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A trivial but efficient numerical proof without assuming that friends of my friends are my friends:

Each pair know a third fellow. There are $$\binom 5 2 = 10$$  pairs $$\equiv 20$$  arrows starting from each to another element ( no self arrow allowed ).

The mean of received arrows is $$2 \frac{\binom 5 2}5 = 2 \frac{10}5 = 4$$

Receiving 4 arrows means the receiver knows the other $$4 \equiv$$  everybody.

It is a mean. If you decrease the arrows received by someone , you must increase the arrows received by another one.

Hence, there is always someone receiving at least 4 arrows. Then there is always someone which is the friend of "everybody".

question 1 Juillet 15
editée 7 Octobre 16
formule dans les commentaires ? $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$
ok
mais il faut la préparer et l'insérer ici entre 'antislash( '  et 'antislash) '  en remplaçant  l'anti-slash par un toucher de clavier standard sur  Alt Gr 8
Alors qu'elle est en wysiwyg dans les questions réponses