Prove that number is not a perfect square

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In summary: This means no such $b$ exists in the integers, either. So, yes, showing it in $\mathbb{Z}_3$ suffices to prove it in the integers.In summary, we have proven that no number of the form $3k-1$ can be a perfect square by showing that in the integers (mod 3), there is no appropriate value of $b$ that would make $a=3k-1$ a perfect square. This technique of using "integers mod n" in number theory allows us to simplify and streamline our proofs by reducing the number of cases we need to check.
  • #1
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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  • #2
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...
 
  • #3
I like Serena said:
Hai! :)

I would try mod 3...

I get that $(3k-1) \mod 3=-1$ or $2$..Is it right?
 
  • #4
evinda said:
I get that $(3k-1) \mod 3=-1$ or $2$..Is it right?

Yes...
 
  • #5
What are the possible squares (mod 3)?

That is, what is:

$0^2 \text{ (mod }3)$
$1^2 \text{ (mod }3)$
$2^2 \text{ (mod }3)$?
 
  • #6
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)

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

That is, what is:

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

$0^2 \text{ (mod }3)=0$
$1^2 \text{ (mod }3)=1$
$2^2 \text{ (mod }3)=1$
 
  • #7
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:
 
  • #8
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?
 
  • #9
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! :eek:
 
  • #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).
 

FAQ: Prove that number is not a perfect square

1. How do you prove that a number is not a perfect square?

To prove that a number is not a perfect square, we can use the property that perfect squares always have an odd number of factors. This means that if a number has an even number of factors, it is not a perfect square.

2. Can you give an example of a number that is not a perfect square?

Yes, for example, the number 6 is not a perfect square because it has an even number of factors: 1, 2, 3, and 6.

3. What is the difference between a perfect square and a square number?

A perfect square is a number that can be expressed as the product of two equal integers, while a square number is simply the result of multiplying an integer by itself.

4. Is there a shortcut to determine if a number is not a perfect square?

Yes, in addition to checking the number of factors, we can also look at the last digit of the number. If the last digit is 2, 3, 7, or 8, then the number is not a perfect square.

5. Can a negative number be a perfect square?

No, a perfect square must be a positive number because it is the result of multiplying two equal integers.

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