Question #71d19

1 Answer
Feb 20, 2017

See below.

Explanation:

Supposing that #p > 3# with p prime, then

#1+1/2+1/3+...+1/(p-1)# has an even number of terms so we can establish an equivalent sum of terms

#1/k + 1/(p-k) = p/((p-k)k)#

then

#1+1/2+1/3+...+1/(p-1)=sum_(k=1)^((p-1)/2) p/((p-k)k)# so finally

#sum_(k=1)^(p-1)1/k equiv 0 mod p#

Now considering #sum_(k=1)^((p-1)/2) 1/((p-k)k)# we have

#(p-k)k equiv -k^2 mod p#.

After reducing the sum to an unique fraction, in the numerator will appear #(p-k)k, (k=1, cdots, (p-1)/2)# as factor into the #(p-1)/2# terms,so,

#sum_(k=1)^((p-1)/2) 1/((p-k)k)equiv-sum_(k=1)^((p-1)/2)k^2 mod p#

but #sum_(k=1)^((p-1)/2)k^2=((p-1)/2((p-1)/2+1)(2((p-1)/2)+1))/6 equiv 0 mod p#

So finally

#sum_(k=1)^(p-1)1/k equiv 0 mod p^2#