Exercițiul 568

E.568. a) Calculați 132+252+352.13^2+25^2+35^2.
b) Arătați că numărul 201920192019^{2019} se poate scrie ca sumă a trei pătrate perfecte.

Olimpiadă, etapa locală, Teleorman, 2019

Răspuns: a) 2019;2019; b) 20192019=(1320191009)2+(2520191009)2+(3520191009)2.2019^{2019}=(13 \cdot 2019^{1009})^2 + (25 \cdot 2019^{1009})^2+ (35 \cdot 2019^{1009})^2.

Soluție:

a) 132+252+352=13^2+25^2+35^2 =
=169+625+1225=2019.=169+625+1225 = 2019.

b) 20192019=201920192018=2019^{2019}=2019 \cdot 2019^{2018}=
(132+252+352)20192018=(13^2+25^2+35^2) \cdot 2019^{2018}=
=132(20191009)2+252(20191009)2+352(20191009)2==13^2 \cdot (2019^{1009})^2 + 25^2 \cdot (2019^{1009})^2 + 35^2 \cdot (2019^{1009})^2=
(1320191009)2+(2520191009)2+(3520191009)2.(13 \cdot 2019^{1009})^2 + (25 \cdot 2019^{1009})^2+ (35 \cdot 2019^{1009})^2.