For $a \in \mathbb Z$, $3\mid a$ if and only if $3\mid a^ 2$

Question

For $a \in \mathbb Z$, $3\mid a$ if and only if $3\mid a^2$?

I don't get how to solve this question.

how would you prove this question?

Answer 1

$3|a$ means $a=3k$ for some $k\in\mathbb Z$, so $a^2=3ka$; i.e., $3|a^2$.

On the other hand, if $3|a^2,$ then by Euclid's lemma $3|a$ or $3|a$; i.e., $3|a$.

Answer 2

$3|a^2$

We can write this as,

$a^2=3k$

Taking square root on both sides

$a=\sqrt{3k}$

$a$ is an integer, so $3k$ must be a perfect square, then $k=3t^2$ for some $t$

$a=\sqrt{9t^2}$

$a=3t$

$3|a$

Answer 3

It is immediate that

$3 \mid a \Longrightarrow 3 \mid a^2, \tag 1$

as has been so eloquently demonstrated by our friend J.W. Tanner in his answer to the question.

On the other hand, suppose that

$3 \mid a^2, \tag 1$

but

$3 \not \mid a; \tag 2$

then since $3$ is prime,

$\gcd(3, a) = 1; \tag 3$

thus by Bezout's identity there exist

$x, y \in \Bbb Z, \tag 4$

with

$3x + ay = 1; \tag 5$

we multiply through by $a$:

$3ax + a^2 y = a; \tag 6$

but now by assumption (1) we have

$3 \mid 3ax, \; 3 \mid a^2 \Longrightarrow 3 \mid a, \tag 7$

contradicting (2); thus

$3 \mid a \tag 8$

and we are done.

Answer 4

It is fun proving things using elementary concepts. Below recourse is made to the binomial theorem and Euclidean division theorem.

See Barry Cipra's answer - it 'lightens the machinery' still further, using two 'step-down' ideas in which the word theorem does not appear.

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For the more interesting direction of the equivalence, assume that $3 \nmid a$. Then applying Euclidean division we can write as true

$\tag 1 a = 3 q + r \quad \text{ where } r = 1 \text{ or } r = 2$

When $r = 1$

$\quad a^2 = (3 q + 1)^2 = 3 \times 3 \times q^2 + 3 \times 2 \times q + 1 = 3 (3 q^2 + 2q) + 1$

and using Euclidean division theory we conclude that $3 \nmid a^2$.

When $r = 2$

$\quad a^2 = (3 q + 2)^2 = 3 \times 3 \times q^2 + 3 \times 4 \times q + 4= 3 (3 q^2 + 4q + 1) + 1$

and, again, using Euclidean division theory we conclude that $3 \nmid a^2$.

Answer 5

Just for fun, let's do this without invoking the fact that $3$ is prime, but instead using the algebraic identity $a^2-1=(a-1)(a+1)$.

It's a general fact for integers that if $d\mid a$ then $d\mid ab$ (i.e., if $a=dm$ then $ab=dm'$ where $m'=mb$), so $3\mid a\implies 3\mid a^2$ is easy.

Another general fact is that any integer $d\ge1$ divides precisely one of any $d$ consecutive integers, so if $3\mid a^2$, then $3\not\mid a^2-1$, which in turn implies $3\not\mid a-1$ and $3\not\mid a+1$ (as the converse of the previous paragraph's general fact). But $a-1$, $a$, and $a+1$ are three consecutive integers, so if $3\not\mid a-1$ and $3\not\mid a+1$, then we must have $3\mid a$. Thus $3\mid a^2\implies 3\mid a$.

Answer 6

The nontrivial part is a special case of Euclid's lemma, since $3$ is prime. The other direction, $3|a\implies 3|a^2$, is trivial.

Euclid's lemma is in fact pretty trivial, too, if you know the fundamental theorem of arithmetic, say.