Question

29²⁰⁰¹ se poate scie ca o suma de trei patrate perfecteDAU COROANA​

Answer 1

Explicație pas cu pas:

$\bf29 ^{2001} =29 ^{2000+1} = 29^{2000} \cdot29 =$

$\bf 29^{2000} \cdot \big(16 + 4 + 9 \big) =$

$\bf 29^{2000} \cdot \big( 4^{2} + 2^{2} + 3^{2} \big) =$

$\bf \big(29^{1000} \big)^{2} \cdot \big( 4^{2} + 2^{2} + 3^{2} \big) =$

$\red{ \boxed{\bf \: \big(29^{1000} \cdot4 \big)^{2} + \big(29^{1000} \cdot2\big)^{2} + \big(29^{1000} \cdot3\big)^{2} \: }}$

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Answer 2

Răspuns:

29^2001 = (29^1000 x 2)^2  +  (29^1000 x 3)^2 + (29^1000 x 4)^2

Explicație pas cu pas:

29^2001 =  

29^(2000 + 1 ) =

29^2000 x 29 =  

29^2000 x (4 + 9 + 16) =

29^2000 x (2^2 + 3^2 + 4^2) =

29^2000 x 2^2 + 29^2000 x 3^2 + 29^2000 x 4^2 =

(29^1000 x 2)^2  +  (29^1000 x 3)^2 + (29^1000 x 4)^2