Q13/010) Let p be a prime > 2, then 1+(1/2)+(1/3)+...+(1/p-1)=j/k, j,k>0 and integers. Show that j is divisible by p

Proof:

Because p is prime so for each m (not -1, not p-1 for these these are their own reciprocals ) there is a reciprocal q such that

mq = 1 mod p or mq = a(m)p + 1 ( we use a(m) for coefficient of m so on and a(m) = a(q))

so we have m = (a(q)p + 1)/q = a(q) p/q + 1/q

or 1/q = m – a(q) p/q

1/m = q – a(m) p/m

 

For each different m there is a different reciprocal.

 

This is for all except 1 and p-1

1/1 = 1

1/(p-1) = 1

 

Adding all the expressions on the left and right we get

1+(1/2)+(1/3)+...+(1/p-1)= ( 1+ 2 + …. + p-1)) + (sum (a(i)/i) p

= p(p-1)/2 + + (sum (a(i)/i) p

Which on adding has p in numerator and not on denominator