# If a prime $p\mid b$ and $a^2=b^3$, then $p^3\mid a$ [duplicate]

I have an exercise that I don't know how to solve. I tried to solve it in many ways, but I didn't get any progress in proving or disproving this... The exercise is:

Prove or disprove: if $p$ is a prime number, if $a$ and $b$ are native numbers and $$a^2 = b^3$$ and if $p \mid b$, then $$p^3 \mid a .$$

If someone has a proof to this exercise I would really appreciate it.

Thanks!

• The key words are "unique factorization into primes." Aug 30, 2014 at 19:10

Hands on:

$p|b$ so that $b=pd$

$p|a^2$ so $p|a$ and $a=pc$ whence

$p^2c^2=p^3d^3$ or $c^2=pd^3$

Now $p|c^2$ so $p|c$ and $c=pe$ whence

$p^2e^2=pd^3$ or $pe^2=d^3$

So that $p|d^3$ so that $p|d$ and $d=pf$ so that

$pe^2=p^3f^3$ or $e^2=p^2f^3$ and $p|e^2$ so that $p|e$ and $e=pg$

Now we have $a=pc=p^2e=p^3g$

Note $p|xy$ implies $p|x$ or $p|y$ is the only property of primes we need. Apply with $x=y=a$ to obtain $p|a^2 \implies p|a$ and then with $x=a, y=a^2$ to obtain $p|a^3 \implies p|a$

Let the highest powers of $p$ in $a,b$ be $A(\ge0),B(\ge1)$ respectively,

So, we have $2A=3B\implies \dfrac{2A}3=B\implies 3|2A\implies 3|A\implies A\ge3$

• Great, Thanks!!
– Roee
Sep 1, 2014 at 14:46