Criterion for a positive integer a>1 to be a square

[Math100]

Homework Statement

Prove that a positive integer $a>1$ is a square if and only if in the canonical form of $a$ all the exponents of the primes are even integers.

Relevant Equations

None.

Proof:

Suppose a positive integer $a>1$ is a square.
Then we have $a=b^2$ for some $b\in\mathbb{Z}$,
where $b=p_{1}^{n_{1}} p_{2}^{n_{2}} \dotsb p_{r}^{n_{r}}$
such that each $n_{i}$ is a positive integer and $p_{i}'s$
are prime for $i=1,2,3,...,r$ with $p_{1}<p_{2}<p_{3}< \dotsb <p_{r}$.
Thus $a=b^2$
$=(p_{1}^{n_{1}} p_{2}^{n_{2}} \dotsb p_{r}^{n_{r}})^2$
$=p_{1}^{2n_{1}} p_{2}^{2n_{2}} \dotsb p_{r}^{2n_{r}}$,
which shows that all the exponents of the primes are even integers in the canonical form of $a$.
Conversely, suppose all the exponents of the primes are even integers in the canonical form of
$a=p_{1}^{n_{1}} p_{2}^{n_{2}} \dotsb p_{r}^{n_{r}}$.
Then we have $n_{i}=2k_{i}$ for some $n_{i}, k_{i} \in\mathbb{Z}$.
Thus $a=p_{1}^{2k_{1}} p_{2}^{2k_{2}} \dotsb p_{r}^{2k_{r}}$
$=(p_{1}^{k_{1}} p_{2}^{k_{2}} \dotsb p_{r}^{k_{r}})^2$,
which shows that a positive integer $a>1$ is a square.
Therefore, a positive integer $a>1$ is a square if and only if in the canonical form of $a$
all the exponents of the primes are even integers.

 

[fresh_42]

That's correct.

 

[PeroK]

There's a tiny flaw. You say $a = b^2$ for some $b \in \mathbb Z$. But, you need $b >1$.

 

[Math100]

There's a tiny flaw. You say $a = b^2$ for some $b \in \mathbb Z$. But, you need $b >1$.

How about if I write $a=b^2$ for some $b\in\mathbb{Z}$ where $b>1$?

 

[PeroK]

How about if I write $a=b^2$ for some $b\in\mathbb{Z}$ where $b>1$?

Yes.