Prove: If $ (a^2 \mid (2b^2))$, then $ (a \mid b). $

Question

Trying to prove the following implication:

$$ (a^2 \mid 2b^2) \implies (a \mid b) $$

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Knowing that I've tried to prove it by using Bézout's Theorem only to find myself lost. My teacher hinted that we can use either prime factorization or prove that $(a\wedge b=a)$.

Answer 1

Hint:

Use valuations: for any prime $p$ and any integer $n \in \mathbb N$, denote by $v_p(n) \in \mathbb N$ the power of $p$ in the prime factorization on $n$. Then, the hypothesis translates as $$v_p(a^2)=2v_p(a)\le v_p(b^2)+v_p(2)=2v_p(b)+v_p(2)\le2v_p(b)+1, $$ so that $\;v_p(a)\le v_p(b)+\frac12$.