"Let $a,\;b\in \mathbb{N}^*$ be such that the number $\frac{a+3}{b}+\frac{b+3}{a}$ is an integer.
If $(a,b)$ is the greatest common divisor of the numbers $a$ and $b$, then show that $(a,b) \leqslant \sqrt{3(a+b)}$."
From the MAGAZINE(aka REVISTE) Gazeta Matematica seria B no.9/2024, page 426, the proposed problem for the 7th grade
ANSWER CiP
A case where the equal sign occurs is
$$a=6,\;b=6.$$
Solution CiP
Let $k=\frac{a+3}{b}+\frac{b+3}{a}\in \mathbb{Z}$, actually $k\in\mathbb{N}^*$.
After calculations, we have the relationship
$$a^2+b^2-kab+3a+3b=0.\tag{1}$$
If $d=(a,b)$ then $a=d\cdot a_1,\;b=d \cdot b_1$ with $(a_1,b_1)=1.$Replacing these in (1) we have
$$d^2 \cdot a_1^2+d^2 \cdot b_1^2-k\cdot da_1\cdot db_1+3d\cdot a_1+3d\cdot b_1=0\;\Leftrightarrow$$
$$da_1^2+db_1^2-dka_1b_1+\underline{3(a_1+b_1)}=0.$$
In the last equation, the underlined term must be divisible by the number $d$, because all other terms are divisible by it. From here we get, with natural numbers
$$d \mid 3(a_1+b_1)\;\Rightarrow\;3(a_1+b_1)=d \cdot c \geqslant d\;\Rightarrow\;$$
$$\Rightarrow\;3\left ( \frac{a}{d}+\frac{b}{d} \right ) \geqslant d\;\Leftrightarrow \; 3(a+b)\geqslant d^2\;\;\Leftrightarrow\;(a,b)\leqslant \sqrt{3(a+b)}.$$
For $a=6$ and $b=6$ we have $\frac{a+3}{b}+\frac{b+3}{a}=\frac{9}{6}+\frac{9}{6}=3$, and
$$d=(6,6)=6=\sqrt{36}=\sqrt{3\cdot(6+6)}.$$
$\blacksquare$
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