JBTST II 2007, Problema 1
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JBTST II 2007, Problema 1
Sa se determine numerele naturale \( n \geq 4 \) cu proprietatea ca \( [\sqrt{n}] + 1 \) divide \( n-1 \) si \( [\sqrt{n}] - 1 \) divide \( n+1 \).
For the easiness of writing, let \( x=\lfloor \sqrt{n} \rfloor \)
\( n=x^2+y,y<2x \)
\( x+1\mid (x+1)(x-1)+y\rightarrow x+1\mid y \)
\( x-1\mid (x+1)(x-1)+y+2\rightarrow x-1\mid y+2 \)
1) \( y=0\rightarrow x\in {2,3}\rightarrow n\in {4,9} \)
2) \( y=x+1\rightarrow x-1\mid x+3\rightarrow x-1\mid 4\rightarrow x\in{2,3,5}\rightarrow n\in{7,13,31} \)
\( n=x^2+y,y<2x \)
\( x+1\mid (x+1)(x-1)+y\rightarrow x+1\mid y \)
\( x-1\mid (x+1)(x-1)+y+2\rightarrow x-1\mid y+2 \)
1) \( y=0\rightarrow x\in {2,3}\rightarrow n\in {4,9} \)
2) \( y=x+1\rightarrow x-1\mid x+3\rightarrow x-1\mid 4\rightarrow x\in{2,3,5}\rightarrow n\in{7,13,31} \)
I can't speak Romanian, so I'll write in English. Hope you understand me!