MHB Prove that number is not a perfect square

[evinda]

Hey! :)
Prove that no number of the form $3k-1$ is a perfect square.
Do I have to use the theorem:
"If n is a positive integer such that n mod(4) is 2 or 3, then n is not a perfect square",or is there also an other way?

 

[I like Serena]

Hey! :)
Prove that no number of the form $3k-1$ is a perfect square.
Do I have to use the theorem:
"If n is a positive integer such that n mod(4) is 2 or 3, then n is not a perfect square",or is there also an other way?

Hai! :)

I would try mod 3...

 

[evinda]

Hai! :)

I would try mod 3...

I get that $(3k-1) \mod 3=-1$ or $2$..Is it right?

 

[I like Serena]

I get that $(3k-1) \mod 3=-1$ or $2$..Is it right?

Yes...

 

[Deveno]

What are the possible squares (mod 3)?

That is, what is:

$0^2 \text{ (mod }3)$
$1^2 \text{ (mod }3)$
$2^2 \text{ (mod }3)$?

 

[evinda]

Yes...

And because of the fact that $-1$ and $2$ are not perfect squares,does this mean that $3k-1$ cannot be a perfect square?Also,how can we know that,at $\mathbb{Z}_m,m\neq 3$,$3k-1 \mod m$ will not be equal to a perfect square?? (Thinking)

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What are the possible squares (mod 3)?

That is, what is:

$0^2 \text{ (mod }3)$
$1^2 \text{ (mod }3)$
$2^2 \text{ (mod }3)$?

$0^2 \text{ (mod }3)=0$
$1^2 \text{ (mod }3)=1$
$2^2 \text{ (mod }3)=1$

 

[I like Serena]

And because of the fact that $-1$ and $2$ are not perfect squares,does this mean that $3k-1$ cannot be a perfect square?

Not in itself.
What we see is that $3k-1 \equiv 2 \pmod 3$.

Furthermore, any square must be of the form $(3r)^2, (3r + 1)^2,$ or $(3r + 2)^2$.
That means that any square is either $0$ or $1 \pmod 3$.
So we have a mismatch.
The left side always has a different remainder than the right hand side.

Also,how can we know that,at $\mathbb{Z}_m,m\neq 3$,$3k-1 \mod m$ will not be equal to a perfect square?? (Thinking)

Is that a new question?

 

[evinda]

Not in itself.
What we see is that $3k-1 \equiv 2 \pmod 3$.

Furthermore, any square must be of the form $(3r)^2, (3r + 1)^2,$ or $(3r + 2)^2$.
That means that any square is either $0$ or $1 \pmod 3$.
So we have a mismatch.
The left side always has a different remainder than the right hand side.

I understand..Thanks a lot! :)

Is that a new question?

No,I was just wondering if it suffices,showing it in $\mathbb{Z}_3$,but I think it does,right?

 

[I like Serena]

No,I was just wondering if it suffices,showing it in $\mathbb{Z}_3$,but I think it does,right?

Yes. It is a complete proof, so it suffices to show it in $\mathbb{Z}_3$.

 

[evinda]

Yes. It is a complete proof, so it suffices to show it in $\mathbb{Z}_3$.

Nice..Thank you!

 

[Deveno]

This is one of the reasons why we use "integers mod n" in number theory.

In the integers (mod n), we have "fewer cases", if we show a certain relationship is impossible in the integers (mod n), sometimes this suffices to show that same relationship is impossible in the integers.

For example, if:

$a = b^2$ in the integers, then:

$a = b^2\text{ (mod }n)$

as well, so if the second is impossible, the first is, as well.

In this example, we have $a = 3k - 1$. Now this is a lot of integers:

$a = \dots,-4,-1,2,5,8,\dots$

and the sheer number of cases to check is quite large (infinite). Going integer-by-integer to see if $a$ is a perfect square seems inefficient.

On the other hand, we have:

$a = 3k - 1 = -1 = -1 + 3 = 2\text{ (mod }3)$

which is just ONE case to check. And in that one case, we see no appropriate $b$ exists (There are only 3 possible values of $b$ mod 3 to test).