08/01/16 23:01:32
>510
p≧2, b[k]≧2 として
c[n] = Π[k=1,n] b[k],
a[n] = p^c[n]
とおく。
a[j]/a[i] -1 = p^(c[j]-c[i]) -1 = p^{c[i](c[j]/c[i] -1)} -1 = a[i]^(c[j]/c[i] -1) -1,
c[j]/c[i] -1 ≧ 1 だから a[i] -1 の倍数。
a[j] -1 = (a[i] -1)(a[j]/a[i]) + a[j]/a[i] -1,
より
gcd(a[i] -1, a[j] -1) = gcd(a[i] -1, (a[j]/a[i]) -1) = (a[j]/a[i]) -1.
ｽﾚﾘﾝｸ(math板:23-29番)
東大入試作問者スレ１３