Is there $a\in \mathbb{Z}$ such that $ a^2\equiv 2 \pmod3$

Question

Is there $a\in \mathbb{Z}$ such that $$ a^2\equiv 2\pmod3$$

I did the following : if the equation is true then one has $(a-1)(a+1)\equiv1 \pmod3$ which implies that necessarly that both $(a-1)$ and $(a+1)$ are not even hence $a$ is an even number hence $$a^2-2\equiv 0\pmod3$$ where $a^2-2$ is even, such that $a^2-2$ is a multiple of $3$ which is impossible!

Is my idea right ?

It's much common to use mod $3$ \pmod3 rather than square brackets as they can mean equivalence class. Your argument is incorrect as $3$ divides $6$ (an even number), for instance. — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Nov 17 '21 at 08:25
And if you do indeed mean mod 3, it is a pretty easy process to just work through all of the relevant examples ($a=0,1,2$) — PrincessEev, Nov 17 '21 at 08:26
@EeveeTrainer or to massively overkill this, we have $(-1/3)=-1$ by quadratic reciprocity :p — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Nov 17 '21 at 08:28
@EeveeTrainer how's working with $a=0,1,2$ would help us prove there's no $a$ such that the equality holds ? and yes I meant mod by $[\cdot]$ — BrianTag, Nov 17 '21 at 08:31
If $a\equiv0 \pmod 3$, then what is the value of $a^2 \pmod 3$ ? If $a\equiv1 \pmod 3$, then what is the value of $a^2 \pmod 3$ ? If $a\equiv2 \pmod 3$, then what is the value of $a^2 \pmod 3$ ? — Jaap Scherphuis, Nov 17 '21 at 08:41
@Brian Parity arguments make sense $!\bmod n$ only when $,n,$ is even, because generally congruences only persist mod factors of the modulus. Thus $,a\equiv b\pmod{2n}\Rightarrow a\equiv b \pmod 2.,$ Therefore, if a congruence has no roots $!\bmod 2,$ then it has none $!\bmod 2n,,$ e.g. see here. — Bill Dubuque, Nov 18 '21 at 05:32
In particular, parity arguments won't help $!\bmod n,$ when $,n,$ is odd, e.g. if $,x=a,$ is a solution to $,x^2 = b\pmod{! n},$ then $,x+n,$ is another solution of opposite parity. Indeed by CRT the system $,x\equiv j\pmod{n},, x\equiv k\pmod{2},$ is solvable for any $,j,k,$ so the parity $,k,$ of $,x,$ implies nothing about the values $,j\equiv x\pmod{n}, - $ they are completely independent generally (i.e. without any further hypotheses). — Bill Dubuque, Nov 18 '21 at 23:09
We can simply check for solutions. If $a\equiv0$ then $a^2\equiv 0$; if $a\equiv 1$ then $a^2\equiv 1$; and if $a\equiv 2$ then $a^2\equiv 4\equiv 1$. Since every integer is congruent to $0$, $1$, or $2$ in modulo $3$ arithmetic, we have exhausted all of the different possibilties, and so we conclude that the equation $a^2\equiv 2$ has no solutions. — Joe, Nov 19 '21 at 17:25

score 2 · Accepted Answer · answered Nov 18 '21 at 23:33

The problem with your disproof in that “evenness” is not something preserved mod 3 (or any other odd number). For example, if $a = 6$, then $a^2 - 2 = 34$ is an even number, but “mod 3” reduces it to 1, an odd number.

I would use the “brute-force” solution suggested by Eevee Trainer's comment. That is, if $a \in \mathbb{Z}$, then one of the following must be true for some integer $k$:

$a = 3k$
$a = 3k + 1$
$a = 3k + 2$

Now, work out the value of $(a^2 - 2) \mod{3}$ for each of these three cases, and see where it leads you.

score 1 · Answer 2 · answered Nov 18 '21 at 23:38

One way to prove this is as follows:

There is a well-known lemma in number theory which is as follows:

Lemma: If $p$ is a prime number with $p\equiv 3 \ (mod \ 4)$ and if $p|a^2+b^2$, then $p|a$ and $p|b$.

Now, since $a^2-2\equiv 0$ $(mod \ 3)$ and $a^2-2\equiv a^2+1$ $(mod \ 3)$, we get that $a^2+1\equiv 0$ $(mod \ 3)$ and thus by using this lemma for your case, we get that $3|a$ and $3|1$, which is impossible and thus it is done.

Is there $a\in \mathbb{Z}$ such that $ a^2\equiv 2 \pmod3$

2 Answers2