S:E18.128 (Vasile Ienuțaș)
Scrieți numărul 2018 2017 {\displaystyle 2018^{2017}} ca sumă de patru pătrate perfecte nenule distincte.
Soluție:
Din 2018 = 3 2 + 4 2 + 12 2 + 43 2 {\displaystyle 2018=3^{2}+4^{2}+12^{2}+43^{2}} se obține
2018 2017 = 2018 ⋅ 2018 2016 = ( 3 2 + 4 2 + 12 2 + 43 2 ) ⋅ ( 2018 1008 ) 2 {\displaystyle 2018^{2017}=2018\cdot 2018^{2016}=\left(3^{2}+4^{2}+12^{2}+43^{2}\right)\cdot \left(2018^{1008}\right)^{2}}
Deci
2018 2017 = ( 3 ⋅ 2018 1008 ) 2 + ( 4 ⋅ 2018 1008 ) 2 + ( 12 ⋅ 2018 1008 ) 2 + ( 43 ⋅ 2018 1008 ) 2 {\displaystyle 2018^{2017}=\left(3\cdot 2018^{1008}\right)^{2}+\left(4\cdot 2018^{1008}\right)^{2}+\left(12\cdot 2018^{1008}\right)^{2}+\left(43\cdot 2018^{1008}\right)^{2}}