Question

29 la puterea 2011 ca suma de 3 patrate perfecte

Answer 1

Explicație pas cu pas:

${29}^{2011} = 29 \cdot {29}^{2010} = (4 + 9 + 16) \cdot { ({29}^{1005}) }^{2} \\ = {2}^{2} \cdot { ({29}^{1005}) }^{2} + {3}^{2} \cdot { ({29}^{1005}) }^{2} + {4}^{2} \cdot { ({29}^{1005}) }^{2} \\ = { ({2 \cdot 29}^{1005}) }^{2} + { ({3 \cdot 29}^{1005}) }^{2} + { ({4 \cdot 29}^{1005}) }^{2}$