$\Bbb{H}_{\Bbb{Q}}$ is only four dimensional division algebra over rationals. How to prove that only four dimensional division algebra (noncommutative) over $\Bbb{Q}$ is rational quaternions?
After a bit of internet research, I am very sure about the above statement, if not please surprise me!
How to proceed at such statement? 
Just letting $D$ be a $4$ dimensional division algebra, is not taking me forward.
 A: It's not true. It would be true if $\mathbb Q$ were replaced by $\mathbb R$ in your statement, and the rational quaternions replaced with the real quaternions.   There is an infinite family of pairwise non-isomorphic quaternion algebras over $\mathbb Q$ which are simple central divison algebras of dimension $4$ over $\mathbb Q$, and of which the rational quaternions are one example. 
Moreover any field extension of $\mathbb Q$ of degree $4$ is a commutative division algebra over $\mathbb Q$ and there are infinitely many of those as well. 
A: I think I will fill out Bruno’s excellent answer with an example and a few facts.
For this discussion, “$\infty$” is a prime, and the completion of $\Bbb Q$ at infinity is $\Bbb R$. The completion of $\Bbb Q$ at an ordinary prime $p$ is the $p$-adic number field $\Bbb Q_p$. Any central division algebra over $\Bbb Q$ of rank $4$ must be “ramified” at evenly many primes. The rank-$4$ division algebra $\Bbb A$ is “unramified at $p$” if $\Bbb A\otimes_{\Bbb Q}\Bbb Q_p$ is isomorphic to the ring of $2$-by-$2$ matrices over $\Bbb Q_p$, and “ramified at $p$” if that tensor product is still a rank-four central division algebra over $\Bbb Q_p$.
If you haven’t seen tensor product $\otimes$ before, don’t panic. It just means using the same formulas for multiplication of the basis elements, but taking the scalars to be in $\Bbb Q_p$ instead of $\Bbb Q$. The regular quaternion division algebra $\Bbb H$ that you know is ramified only at $\infty$ and $2$. This means that if you use $\Bbb R$-coefficients, you still get a division algebra (as you know), but you also get a division algebra if you allow $\Bbb Q_2$-coefficients instead. For any other prime $p$, if you allow $\Bbb Q_p$-coefficients, the resulting noncommutative algebra will be isomorphic to the ring of $2$-by-$2$-matrices over $\Bbb Q_p$.
Oh yeah, you say, what about a rank-four division algebra ramified only at $3$ and $5$? I’ll give you the constants, but I’m afraid I’ll have to leave it up to you to check that my claims are all right. The basis is $\{1,r,s,t\}$, where $r$ behaves like a $\sqrt3$, $s$ like a $\sqrt5$, and $t$ behaves like a $\sqrt{-15}$. Here’s the rules for multiplying the units:
$r^2=3$, $s^2=5$, $t^2=-15$, $rs=t=-sr$, $rt=3s=-tr$, and $ts=15r=-st\,$.
The important thing is the “reduced norm form”, which is what you get when you start with $a+br+cs+dt$, which generates a quadratic subfield of the algebra, and look at its field-theoretic norm (the constant term in its characteristic polynomial). This is
$$
Q(a,b,c,d)=a^2-3b^2-5c^2+15d^2\,,
$$
and you can check, just by looking at integer values of $a,b,c,d$ modulo $9$, that there are no nontrivial rational zeros $(a,b,c,d)$ of $Q$. It is this fact that guarantees that the quaternion algebra defined by the the relations above among the basis elements is indeed a division algebra. I don’t have the courage to verify that this algebra is unramified (“splits”) at all other primes than $3$ and $5$, but I’m pretty sure that this is the case.
Oh, yes, and one other fact about central division algebras over $\Bbb Q$ that you would not have expected just from knowing about $\Bbb H$ as an $\Bbb R$-division algebra: for every $n>1$, there are central division algebras over $\Bbb Q$ of rank $n^2$. 
