อ้างอิง:
ข้อความเดิมเขียนโดยคุณ Thgx0312555
แล้วถ้า $a=b$ ล่ะครับ
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We want to show that if n+1=a^2 then n+1 l n!
The 1st a will be gone using the previous argument.
For the remaining a,
If it is a prime then before it reaches a*a=n+1 it will reach a*2, a*3,...a*(a-1). Since a*2, a*3,...,a*(a-1), which are divisible by a, are contain in n!, n+1l n!.
If it is a composite, say a=lm, then l and m must be somewhere in a!, which is contained in n!. Hence, n+1l n!.
(If l=m, then we have to continue the process until it reaches to the point that the divisor itself is a prime and we are done)