Daca numarul 9N se scrie ca si o suma de 2 patrate perfecte ( N natural ) aratati ca 10N are aceeasi proprietate .
Mihai Opincariu G.M.B. 2009
O problema simpla
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opincariumihai
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O problema simpla
Last edited by opincariumihai on Wed Jun 24, 2009 11:28 am, edited 1 time in total.
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Claudiu Mindrila
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Solutie.
Este cunoscut ca daca \( x\in\mathbb{Z}\Longrightarrow x\equiv0,\ 1\left(\text{mod 3}\right) \). Atunci , cum \( 9N=\alpha^{2}+\beta^{2}\left(\alpha.\ \beta\in\mathbb{Z}\right) \) deducem ca \( \alpha,\ \beta\equiv0\left(\text{mod 3}\right)\Longrightarrow\alpha=3\alpha_{1},\ \beta=3\beta_{1}\left(\alpha_{1},\ \beta_{1}\in\mathbb{Z}\right) \), deci \( N=\alpha_{1}^{2}+\beta_{1}^{2} \).
Dar
\( 10N=\left(3^{2}+1^{2}\right)\left(\alpha_{1}^{2}+\beta^{2}\right)=\left(3\alpha_{1}+\beta_{1}\right)^{2}+\left(3\beta_{1}-\alpha_{1}\right)^{2} \), conform identitatii lui Lagrange.
Este cunoscut ca daca \( x\in\mathbb{Z}\Longrightarrow x\equiv0,\ 1\left(\text{mod 3}\right) \). Atunci , cum \( 9N=\alpha^{2}+\beta^{2}\left(\alpha.\ \beta\in\mathbb{Z}\right) \) deducem ca \( \alpha,\ \beta\equiv0\left(\text{mod 3}\right)\Longrightarrow\alpha=3\alpha_{1},\ \beta=3\beta_{1}\left(\alpha_{1},\ \beta_{1}\in\mathbb{Z}\right) \), deci \( N=\alpha_{1}^{2}+\beta_{1}^{2} \).
Dar
\( 10N=\left(3^{2}+1^{2}\right)\left(\alpha_{1}^{2}+\beta^{2}\right)=\left(3\alpha_{1}+\beta_{1}\right)^{2}+\left(3\beta_{1}-\alpha_{1}\right)^{2} \), conform identitatii lui Lagrange.
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