Since ALINA wished me "Have a good trip ! ... Take care on the road !" as I wished (see the editorial of Friday, July 3rd, evening), I can also focus a little on the other beauty, mathematics.
From the Exercise Supplement today (page 8).
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loc imagine ; inserat LINK
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"S:E26.162. Let $a$ and $b $ be two nonzero natural numbers. Show that the
statements :
a) $a\;\; divides \;\;b^2+1\;\;\;and\;\;\;b\;\;divides\;\;a^2+1,$
b) $ab\;\;divides\;\;a^2+b^2+1.$
are equivalent.
{Authors : } Romanța GHIȚĂ and Ioan GHIȚĂ, Blaj"
ANSWER CiP
An example is $5\mid 13^2+1\;\;and\;\;13\mid 5^2+1\;\;\;\Leftrightarrow\;\;65\mid 5^2+13^2+1$
Solution CiP
$\underline{a)\;\Rightarrow\;b)}$
We have the hypothesis that
$a\mid b^2+1 \tag{1}$
and
$b\mid a^2+1 \tag{2}$
are both true.
If $\fbox{$a=b$}$ , then $a\mid a^2+1$ is only possible when $a=1$. Then, obviously
$ab=1\mid 1^2+1^2+1$ therefore b) takes place.
The case $\fbox{$a\neq b$}$ still remains to be discussed. First, under the given conditions it follows that the greatest common divisor of $a$ and $b$ is 1 :
$(a,b)=1 \tag{3}$
$\square$
$b)\;\Rightarrow\;a)$
$\square\;\;\square$
(in construction)