Question

1.对于每个正整数n,让f(n)代表最小的正整数s,并且1+2+3+...+(s-1) +s 的值能被n整除.举个例子,

Answer 1

a) 1+2+3+...+(s-1) +s= s(s+1)/2 = 36

Consider all the divisors of 36,and notice it is impossible for n to be any odd number less than 9,since simply let s=n2007.

Thus completes the proof.

(c) Prove:f(c)=f(c+1) and f(c)=f(c+2) have no odd solutions.

Pf:(c,c+1)=1

if f(c)=f(c+1)=s,then c|s(s+1)/2,and c+1|s(s+1)/2

thus,c(c+1)|s(s+1)/2

c(c+1)c

it contradicts with "Lemma 1:if m is an odd prime,f(m)