Prove that number is not a perfect square

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Discussion Overview

The discussion revolves around proving that no number of the form $3k-1$ is a perfect square. Participants explore various approaches, including modular arithmetic, specifically mod 3, and the implications of certain theorems related to perfect squares.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • Some participants propose using the theorem that states if a positive integer $n$ is such that $n \mod(4)$ is 2 or 3, then $n$ is not a perfect square.
  • Others suggest examining the expression $3k-1$ under mod 3, noting that $(3k-1) \mod 3$ results in $-1$ or $2$.
  • One participant questions the possible squares modulo 3, specifically asking for the values of $0^2$, $1^2$, and $2^2$ mod 3.
  • It is noted that $0^2 \mod 3 = 0$, $1^2 \mod 3 = 1$, and $2^2 \mod 3 = 1.
  • Some participants argue that since $-1$ and $2$ are not perfect squares, this suggests that $3k-1$ cannot be a perfect square, while others clarify that this alone does not suffice as proof.
  • Further discussion highlights that any square must be of the form $(3r)^2$, $(3r + 1)^2$, or $(3r + 2)^2$, leading to the conclusion that any square is either $0$ or $1 \mod 3$, indicating a mismatch with $3k-1$.
  • Participants also consider whether demonstrating the result in $\mathbb{Z}_3$ is sufficient for a complete proof.
  • One participant explains the utility of using integers mod n in number theory to simplify the proof process by reducing the number of cases to check.

Areas of Agreement / Disagreement

Participants generally agree on the approach of using modular arithmetic to analyze the expression $3k-1$, but there is no consensus on whether the arguments presented constitute a complete proof. Some participants believe it suffices, while others express uncertainty.

Contextual Notes

The discussion includes various assumptions about the properties of perfect squares and modular arithmetic, but these assumptions are not universally accepted or resolved within the thread.

evinda
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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?
 
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evinda said:
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...
 
I like Serena said:
Hai! :)

I would try mod 3...

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

Yes...
 
What are the possible squares (mod 3)?

That is, what is:

$0^2 \text{ (mod }3)$
$1^2 \text{ (mod }3)$
$2^2 \text{ (mod }3)$?
 
I like Serena said:
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)

- - - Updated - - -

Deveno said:
What are the possible squares (mod 3)?

That is, what is:

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

$0^2 \text{ (mod }3)=0$
$1^2 \text{ (mod }3)=1$
$2^2 \text{ (mod }3)=1$
 
evinda said:
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? :confused:
 
I like Serena said:
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! :)

I like Serena said:
Is that a new question? :confused:

No,I was just wondering if it suffices,showing it in $\mathbb{Z}_3$,but I think it does,right?
 
evinda said:
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 $\mathbb{Z}_3$.
 
  • #10
I like Serena said:
Yes. It is a complete proof, so it suffices to show it in $\mathbb{Z}_3$.

Nice..Thank you! :o
 
  • #11
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).
 

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