Question

2015 la puterea 2015 sa fie scris ca suma a 6 p.p. (pătrate perfecte )

Answer 1

[tex]2015^{2015}=2015\cdot (2015^{1007})^2=\\ =(1^2+2^2+6^2+22^2+23^2+31^2)\cdot (2015^{1007})^2=\\ =(2015^{1007})^2+(2\cdot2015^{1007})^2+(6\cdot 2015^{1007})^2+...+(31\cdot 2015^{1007})^2[/tex]