If $d|x^2$ how to show that $d|x$ using prime factorization method? I can write $x={p_1}^{a_1}.{p_2}^{a_2}......{p_r}^{a_r} $ . Then $x^2={p_1}^{2a_1}.{p_2}^{2a_2}...{p_r}^{2a_r}$. If $d={p_1}^{b_1}.{p_2}^{b_2}...{p_r}^{b_r}$ . How can I proceed further?

  • $\begingroup$ Not true - $8|4^2$, but $8$ does not divide $4$ .., maybe you want $d$ to be prime? $\endgroup$ – Nicky Hekster May 21 '14 at 10:56
  • $\begingroup$ yes I agree. what will be if d is prime $\endgroup$ – user152418 May 21 '14 at 11:01
  • $\begingroup$ If a prime divides $ab$, then it divides $a$ or it divides $b$ --- this is a step in one of the common proofs of unique factorization. It follows that if a prime divides $x^2$ then it divides $x$. $\endgroup$ – Gerry Myerson May 26 '14 at 12:58

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