Is there an easy way to prove that $3a^2+2=b^2$ does not have any rational solutions? I am a physicist who needs to prove (for his research) that there are no $a,b\in\mathbb{Q}$ such that $3a^2+2=b^2$. Is there an easy way to do this?
 A: If there is a rational solution $(a,b)$, it can be expressed as $a=\frac{A}{z}$, $b=\frac{B}{z}$ where $A$ and $B$ are integers, and $z$ is a non-zero integer. Clearing denominators, we get 
$$3A^2+2z^2=B^2.$$
If the above equation has solutions with $B\ne 0$, there is a triple of integers, not all $0$, such that $B$ is positive, $3A^2+2z^2=B^2$, and $B$ is the smallest positive integer for which the equation $3x^2+2y^2=B^2$ has a solution. 
Note that $B$ must be divisible by $3$. For if it is not, we have $2z^2\equiv 1\pmod{3}$, which is impossible. 
Let $B=3B_1$. Since $3$ divides $3A^2$, it follows that $3$ divides $2z^2$, and therefore $3$ divides $z$. So $z=3y$ for some integer $y$. 
Then $9$ divides $3A^2$, so $3$ divides $A$, say $A=3x$. Substituting and dividing through by $9$, we get
$$3x^2+2y^2=B_1^2.$$
This contradicts the choice of $B$ as the smallest positive integer for which the equation $3x^2+2y^2=B^2$ has a solution. 
Remark: The same argument can be rewritten in various ways. One can make it look like one of the standard proofs of the irrationality of $\sqrt{2}$, by assuming that $a$ can be expressed as $\frac{A}{z}$, $b$ as $\frac{B}{z}$, where $z\gt 0$ is as small as possible. 
Or else we can rewrite the proof as a Fermat Infinite Descent argument. We showed that if there is a solution $A,z,B$ with $B\ne 0$, then there is a solution $A_1,z_1,B_1$ with $B_1\lt B$. Continue. We obtain an infinite descending chain $B\gt B_1\gt B_2\gt \cdots$, which is impossible.  
A: Hint $ $ Suppose $\,3a^2\!+2 = b^2$ for reduced $a,b\in\Bbb Q.\,$ If $\,3\,$ divides the denominator of $a$ or $b$ then it divides both, contra $3$ occurs to even power in denom of $\,b^2$ vs. odd power in denom of $\,3a^2\!+2.\,$ Thus both denoms are coprime to $\,3\,$ so we can reduce mod $3,\,$ yielding $\,2\equiv b^2,\,$ contradiction.
A: Another way around:
As André Nicolas showed, the original equation is equivalent with $3A^2+2z^2=B^2,$ to be solved in integers.
Rewrite it as $A^2+z^2=\frac{B^2-A^2}{2}.$ By one of Fermat's theorem, if it is solvable, then $\frac{B^2-A^2}{2}\equiv1\pmod4,$ hence $B^2-A^2\equiv2\pmod8.$
However, it is easy to verify that the square of an integer is congruent modulo $8,$ either to $0,$ to $1,$ or to $4.$ Thus the above congruence equation is not solvable in integers. $\square$
Hope this helps.
P.S. I think this is also "Fermatian," though not by way of descent.
