> 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 <n.
Then
    s = sum_{k in S} a_k  +  sum_{k in T} a_k
and
    phi(s) = sum_{k in S}phi(a_k)  +  sum_{k in T} phi(a_k),
i.e.
    s = sum_{k in S}-a_k  +  sum_{k in T}a_k.
Combining these equations, we obtain
    sum_{k in T} a_k = (s+s)/2 = s.
Again, this contradicts the minimality of n.
qed.

Hagen von Eitzen

-----------------------------------------------------
To show that a linear combination of four square roots (with the specified
conditions) is impossible, I will show that its existence would imply the
existence of a linear combination of between one and three (inclusive)
square roots (also with the specified conditions). Given impossibility
proofs for cases one through three, we can conclude that no such linear
combination of four square roots exists.

Suppose such a linear combination exists:

a sqrt(w) + b sqrt(x) + c sqrt(y) + d sqrt(z) = r   ...(1)
a sqrt(w) + b sqrt(x) + c sqrt(y) = r - d sqrt(z)

(squaring both sides)

a^2w + b^2x + c^2y + 2ab sqrt(wx) + 2bc sqrt(xy) + 2ac sqrt(wy) = r^2 -
d^2z - 2dr sqrt(z)

(rearranging)

2ab sqrt(wx) + 2bc sqrt(xy) + 2ac sqrt(wy) + 2dr sqrt(z) = r^2 - d^2z -
a^2w - b^2x - c^2y

Does this new linear combination meet the same requirements as the original?

- The coefficients are positive integers (yyes)
- The combination is equal to a positive innteger (yes; the left-hand side is
clearly positive and the right-hand side is clearly an integer)
- The square roots are all irrational (maybbe)

Case 1: Not all the square roots are irrational.

If the square root of a positive integer is not irrational, it must be
integral. Move all the integral terms to the right-hand side. This results
in a linear combination, under the necessary conditions, with between one
and three (inclusive) square roots, which is an impossibility. This
completes the proof for this case.

Case 2: All the square roots are irrational.

Now for the fun part. With some made-up notation, the previous lines can be
summarised as:

<w, x, y, z> = r   ==>   <wx, xy, wy, z> = s

Since the new combination meets the original requirements, the new
combination implies the existence of a third combination:

<wx, xy, z, wy> = s   ==>   <wx^2y, xyz, wxz, wy> = 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:

<wx^2y, wy, xyz, wxz> = u   ==>   <w^2x^2y^2, wxy^2z, wx^3yz, wxz> = 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