A Divisibility Problem That Catches Us in the Morning

                It is Problem S:E24.359 from SGMB magazine (12/2024 at page 8), proposed for 8th grade, by a well-known author.

In translation:

              "Let $x$ and $y$ be nonzero natural numbers for which $13\mid (3x-2y)$. Show 

                 that the fraction

$\frac{31^{8(x+y)+1}}{x^2+y^2}$ is reducible."

This problem led me to make the Post here. 

ANSWER CiP

The given fraction can always be simplified by 13 

               Solution CiP

               Lemma 1  $13\mid (3x-2y) \Rightarrow 13\mid (5x+y)$ and $13\mid (x-5y).$

          Proof of  L1  We observe relationships:

$$2\cdot (5x+y)=13\cdot x-(3x-2y)\;\;;\;\;3\cdot (x-5y)=(3x-2y)-13 \cdot y.$$

From the hypothesis $13\mid (3x-2y)$ it then follows that

$$13 \mid 2\cdot (5x+y)\; and\; 13\mid 3\cdot (x-5y)$$

and, because $(13\;,2)=1=(13\;,3)$, it must $13\mid (5x+y) \;and\; 13 \mid (x-5y)$.

qed_L1 $\square$

          Remark CiP I found the property in the mentioned post. <end Rem>

               Lemma 2  $13\mid (3x-2y) \Rightarrow 13\mid (x^2+y^2).$

          Proof of L2  We have the equation, easy to verify by calculations

$$(5x+y)^2+(x-5y)^2=26 \cdot (x^2+y^2).\tag{1}$$

From the hypothesis $13\mid (3x-2y)$ we have, applying Lemma 1 :

 $13^2 \mid (5x+y)^2\;and\;13^2 \mid (x-5y)^2\;\;\overset{(1)}{\Rightarrow} 13^2\mid 26\cdot (x^2+y^2)\Rightarrow 13\mid 2\cdot(x^2+y^2)$

and, because $(13\;,2)=1$ it must
$$13\mid (x^2+y^2). \tag{2}$$

qed_L2 $\square$

          In the following equations the number $k$ has different values, which we are not interested in:

$31=13\cdot k+5\Rightarrow 31^2=31\cdot(13k+5)=13k+155=13k+12\cdot 13-1=13k-1\Rightarrow$

$\Rightarrow 31^3=31\cdot (13k-1)=13k-26-5=13k-5\Rightarrow 31^4=31\cdot(3k-5)=13k-12\cdot 13+1=13k+1.$

For $m=2(x+y)$ we obtain 

$31^{8(x+y)+1}-18=31^{4\cdot m+1}-18=31^{4\cdot m}\cdot 31-18=(31^4)^m\cdot 31-18=$

$=(13k+1)^m\cdot 31-18=(13k+1) \cdot 31-18=13k+31-18=13k+13=13\cdot k.$

So the numerator of the fraction in the statement is divisible by $13$. This together with (2) proves the answer.

$\blacksquare$