Simple proof that if $a^2$ is divisible by a prime $b$ then $a$ is divisible by $b$. I know it's obvious that if $a^2$ is divisible by $b$, given that $b$ is prime  ,then $a$ is divisible by $b$. Is there any way to prove this without going into the fundamental theorem of arithmetic? I was told it can be proved starting from the fact that every number is the product of distinct primes, but is there a simpler way? 
 A: This could be another expression of Euclids lemma, which states that if $p$ is a prime and divides the product $ab$, then $p$ divides $a$ or $p$ divides $b$. Hence if $p$ divides $a\cdot a$ then $p$ divides $a$. 
A: It is true. Here is a proof without prime decomposition, using only the euclidean division and the definition of a prime number, along the lines of a proof I gave of Euclid'slemma.
Let $E=\bigl\{x\in\mathbf N^*\:;\: b\mid xa\bigr\}$. $E$ is a non-empty subset of $\mathbf N$, since $a$ and $b$ belong to $E$.Hence it has a smallest element, $x_0$.
Claim: $x_0$ divides all elements in $E$.
Indeed, let $x\in E$. The euclidean division by $x_0$ gives $\, x=qx_0+r$ for some $q, r$ ($0\le r<x_0$). Now $b$ divides $xa$ and $x_0a$ by hypothesis, hence it divides $(x-qx_0)a=ra$, which means $r\in E$, unless $r=0$. As $x_0$  is the smallest element in $E$, $r$ can't be in $E$, thus $r=0$, and $x_0$ divides $x$.
In particular, $x_0$ divides $b\,$; as $b$ is prime, $x_0=0$ or $x_0=1$.


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*If $x_0=b$, this implies $b$ divides all elements in $E$, amongst which $a$.

*If $x_0=1$, the definition of $E$ implies $b$ divides $1\cdot a=a$.

