Find $m,\alpha>0,\alpha\in\mathbb Z[\sqrt{m}]$ s.t. $\forall p,q,r,n\in\mathbb Z[\sqrt{m}]$, $p^2+q^2+r^2\neq\alpha n^2$ Let us assume that $\alpha, p,q,r,n\in\mathbb Z[\sqrt{m}]$ and $m\in\mathbb{N}$ is a square free integer.
is it possible to choose $\alpha,m$  such that the following equation is $never$ satisfied?
$$p^2+q^2+r^2=\alpha n^2$$
Note that $\alpha>0$ , $n\neq0$. 
 A: First of all, division by $n^2$ shows the question is equivalent to asking if there is any $\alpha > 0$ in ${\mathbf Z}[\sqrt{m}]$ such that $\alpha$ is not a sum of three squares in the field ${\mathbf Q}(\sqrt{m})$. (Any finite set of numbers in ${\mathbf Q}(\sqrt{m})$ can be written as ratios from ${\mathbf Z}[\sqrt{m}]$ with a common denominator in ${\mathbf Z}[\sqrt{m}]$.)  So you seek $\alpha > 0$ in ${\mathbf Z}[\sqrt{m}]$ such that you can't write $\alpha = x^2 + y^2 + z^2$ for any $x, y, z \in {\mathbf Q}(\sqrt{m})$. 
There is a trivial solution: use $\alpha = \sqrt{m}$. 
Indeed, writing $\sigma$ for the conjugation on ${\mathbf Q}(\sqrt{m})$, so $\sigma(a + b\sqrt{m}) = a - b\sqrt{m}$ ($a, b \in \mathbf Q$), if $\alpha = x^2 + y^2 + z^2$ then $\sigma(\alpha) = (\sigma(x))^2 + (\sigma(y))^2 + (\sigma(z))^2$. Thus not only must $\alpha$ be positive in order to be a sum of three squares, but necessarily also $\sigma(\alpha) > 0$ since it too is a sum of three squares if $\alpha$ is. 
But $\sqrt{m} > 0$ while $\sigma(\sqrt{m}) = -\sqrt{m} < 0$.  
More generally, any element of ${\mathbf Z}[\sqrt{m}]$ that is not totally positive (meaning both it and its conjugate are positive) can't be written as a sum of three squares, or any finite number of squares, in ${\mathbf Q}(\sqrt{m})$.
While this is where an answer to the exact question that was asked ends, it suggests the question was not properly posed since an example is so easy to write.
Suppose we assume that $\alpha$ is totally positive (that is, $\alpha > 0$ and $\sigma(\alpha) > 0$). Then the easy examples don't occur for such $\alpha$. Are there any examples where $\alpha$ is totally positive? Yes, but justifying them as examples will be harder. I'll state an example right now: $7 + \sqrt{17}$. It is totally positive but it is not a sum of three squares in ${\mathbf Q}(\sqrt{17})$. To prove that is essentially the content of the rest of this answer.
The Hasse-Minkowski theorem says that a set of necessary and sufficient conditions for an element $\alpha$ of a quadratic field $K$ (or, more generally, any number field) to be a sum of three squares in $K$ is that it can be written as a sum of three squares in every completion $K_v$ of $K$. We don't even need to suppose $\alpha \in {\mathbf Z}[\sqrt{m}]$; the result is true for every $\alpha \in K$. 
To figure out what this means for the problem at hand we need to look at the different completions of $K$. They can be divided into a few types, and in some of them "easily" every element is a sum of three squares and in others this is not assured. That is where obstructions to representability will come from.
(1) $K_v = {\mathbf C}$. Every element in ${\mathbf C}$ is already a square in ${\mathbf C}$, so in these completions there is no obstruction to writing $\alpha$ as a sum of three squares. (In fact, though, this case is vacuous for the field of interest since ${\mathbf Q}(\sqrt{m})$ for $m > 0$ has no complex completions.) 
(2) $K_v = {\mathbf R}$. The nonzero sums of three squares in ${\mathbf R}$ are the positive elements, and the field $K$ has two completions isomorphic to ${\mathbf R}$, corresponding to the two embeddings of $K = {\mathbf Q}(\sqrt{m})$ into $\mathbf R$, which are the identity and conjugation. To say $\alpha$ is positive in both real completions of $K$ means $\alpha$ is totally positive, a condition we met already.
(3) $K_v$ is a $p$-adic completion for an odd prime $p$ (that is, $K_v = K_{\mathfrak p}$ for a prime ideal ${\mathfrak p}$ of the ring of integers of $K$ with odd residue field characteristic $p$).  Here all elements are sums of three squares because in fact all elements are sums of two squares (ultimately from the fact that in a finite field of odd characteristic every element is a sum of two squares, and then use Hensel's lemma). So as with the complex completions of $K$ we meet no obstructions here.
(4) $K_v$ is a $2$-adic completion, meaning $K_v = K_{\mathfrak p}$ where ${\mathfrak p}$ is a prime ideal in the integers of $K$ that has residue field characteristic $2$.  Here there can be an obstruction: not all elements of a $2$-adic field are necessarily sums of three squares. Off the top of my head I know the constraint for ${\mathbf Q}_2$, so I'll record it in an example.
Example: In ${\mathbf Q}_2$, if an element $\alpha$ of ${\mathbf Z}_2$ is a sum of three squares $x^2 + y^2 + z^2$ with  $x, y, z \in {\mathbf Q}_2$ then $x$, $y$, and $z$ must be in ${\mathbf Z}_2$. From this one can show with some work that an element of ${\mathbf Z}_2$ is a sum of three squares in ${\mathbf Q}_2$ iff it is not of the form $4^m(7 + 8\beta)$ where $m \geq 0$ and $\beta \in {\mathbf Z}_2$. This, by the way, is the source of the classical obstruction to an integer being a sum of three squares in ${\mathbf Z}$, since ${\mathbf Q}$ has one completion being ${\mathbf Q}_2$.
Putting this all together, we meet obstructions from the Hasse-Minkowski theorem only in cases (2) and (4), so we can say: an element $\alpha$ in a real  quadratic field $K$ is a sum of three squares in $K$ if and only if (i) it is totally positive and (ii) it is a sum of three squares in each $2$-adic completion of $K$.
Each real quadratic field has at most two $2$-adic completions. If they are both ${\mathbf Q}_2$ then we can apply the above example to provide instances where a totally positive element of $K$ is not a sum of three squares in $K$ because of a $2$-adic obstruction rather than a real obstruction (real in the sense of real numbers, not in the sense of genuine).
Example: Let $K = {\mathbf Q}(\sqrt{17})$. The prime $2$ splits completely in $K$ and both $2$-adic completions of $K$ are isomorphic to ${\mathbf Q}_2$; these completions correspond to the two different embeddings of $K$ into ${\mathbf Q}_2$, as $17$ is a square in ${\mathbf Q}_2$ (it is $1 \bmod 8$). We seek an $\alpha$ in ${\mathbf Z}[\sqrt{17}]$ that is totally positive but in one of the $2$-adic completions of $K$ we have $\alpha = 4^m(7 + 8\beta)$. For $\alpha$ I  tried the numbers $a + \sqrt{17}$ where $a \geq 5$ to make $\alpha$ totally positive and I found an example when $a = 7$: $7 + \sqrt{17}$ is totally positive but it is not a sum of three squares in $\mathbf Q(\sqrt{17})$ because it is not a sum of three squares in one of the $2$-adic completions of $\mathbf Q(\sqrt{17})$.
Here are the details.  Using PARI, the two square roots of $17$ in ${\mathbf Q}_2$ start out as 
$$
1 + 2 + 2^2 + 2^4 + 2^8 + 2^{11} + \cdots, \ \ \ 1 + 2^3 + 2^5 + 2^6 + 2^7 + 2^9 + \cdots.
$$
Call the first square root $u$, so the second one is $-u$ (they are negatives of each other).  It turns out that $5 \pm u$, $6 \pm u$, and $7 + u$ don't have the form $4^m(7 + 8\beta)$, but $7 - u$ does have this form:
$$
7 - u = 2^4 + 2^5 + 2^6 + 2^7 + 2^9 + \cdots = 16(1 + 2 + 2^2 + 2^3 + 2^5 + \cdots) = 16(7 + 8\beta)
$$
where $\beta = 1 + 2^2 + \cdots$. Therefore $7+\sqrt{17}$ is not a sum of three squares in ${\mathbf Q}_2$, because one of its embeddings into that field has the form $4^m(7+8\beta)$. So $7+\sqrt{17}$ is not a sum of three squares in ${\mathbf Q}(\sqrt{17})$, and that proves we can't solve $p^2 + q^2 + r^2 = (7+\sqrt{17})n^2$ for $p, q, r, n \in {\mathbf Z}[\sqrt{17}]$ with $n \not= 0$. 
A: This is the Hasse part of Hasse-Minkowski, and I've never studied it. Instead, I will just point out how to get anisotropic forms in four variables in the ordinary integers. 
Consider $$ f(w,x,y,z) =  w^2 -2 x^2 + 3 y^2 - 6 z^2.$$ We see that
$$ f(\sqrt 2 , 1,0,0) = 0.  $$ However, it is not possible to represent $0$ in integers not all $0.$ The very simple fact underlying that statement is this: IF
$$   w^2 -2 x^2 + 3 y^2 - 6 z^2 \equiv 0 \pmod 9,$$
THEN $w,x,y,z$ are all divisible by $3.$
LEMMA: if $u^2 - 2 v^2 \equiv 0 \pmod 3,$ then $u,v$ are divisible by $3$ and
$u^2 - 2 v^2 \equiv 0 \pmod 9.$ PROOF: assume $v$ not divisible by $3.$ Then it has a multiplicative inverse in the field $\mathbb Z / 3 \mathbb Z.$ So we have
$$u^2 - 2 v^2 \equiv 0 \pmod 3,$$
$$u^2 \equiv 2 v^2  \pmod 3,$$
$$ \frac{u^2}{v^2} \equiv 2   \pmod 3,$$
$$ \left( \frac{u}{v} \right)^2 \equiv 2   \pmod 3.$$ This is a contradiction, as $2$ is not a quadratic residue $\pmod 3.$ So, the assumption was wrong, actually $v$ is divisible by $3,$ there $u^2$ and $u$ itself are divisible by $3$ as well.  $$ \bigcirc  \bigcirc  \bigcirc  \bigcirc   $$
Return to:
THEOREM: IF
$$   w^2 -2 x^2 + 3 y^2 - 6 z^2 \equiv 0 \pmod 9,$$
THEN $w,x,y,z$ are all divisible by $3.$
We find, first, that $w^2 - 2 x^2$ must be divisible by $3.$ So, by the Lemma, both $w,x$ are divisible by$3$ and $w^2 - 2 x^2$ is divisible by $9.$ Therefore
$3 y^2 - 6 z^2$ is divisible by $9$ and $y^2 - 2 z^2$ is divisible by $3.$ Repeating the Lemma, we find $y,z$ are also divisible by $3.$ $$ \bigcirc  \bigcirc  \bigcirc  \bigcirc   $$
So, this is the special form that Fermat's "infinite descent" takes in the special case of quadratic forms. IF THERE WERE a nontrivial solution in integers to $f(w,x,y,z) = 0,$ we could find $g = \gcd(w,x,y,z)$ and confirm
$$ f(w/g, x/g, y/g, z/g ) = f(w,x,y,z) / g^2 = 0.  $$ We would also have
$$  \gcd(w/g, x/g, y/g, z/g) =1. $$
However, earlier, we proved that, for any solution, the  four numbers are divisible by $3$ and the gcd cannot be $1.$  $$ \bigcirc  \bigcirc  \bigcirc  \bigcirc   $$
For anyone interested in this topic, highly recommend CASSELS. The generalization to number field coefficients is lots of work, to put it mildly.   
