Prime factors of Φ3(x)\Phi_3(x) of the same form

We parameterize solutions to the equality $\Phi_3(x)=\Phi_3(a_1)\Phi_3(a_2)\cdots\Phi_3(a_n)$ when each $\Phi_3(a_i)$ is prime. Our focus is on the special cases when $n=2,3,4$, as this analysis simplifies and extends bounds on the total number of prime factors of an odd perfect number

On a theorem of Kanold on odd perfect numbers

We shall prove that if $N=p^\alpha q_1^{2\beta_1} q_2^{2\beta_2} \cdots q_{r-1}^{2\beta_{r-1}}$ is an odd perfect number such that $p, q_1, \ldots, q_{r-1}$ are distinct primes, $p\equiv\alpha\equiv 1\mod{4}$ and $t$ divides $2\beta_i+1$ for all $i=1, 2, \ldots, r-1$, then $t^5$ divides $N$, improving an eighty-year old result of Kanold.Comment: 8 pages, the author's final version of the paper of the same title in Indian J. Pure Appl. Math, which is available at https://doi.org/10.1007/s13226-023-00530-