Exercise · ZW-BEZOUT-S-IDENTITY-008 · ★★☆ Intermediate
Can \( 8x+12y \) equal 2 for some integers \( x,y \)?
Choose one option.
A No, all its values are multiples of \( 4 \) B No, because \( 8x+12y \) only takes odd values C Yes, because 2 divides both 8 and 12 D Yes, with \( x=1 \) and \( y=-1 \) we get 2
Show answer A. No, all its values are multiples of \( 4 \)
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Explanation The values of \( 8x+12y \) are exactly the multiples of \( \gcd(8,12)=4 \). So 2 is never reached, even though 2 divides both 8 and 12.
Bézout's identity states that, for integers \( a \) and \( b \) not both zero, there exist integers \( x \) and \( y \) such that \( ax+by=\gcd(a,b) \).
\[ax+by=\gcd(a,b) \text{ for some } x,y\in\mathbb{Z}\]
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