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aratati ca nr natural a=13^n,unde n este nr natural nenul, se poate scrie ca suma de doua patrate perfecte

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[tex]Daca~n=2k~atunci~13^n=13^{2k}=0^2+(13^k)^2\\ Daca~n=2k+1~atunci~13^n=13^{2k+1}=13\cdot13^{2k}=\\ =(2^2+3^2)\cdot(13^k)^2=(2\cdot13^k)^2+(3\cdot13^k)^2[/tex]


I) n = par ⇒ n = 2k, k∈ℕ

(5,  12,  13) - triplet pitagoreic ⇔ 13² = 5² + 12²

[tex]\it 13^n = 13^{2k} = 13^2\cdot13^{2k-2}= (5^2+12^2)(13^{k-1})^2= \\\;\\ (5\cdot13^{k-1})^2+(12\cdot13^{k-1})^2[/tex]

II) n=impar ⇒ n = 2k+1, k∈ℕ

13 = 4 + 9 = 2² + 3²

[tex]\it 13^n=13^{2k+1} = 13\cdot13^{2k} = (2^2+3^2)(13^k)^2 = \\\;\\ =(2\cdot13^k)^2 + (3\cdot13^k)^2[/tex]
 


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