> Johan de Ruiter: "Last night I was wondering whether any integer can
be written as a linear combination of a finite number of noninteger
squareroots of integers
> where all coefficients are integers. Maybe it's trivial, but I wasn't
able to find a solution yet." I wasn't able to find a trivial proof
either, beyond proofs for 2 or 3
> square roots. Is there a clever impossibility proof?
There is no such thing, apart from trivial things like
0 = sqrt(2)+sqrt(8)-sqrt(18).
It is trivial--with sufficient theory.
Let n>0 and let a_k, k=1,...,n be algebraic numbers with (a_k)^2 in the
field Q of rationals
but none of the a_k is rational and assume that
s:=sum a_k is rational.
Moreover, assume that a similar setup is not possible with a smaller
value of n.
Trivially, n>1 (we cannot have rational s=a_1 irrational).
Let b:=a_2/a_1.
Now, b might be rational. Then we could replace two summands
a_1+a_2
by one
a'_1 := (b+1)*a_1
where a'_1 is irrational but (a'_1)^2 = (b+1)^2*a_1^2 is
rational--contradicting the minimality of n.
Hence, b is irrational.
But clearly b^2 = (a_2)^2/(a_1)^2 is rational.
Let F=Q(a_1,a_2,b) = Q(a_1,a_2) be the smallest subfield of C containing
both a_1 and a_2 (and hence b).
Then F is Galois over Q.
The mapping b -> -b can be extended to an automorphism phi of F.
(short sentences with all theory hidden in them)
Since phi(a_1)^2=phi(a_1^2), either phi(a_1)=a_1 or phi(a_1)=-a_1.
Similarly, phi(a_2)=a_2 or phi(a_2)=-a_2.
But since phi(a_2/a_1) = -a_2/a_1, we cannot have the same sign in both
cases.
Without loss of generality,
phi(a_1) = -a_1,
phi(a_2) = +a_2.
Now, E:=Q(a_1,...,a_n) is also Galois over F (and Q).
Hence phi can be extended to an automorphism psi of E.
For each k, either psi(a_k)=a_k or psi(a_k)=-a_k (since we must have
psi(a_k)^2=a_k^2).
Let S be the subset of {1,...,n} such that psi(a_k)=-a_k and T the
complement.
By assumption, 1 is in S and 2 is in T, i.e. #T is >0 and = r ==> = s
Since the new combination meets the original requirements, the new
combination implies the existence of a third combination:
= s ==> = t
Are any of the square roots here integral? If so, this implies the same
impossibility as before. If not, this implies the existence of a fourth
combination:
= u ==> = v
Since w^2x^2y^2 is a perfect square, at least one integral square root must
appear in the expression. Therefore, this *must* simplify to a case with
between one and three (inclusive) square roots. This completes the proof.
J K McLean