_i+1}-1}{j_i-1})}$$ is singly even (of form $2m$ where $m$ is odd).

Here, $g$ is the number of terms in set $j$, which is the set of prime factors of an odd integer $n$ that are the sums of two squares. Every $L_i$ is the corresponding exponent of $j_i$ in the prime factorization of $n$.

Here is what I know about these:

  • Every $j_i$ is odd.
  • Every $L_i$ except for $L_1$ is even.
  • I do not know the parity of $g$.

Here is what I have tried so far:

Since every $j_i$ is odd and the sum of two squares, it must be of form $4a+8b+1$ (Euler). In the numerator of the pi notation, we have $j_i^{L_i+1}-1$. A sum of two squares raised to any power is a sum of two squares, so the numerator is of form $4a+8b$. In the denominator, we have $j_i-1$, which must be of form $4c+8d$. Thus we have $\frac{4a+8b}{4c+8d}$, which can be simplified to $\frac{a+2b}{c+2d}$.

I'm not sure where to go from there.



Since there are $L_i+1$ terms in that addition, and they are each odd, the parity of the addition is the parity of $L_i+1$. For all even $L_i$, this means that $L_i+1$ is odd, and thus this addition is odd. For all odd $L_i$, $L_i+1$ is even, and thus this addition is even. This means that $$\prod_{i=1}^{g}{(\frac{j_i^{L_i+1}-1}{j_i-1})}$$ is singly even, and you can't disprove the conjecture.


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