Let for integers $n\geq 1$ the Euler's totient function $\varphi(n)$. On the other hand one knows the so-called Bernoulli inequality, see for example this MathWorld. This morning I wondered about two puzzles that I've created involving previous inequality and two well known sequences of integers (in this post the arithmetic function $\varphi(n)$). These puzzles do not have a context in this inequality,
my creation or definition of the puzzles was artificious and has no a special mathematical meaning, but I am curious about if it is possible to deduce a solution.
Definition. I define the sequence of integers $n\geq 1$ such that $\varphi(1+n^2)$ divides $\varphi\left((1+n)^n\right)$.
Computational fact. Our sequence starts with $$1,3,5,9,13,25,45,47,\ldots$$ and seems that there isn't in The On-Line Encyclopedia of Integer Sequences.
Question. Is it possible to deduce that previous sequence has infinitely many terms? What is you reasoning or heuristic? Many thanks.
I think that it is very difficult. I've observed that in our sequence there are primes, and also some perfect squares.