Determinati \( \overline{abcd} \), \( a,c \neq 0 \), pentru care \( \frac{\sqrt{abcd}}{\sqrt{ab}+\sqrt{cd}} \in \mathbb{Q} \).
Gheroghe Iurea, lista scurta 2008
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Claudiu Mindrila
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Fie \( d=(\overline{ab},\overline{cd}) \) atunci \( \overline{ab}=d\cdot x \) , \( \overline{cd}=d\cdot y \) , \( (x,y)=1 \)
\( \frac{\sqrt{\overline{abcd}}}{\sqrt{\overline{ab}}+\sqrt{\overline{cd}}}=\frac{\sqrt{100\overline{ab}+\overline{cd}}}{\sqrt{\overline{ab}}+\sqrt{\overline{cd}}}=\frac{\sqrt{100x+y}}{\sqrt{x}+\sqrt{y}}\in \mathbb{Q} \)
1) Daca x=y atunci ab=cd \( \Rightarrow \) \( \frac{\sqrt{101}}{2}\in\mathbb{Q} \) fals.
2) Daca \( x\neq y \) atunci \( \sqrt{100x+y}(\sqrt{x}-\sqrt{y})\in\mathbb{Q} \) (am rationalizat)
\( \Rightarrow \) \( \sqrt{xy}\in\mathbb{Q} \) (am ridicat la patrat) \( \Rightarrow \) \( x=m^2 , y=n^2 \) , m ,n cifre.
\( \Rightarrow \) \( 100x+y=p^2 \) deci \( (10m)^2+n^2=p^2 \) (ecuatie pitagorica)
\( \Rightarrow \) \( \left{\begin10m=2\alpha\beta\\n=\alpha^2-\beta^2\\p=\alpha^2+\beta^2 \) sau invers.
Singura situatie care convine este \( \alpha=5, \beta=4\Rightarrow n=9, m=4, x=16, y=81, d=1 \Rightarrow\overline{abcd}=\overline{1681}. \)
\( \frac{\sqrt{\overline{abcd}}}{\sqrt{\overline{ab}}+\sqrt{\overline{cd}}}=\frac{\sqrt{100\overline{ab}+\overline{cd}}}{\sqrt{\overline{ab}}+\sqrt{\overline{cd}}}=\frac{\sqrt{100x+y}}{\sqrt{x}+\sqrt{y}}\in \mathbb{Q} \)
1) Daca x=y atunci ab=cd \( \Rightarrow \) \( \frac{\sqrt{101}}{2}\in\mathbb{Q} \) fals.
2) Daca \( x\neq y \) atunci \( \sqrt{100x+y}(\sqrt{x}-\sqrt{y})\in\mathbb{Q} \) (am rationalizat)
\( \Rightarrow \) \( \sqrt{xy}\in\mathbb{Q} \) (am ridicat la patrat) \( \Rightarrow \) \( x=m^2 , y=n^2 \) , m ,n cifre.
\( \Rightarrow \) \( 100x+y=p^2 \) deci \( (10m)^2+n^2=p^2 \) (ecuatie pitagorica)
\( \Rightarrow \) \( \left{\begin10m=2\alpha\beta\\n=\alpha^2-\beta^2\\p=\alpha^2+\beta^2 \) sau invers.
Singura situatie care convine este \( \alpha=5, \beta=4\Rightarrow n=9, m=4, x=16, y=81, d=1 \Rightarrow\overline{abcd}=\overline{1681}. \)