Question

Fie x, y, z, apartine lui N astfel incat x^2+y^2=z^2. Aratati ca 3 divide pe x ori y.

Answer 1

$\displaystyle\bf\\x^2+y^2=z^2,~x,~y,~z~\in\mathbb{N},~3~\bigg|xy~?\\3~\bigg|xy \Leftrightarrow 3~\bigg|x,~3~\bigg|y~sau~3~\bigg|x~si~y~simultan.\\consideram~egalitatea~modulo~3,~cu~observatia~ca~un~patrat~perfect~\\este~congruent~cu~0~sau~1~modulo~3,~deci~avem~2~cazuri.\\\\\boxed{\bf cazul~1~:~z^2 \equiv 0 (mod3)}~.\\\\z^2 \equiv 0 (mod3) \implies x^2+y^2 \equiv 0(mod3),~de~aici~scoatem~doua~cazuri.\\1.~daca~x^2\equiv2(mod3)~sau~y^2\equiv2(mod3)~este~imposibil.$

$\displaystyle\bf\\asadar~in~cazul~2~aveam~ca~cel~putin~unul~dintre~cele~doua~patrate~\\este~congruent~cu~0(mod3),~dar~evident~ca~amandoua~vor~fi~congruente~\\cu~0(mod3).\\de~aici~ne~rezulta~evident~ca~3\bigg|xy.\\\\\boxed{\bf cazul~2~:~z^2 \equiv 1(mod3)}~.\\\\z^2 \equiv 1(mod3) \implies x^2+y^2\equiv1(mod3),~dar~aici~este~evident~ca~unul\\dintre~patrate~va~fi~congruent~cu~0(mod3),~iar~celalalt~patrat~cu~1(mod3),\\de~unde~rezulta~ca~3\bigg|xy.\\(q.e.d)$