Determine the nonzero natural numbers $a$ and $b$ for which the number $n=\sqrt{a^{2}+6b}+\sqrt{b^{2} +6a}$ is rational.

Question

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.

Answer 1

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).

Answer 2

Edit:
Proof that both $a$ and $b$ are even. Suppose both are odd, then $a^2+6b \equiv 3 \pmod 4$, but all squares are either $0$ or $1$ $\pmod 4$.
Suppose one is odd, and the other is even. Without loss of generality, $a$ is even, and $b$ is odd. Then, $a^2+6b \equiv 2 \pmod 4$, but all squares are either $0$ or $1$ $\pmod 4$.

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We want to find all $a,b \in \Bbb N$ such that $a^2+6b$ and $b^2+6a$ are perfect squares. Using modular arithmetic, it is not hard to deduce that both $a,b$ are even. Let $2x = a, 2y = b$. Then, $x^2+3y$ and $y^2+3x$ are perfect squares.

Suppose $x \ge y$. Since $x^2+3y$ is a perfect square greater than $x^2$, it is at least $(x+1)^2$.

So $(x+1)^2\le x^2+3y < x^2+4x+4 = (x+2)^2$. So, $3y = 2x+1$. Putting this in $y^2+3x = k^2$, we get $2y^2+9y - 3 - 2k^2 = 0$, hence $$y = \frac{-9+\sqrt{105+16k^2}}4$$ which has a discriminant $105+16k^2 = 105+(4k)^2$. Since this has to be a perfect square, write $105 = (m-4k)(m+4k)$. Both $k,m$ are positive integers, so there are only four cases to check: 1 x 105, 3 x 35, 5 x 21, 7 x 15 (Moreover, $m \equiv 1 \pmod 4$, so two cases can be eliminated directly). You get $(x,y) = (1,1)$ or $(16,11)$. Removing condition $x \ge y$, another solution $(11,16)$ is obtained. So, $$\boxed{(a,b) \in \{(2,2),(32,22),(22,32)\}}$$