'Neața NEAȚĂ !! again... Problem S:E24.314

           Proposed for 8th grade by the same author; in SGMB 11/2024.

     In translation:

          "If $x$ and $y$ are nonzero integers such that $13(x^2+y^2)=(2x+3y)^2$, show

          that the fraction $\frac{3x+11y}{x^4+y^4+15x^2y^2}$ is reducible."

 

ANSWER CiP

$x=2 \cdot k,\;y=3 \cdot k,\;\;k \in \mathbb{Z}^*$ and 

the fraction always simplifies (at least) to $13$

                    Solution CiP

          The given condition about the numbers $x$ and $y$ is successively equivalent to

$13 \cdot x^2+13 \cdot y^2=4 \cdot x^2+12 \cdot xy+9 \cdot y^2\;\;\Leftrightarrow\;\;9x^2-12xy+4y^2=0\;\;\Leftrightarrow$

$$\Leftrightarrow\;\;(3x-2y)^2=0\;\;\Leftrightarrow\;\;3 \cdot x=2 \cdot y\;.$$

We see from the last equation that $2 \mid 3\cdot x$, but(EUCLD's Lemma) $2$ and $3$ being relatively prime, the result is that $2 \mid x$, so $x=2\cdot k$  fore some  $0 \neq k\in \mathbb{Z}$ and then 

$$3\cdot x=2\cdot y \;\Rightarrow\;\;3 \cdot 2k=2 \cdot y\;\;\Rightarrow y=3k\;.$$

          In this case the given fraction is

$$\frac{3\cdot 2k+11 \cdot 3k}{16k^4+81k^4+15 \cdot 4k^2\cdot 9k^2}=\frac{39 \cdot k}{637 \cdot k^4}=\frac{13\cdot 3k}{13 \cdot 49k^4}$$

and we get the answer.

$\blacksquare$