'Neața NEAȚĂ ! (meaning "Good morning...")

          Ion NEAȚĂ from Slatina is also the author of Problem S:E24.271, published in the same issue of GMB, but this time in the Exercise Supplement (on page 7, proposed for 8th grade).

In translation:

  "S:E24.271  Knowing that $x,\;y\;$ and $A=\sqrt{4x^2+2y+6}+\sqrt{4y^2+2x+11}$ 

      are natural numbers, show that A is prime."

Answer CiP

$$x=1,\;\;y=3,\;\;\;A=11$$

Solution CiP

(obtained from a collaborator on AOPS)

          We first show the following:

              Lemma If  $a\;$ and $\;b\;$ are natural numbers s.t. $\sqrt{a}+\sqrt{b} \in \mathbb{N}\;$

                            then $a$ and $b$ are perfect squares.

               Proof (CiP) It is known that "the square root of $X$ is rational if and only if X is a rational number that can be represented as a ratio of two perfect squares" (see Wikipedia).

Now, if $\sqrt{a}+\sqrt{b}=m\in \mathbb{N}$ then $\sqrt{b}=m-\sqrt{a}$ and squaring it gives

$$\sqrt{a}=\frac{m^2+a-b}{2m}\in \mathbb{Q}$$

(the excluded case $m=0$ is trivialy). Hence $a$ is perfect square; but then $\sqrt{b}=m-\sqrt{a} \in \mathbb{Q}$ and so $b$ is a perfect square too.

$\square$ Lemma

          According to the Lemma, $A=\sqrt{4x^2+2y+6}+\sqrt{4y^2+2x+11} \in \mathbb{N}\;\Rightarrow$

$4x^2+2y+6=$ perfect square

$4y^2+2x+11=$ perfect square

so we have the inequalities (the first perfect square being an even number)

$$4x^2+2y+6 \geqslant (2x+2)^2,\;\;\;4y^2+2x+11 \geqslant (2y+1)^2$$

Adding the two inequalities

$$(4x^2+2y+6)+(4y^2+2x+11)\geqslant (4x^2+8x+4)+(4y^2+4y+1)\;\;\Leftrightarrow$$

$$\Leftrightarrow\;12\geqslant 6x+2y\;\;\;\Rightarrow \;\;\;\;y\leqslant 3(2-x).$$

Obvious $x \ngtr 2$. If $x=2\;\Rightarrow\;y=0$, but $4x^2+2y+6=22\neq$perfect square.

If $x=1\;\Rightarrow\;y\leqslant 3\;$ and $\;4x^2+2y+6=2y+10\leqslant 16$ it is a perfect square only for $y=3$.

If $x=0\;\Rightarrow\;y\leqslant 6\;$and $\;4x^2+2y+6=2y+6\leqslant 18$; it is possible only $y=5$ but then $4y^2+2x+11=100+11=11\neq$perfect square.

We got the answer.

$\blacksquare$