Is √1+√2+⋯+√n\sqrt1+\sqrt2+\dots+\sqrt n ever an integer?

Related: Can a sum of square roots be an integer?

Except for the obvious cases n=0,1, are there any values of n such that nk=1k 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 σG is fully determined by a sequence of k signs si{+1,1}, σ(pi)=sipi, 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=n=1K 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 σG, so this is not the case.

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

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