How to prove that $x^2≡2(\bmod 3)$ is not a complete square

Question

Let $m$ be the product of first n primes (n > 1) , in the following expression :

$$m=2⋅3…p_n$$

I want to prove that $(m-1)$ is not a complete square.

I found two ways that might prove this . My problem is with the SECOND way .

First solution (seems to be working) :

The first way that I used is this :

Proof by negation : assume that $m-1$ is a complete square , i.e. $m-1 = x^2$ , then

$m=x^2+1=x^2-(-1)=(x-(-1))(x+(-1))=(x+1)(x-1)$

So we have either :

1. $(x+1)$ is even and $(x-1)$ is even

2. $(x+1)$ is even and $(x-1)$ is odd

3. $(x-1)$ is even and $(x+1)$ is odd

First case : $(x+1)$ is even and $(x-1)$ is even , then $m$ looks like this :

$m=2⋅otherNumbersA⋅2⋅otherNumbersB$

If we disregard $2$ then $m$ is a multiplication of $n-1$ prime numbers , then

$m$ is a multiplication of : $2 \cdot bigPrimeNumber$ . Contradiction .

The other two cases are just the same .

Second solution (my problem) :

What I'm interested in is the following solution (that I'm stuck in) :

Proof by negation : assume that : $m-1 = x^2$ and $m=2⋅3…p_n$ , means that $m$ divides by 3 , so we can write : $m-1≡2(mod 3)$ , which means that :

$m-1≡2(mod 3) ===> (m-1)-2=3q , q\in N ===> m-3=3q=m=3(1+q)$

Meaning :

$m-1=x^2$

$m-1≡2(mod 3)$

$x^2≡2(mod 3)$

How do I continue from here ? how can I use : $x^2≡2(mod 3)$ to reach a contradiction ?

Thanks

Answer 1

For any integer $\displaystyle x, x\equiv0, 1,2\pmod3$

$\displaystyle\implies x^2\equiv0,1^2\equiv1,2^2\equiv1\pmod3$

Answer 2

The follow-on to the previous answer is that since $x^2 \equiv 2 \mod 3$ has no solution, and $m-1 \equiv 2 \mod 3$, there cannot be a number $x$ such that $m-1 = x^2$.

It is worth pointing out that the first "solution" given is completely bogus because it contains a mistake in the first equation, which ends up reading $[x^2 + 1 ]= m = (x+1)(x-1) [= x^2 - 1]$.

Answer 3

All squares modulo $3$ are: $\{0,1\} $.

Answer 4

If there is a solution to $x^2 \equiv 2$ (mod $3$), then $x$ must be $0$, $1$ or $2$ (mod $3$). Now you can try each case and see that none of these will give you a solution:

• If $x \equiv 1,2$ (mod $3$) then $x^2 \equiv$ $1$ (mod $3$)
• If $x \equiv 0$ (mod $3$) then $x^2 \equiv$ $0$ (mod $3$)

So in none of these cases do we have $x^2 \equiv 2$ (mod $3$)

Answer 5

IT IS CLEAR THAT ONE CAN WRITE ANY INTEGER IN THE FORM OF - 3K OR 3K+1 OR 3K+2 AS 0,1,2 ARE THE RESIDUES SET FOR 3. THUS WHEN WE SQUARE IT AND DIVIDE THE EXPRESSION BY 3- WE GET ONLY 0 AND 1 AS THE REMAINDER- IE;- FOR 3K IT IS OBVIOUSLY 0, FOR THE OTHER TWO IT IS 1. THUS WE CAN NEVER WRITE IT AS X^2 CONGRUENT TO 2 ( MOD 3)