Partea intreaga

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alex2008
Leibniz
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Partea intreaga

Post by alex2008 »

Sa se afle \( a,b,c\in \mathbb{N}^* \) astfel incat \( [\frac{a^2}{2b}]+[\frac{b^2}{2c}]+[\frac{c^2}{2a}]=0 \) .
. A snake that slithers on the ground can only dream of flying through the air.
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Marius Mainea
Gauss
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Post by Marius Mainea »

\( a=b=c=1 \)

Intradevar,

Din \( 0\le \frac{a^2}{2b}<1 \) rezulta \( a^2+1\le 2b \) si analoagele, de unde prin adunare,

\( (a-1)^2+(b-1)^2+(c-1)^2\le 0 \) si de aici concluzia.
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