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Note 4

Assume that \(a|n\) and \(b|n\), and there exist integers \(x\) and \(y\) that \(ax+by=1\).

Prove: \(ab|n\)

Note by Jason Snow
1 year, 10 months ago

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By Bézout's identity, if there exists integer \((x,y)\) such that it satisfy \(ax+by = 1\), then \((a,b)\) shall have the GCD as 1 (i.e. A and B is coprime)

If \(a|n\) and \(b|n\), then for nonzero integer \(i, j\), it will be \(n = ai\) and \(n = bj\) or \(n^2 = abij\), which means

\[n^2\times\frac{1}{ab} = ij\]

Which proves that \(ab|n\), by closed-form of integer multiplication.

Kay Xspre - 1 year, 9 months ago

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