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scrieți numărul A=38^2013 ca suma de trei pătrate perfecte

vă rog frumos dacă puteți să mă ajutați! dau coroană!



Răspuns :

 

[tex]\displaystyle\bf\\A=38^{2013}\\\\A=38^{1+2012}\\\\A=38\times38^{2012}\\\\38=4+9+25 = 2^2+3^2+5^2\\\\A=(4+9+25)\times38^{2012}\\\\A=4\times38^{2012}+9\times38^{2012}+25\times38^{2012}\\\\A=2^2\times38^{1006\times2}+3^2\times38^{1006\times2}+5^2\times38^{1006\times2}\\\\A=2^2\times\Big(38^{1006}\Big)^2+3^2\times\Big(38^{1006}\Big)^2+5^2\times\Big(38^{1006}\Big)^2\\\\\\\boxed{\bf~A=\Big(2\times38^{1006}\Big)^2+\Big(3\times38^{1006}\Big)^2+\Big(5\times38^{1006}\Big)^2}\\\\\\A=pp+pp+pp[/tex]

 

 

Răspuns:

38^2013=38^2012×38 ;   38=2^2+3^2+5^2=4+9+25;    inlocuim in prima relatie ;38^2012(2^2+3^2+5^2)=38^2012×2^2+38^2012×3^2+38^2012×5^2

Explicație pas cu pas: