# If $a^2$ divides $b^3$, then $a$ divides $b$.

I want to prove or provide a counterexample to the following statement: $a^2|b^3 \Rightarrow a|b$. I know that $a^k|b^k \Rightarrow a|b$. My thought is that, e.g in the case of $k = 3$, where we have $a^3 = p_1^{3\alpha_1} \cdots p_k^{3\alpha_k}$ and $b^3 = p_1^{3\beta_1} \cdots p_k^{3\beta_k}$ where $\alpha_i, \beta_i \geq 0$ (equal to $0$ if one of the primes $p_i$ isn't in the factorization), we know that $a|b$ because $a^3|b^3 \Rightarrow 3\alpha_i \leq 3\beta_i \Rightarrow \alpha_i \leq \beta_i$ and so $a|b$.

In this case, where we have $a^2|b^3$, it does not seem to follow that $2\alpha_i \leq 3\beta_i \Rightarrow \alpha_i \leq \beta_i$. Hence, I believe the statement is false. I'm unsure of how to approach finding a counterexample, however.

• try some little numbers. Commented Jan 30, 2015 at 3:43
• Hint: what about $a = \alpha^{3}, b=\alpha^{2}$? Commented Jan 30, 2015 at 3:46
• How about if $a^2=b^3$? That's the extreme case of $a^2\mid b^3$. Commented Jan 30, 2015 at 3:46
• Choose any $\,\alpha,\beta\,$ so $\,\beta < \alpha \le \frac{3}2\beta\ \$ Commented Jan 30, 2015 at 4:52

Counterexample: Suppose we are dealing with $a=16$ and $b=8$. Then $a^2=16^2=256$ and $b^3=8^3=512$. Then we have that $a^2\mid b^3$ but $16 \not\mid 8$.