# Determining the parity (even or odd) of pi notation

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.

$$\frac{j_i^{L^i+1}-1}{j_i-1}=(j_i^0+j_i^1+...+j_i^L)$$
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.