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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, 8 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. · 1 year, 8 months ago
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