Except for the obvious cases n=0,1, are there any values of n such that ∑nk=1√k is an integer? How does one even approach such a problem? (This is not homework – just a problem I thought up.)

**Answer**

No, it is not an integer.

Let p1=2<p2<p3<⋯<pk be all the primes ≤n. It is known that K=Q(√p1,√p2,…,√pk) is a Galois extension of the rationals of degree 2k.

The Galois group G is an elementary abelian 2-group. An automorphism σ∈G is fully determined by a sequence of k signs si∈{+1,−1}, σ(√pi)=si√pi, i=1,2,…,k.

See this answer/question for a proof of the dimension of this field extension. There are then several ways of getting the Galois theoretic claims. For example we can view K as a compositum of linearly disjoint quadratic Galois extensions, or we can use the basis given there to verify that all the above maps σ are distinct automorphisms.

For the sum Sn=∑nℓ=1√ℓ∈K to be a rational number, it has to be fixed by all the automorphisms in G. This is one of the basic ideas of Galois correspondence. But clearly σ(Sn)<Sn for all the non-identity automorphisms σ∈G, so this is not the case.

**Attribution***Source : Link , Question Author : Mario Carneiro , Answer Author : Community*