If $d$ and $y$ are positive integers and I know that $d^{2}|2y^{2}$ then $d^2|2$ (i.e $d=1$) or $d^2|y^2$ .
In the case that $d^2|y^2$ does that imply that $d|y$ for all $d,y$ ?
Thank you.
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$\begingroup$ Try to represent $d^2$ and $y^2$ as products of prime factors, compare it with prime factorization of $d$ and $y$ and this will be enough, I hope, to get an answer. $\endgroup$– Andrei RykhalskiOct 19, 2014 at 15:20
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2 Answers
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Hint:
Write out as definition of divisibility: $$d^2|y^2\implies y^2=ad^2$$ Taking the positive square root we arrive at $y=d\sqrt{a}$ therefore $d|y$ if and only if $a$ is square. Now show that, for $y$ to be square, $a$ must be square in the first place. Thus completing the proof.
answered Oct 19, 2014 at 15:20
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The answer is YES and you can think of it using the prime factorization.
answered Oct 19, 2014 at 15:18