> 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