Wilson perfect powers
Wilson perfect powers
Determinare tutti i primi $p$ tali che per un qualche $n\in\mathbb N$ valga $$(p-1)!+1=p^n$$
Imagination is more important than knowledge. For knowledge is limited, whereas imagination embraces the entire world, stimulating progress, giving birth to evolution (A. Einstein)
- Troleito br00tal
- Messaggi: 683
- Iscritto il: 16 mag 2012, 22:25
Re: Wilson perfect powers
Supponiamo $p>5$. Allora $2 \cdot \frac{p-1}{2} \cdot (p-1) | (p-1)!$, poiché $\frac{p-1}{2} \not = 2$. Quindi $(p-1)^2|(p-1)!=p^n-1 \rightarrow p-1|1+...+p^{n-1} \rightarrow p-1|n$. Quindi $n \ge p-1$. Ma ovviamente $p^{p-1}>(p-1)!+1$, poiché $p>i$ per ogni $0<i<p$. Quindi $p \le 5$. Ma allora risolvono solo $(p,n)=(2,1);(3,1);(5,2)$.
Re: Wilson perfect powers
Sintetico ma efficace!
Imagination is more important than knowledge. For knowledge is limited, whereas imagination embraces the entire world, stimulating progress, giving birth to evolution (A. Einstein)