Rational solutions $(a,b)$ to the equation $a\sqrt{2}+b\sqrt{3} = 2\sqrt{a} + 3\sqrt{b}$ Find all rational solutions $(a,b)$ to the equation
$$a\sqrt{2}+b\sqrt{3} = 2\sqrt{a} + 3\sqrt{b}.$$
I can see that we have the solutions $(0,0), (2,0), (0,3), (3,2), (2,3)$, and I suspect that there are no more.
I tried to do the thing where you square both sides, rearrange terms, square both sides again, but it got messy.
Edited: Also, I think I recall that all distinct square roots of square-free numbers are linearly independent over the rationals. Is this true? This might lead to a more direct way of proving it.
 A: Actually, not that messy. Squaring both sides, then isolating the only term that is irrational in the unknowns and squaring again gives
$$(2a^2+3b^2+2\sqrt6ab-4a-9b)^2=144ab$$
Since $\sqrt6$ is irrational, this can only be true if either $ab=0$ or
$$2a^2+3b^2-4a-9b=0$$
This is the equation of an ellipse centered at $(a=1,b=\frac32).$
Substituting this into the previous equality yields $24a^2b^2=144ab,$ or (still assuming $ab\neq0$) $ab=6.$
This hyperbola intersects the ellipse in only two points: the nonzero solutions we already had.
A: We can use the interesting
Proposition. The square roots of the square-free naturals are $\Bbb Q$-linearly independant.
Write $a=c^2n$, $b=d^2m$ with $c,d\in\Bbb Q$ and $n,m$ square-free (note that $a,b$ cannot be negative anyway). Then we have
$$ a\sqrt 2+b\sqrt 3-2c\sqrt n-3d\sqrt m=0.$$
By the proposition, the following conclusions can be made casewise:


*

*$a=b=0$.

*$a=0$, $b\ne 0$. Then $b\sqrt 3-3d\sqrt m=0$ implies $m=3$ and $3d=b$, so $b=3$

*$b=0$, $a\ne 0$. Then $a\sqrt 2-2c\sqrt n=0$ implies $n=2$ and $2c=a$, so $a=2$.

*$a\ne 0,b\ne 0$. Then either $n=2$, $m=3$, $2c=a$, $3d=b$; this means $a=2$, $b=3$. Or $n=3$, $m=2$, $2c=b$, $3d=a$; this means $3b^2=12c^2=4a$ and $2a^2=18d^2=9b$, so $72b=16a^2=9b^4$, $b^3=8$, $b=2$ and $a=3$.

A: Your given five solutions can be calculated at the sight. Searching about others, one has $$2a^2+3b^2+2ab\sqrt 6=4a+9b+12\sqrt{ab}$$ in which trying to separate rational and irrational we have
$$2a^2+3b^2=4a+9b\iff 2ab\sqrt 6=12\sqrt {ab}\iff ab=6$$ this curves, ellipse and hyperbola respectively, have no rational common points.Therefore I agree with you there are just five solutions (very well stablished by Hagen Von Eitzen above).

