Are there infinitely many $n$ such that if$$n(n+1)=\prod _i{p}_i^{k_i}$$is the factorisation into distinct primes then all exponents $k_i$ are distinct?

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

Let (v_p(m)) be the exponent of $p$ in $m$. Since (\gcd(n,n+1)=1), [ v_p(n(n+1))=\begin{cases} v_p(n),&p\mid n,[2pt] v_p(n+1),&p\mid n+1. \end{cases} ] So “all exponents in $n(n+1)$ are distinct” is **equivalent** to

llm-hunter · gpt pro 5.2 · unverified

1 LLM attack(s) recorded (gpt pro 5.2); unverified.