Question
Determine the nonzero natural numbers $a$ and $b$ for which the number $n=\sqrt{a^{2}+6b}+\sqrt{b^{2} +6a}$ is rational.
My idea
First of all because $a$ and $b$ are natural this means that also $a^{2}+6b$ and $b^{2} +6a$.
We will note $a^{2}+6b=x$ and $b^{2} +6a=y$ and we know that $x, y$ are natural.
So this means we have $\sqrt{x}+\sqrt{y}=N$, where $N$ is a rational number.
$\sqrt{x}=N-\sqrt{y}$
$x=(N-\sqrt y)^2=N^2-2N\sqrt{y}+y$
$\sqrt{y}=\frac{N^2+b-a}{2N}$ which is rational, so this means that $y$ has the form $k^2$.
Analogously, $x$ also has the form $k^2$.
From here I dont know what to do… I thought of doing modular arithmetic, but I don't know where to start.
Best Answer
EDIT: I'm using a lemma: the sum of two squareroots of natural numbers can be rational only when they both are in fact natural numbers (see here)
Let $a \geq b$ (due to symmetry we can assume this, then all solutions will be permutations of the ones we find).
Denote $q = \frac{b}{a}$. Now $0<q\leq 1$.
Let's complete a square under the first squareroot:
$$ a^2 + 6b = a^2 + 2\cdot 3qa + (3q)^2 - (3q)^2 = (a+3q)^2 - 9q^2 $$
Since this must be a square and bounded below by $a^2+1$ (since $b>0$) and from above (stricly since $q>0$) by $(a+3)^2$, it can only be either $(a+1)^2$ or $(a+2)^2$. Let's study these cases.
Case $(a+1)^2$
$$a^2+6b = (a+1)^2 = a^2+2a+1$$ So $a = 3b - \frac{1}{2}$ which is impossible as this isn't a natural number.
Case $(a+2)^2$
$$a^2+6b = (a+2)^2 = a^2+4a+4$$ So $a = \frac{3}{2}b - 1$. Let's plug this into the second square root and see what we get.
$$b^2 + 6a = b^2 + 6\left(\frac{3b}{2} - 1\right) = b^2 + 9b - 6 $$
Now the square must be $(b+m)^2$ for some $m$. What can $m$ be?
If $m>4$, we have $ (b+m)^2 - (b^2 + 9b - 6) = (2m-9)b + m^2 + 6 > 0$, so $m>4$ won't work.
The case $m=4$ works and gives $b=22$.
How about $m<4$. Well, there we get the solution $b=2$.
I was a bit hasty here in the case $m<4$, since the negative values could also give solutions. But we can write $(b+m)^2 = b^2 + 9b - 6$ and get
$$b = \frac{m^2+6}{9-2m} = -\frac{m}{2} + \frac{105-9(9-2m)}{4(9-2m)}$$
so $m$ must be even and $9-2m$ a divisor of $105$. So it actually goes like in the other answer and I think that is simpler.
Answer: $a=b=2$ or $a=32, b = 22$ (or other way around).