A nice DIVISIBILITY problem //

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)