Show That $9$ Doesn't Divide $a^2-3$

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In summary, to show that $9 \nmid (a^2 - 3)$ for any integer $a$, we can start by expressing $a^2-3$ as $9r+k$, where $k \ne 0$ and $r$ is some integer. From there, we can use congruence mod 9 to compute all possible values of $a^2-3$. The resulting set will not contain any multiples of 9, proving that $9 \nmid (a^2 - 3)$.
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tmt1
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For any integer $a$, show that $$9 \nmid (a^2 -3) $$

I start with

$a^2 - 3 = 9r + k$, where $k \ne 0$ and $r$ is some integer but I'm unsure how to proceed
 
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  • #2
Hi,
I must confess it took me longer to understand your starting point than to solve the problem.

[FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]9[/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]k[/FONT], where [FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]≠[/FONT][FONT=MathJax_Main]0[/FONT] and [FONT=MathJax_Math]r[/FONT] is some integer

I assume what you meant was:
For any integer $a$, if $a^2-3=9r+k$ for some integers $r$ and $k$, then 9 does not divide $r$. Basically, then you're talking about congruence mod 9.

For any integer $x$, 9 divides $x$ iff $x\equiv 0\pmod{9}$

So now compute all possible values mod 9 of $a^2-3$. You should quickly get the set $$\{0-3\pmod{9},\,1-3\pmod{9},\,4-3\pmod{9},\,7-3\pmod{9}\}$$

Quickly finish from here.
 

FAQ: Show That $9$ Doesn't Divide $a^2-3$

How do you show that 9 doesn't divide a2-3?

To show that 9 doesn't divide a2-3, we can use proof by contradiction. We assume that 9 does divide a2-3 and then show that it leads to a contradiction. This will prove that our assumption was false and therefore 9 does not divide a2-3.

Can you give an example to demonstrate that 9 doesn't divide a2-3?

Sure, let's take a=4. Then a2-3 = 16-3 = 13. Since 13 is not divisible by 9, we have shown that 9 doesn't divide a2-3 for a=4.

Is there a general method for showing that 9 doesn't divide a2-3 for any value of a?

Yes, there is a general method called modular arithmetic. We can use the fact that if a number (in this case a2-3) leaves the same remainder when divided by 9 as another number (in this case 0), then the two numbers are divisible by 9. By manipulating the equation a2-3 ≡ 0 (mod 9), we can show that this is not possible for any value of a, thus proving that 9 doesn't divide a2-3.

Can you explain why it is important to show that 9 doesn't divide a2-3?

Showing that 9 doesn't divide a2-3 is important because it helps us understand the properties of numbers and how they interact with each other. It also allows us to make conclusions about other numbers that may have similar properties to a2-3.

Are there any real-world applications of showing that 9 doesn't divide a2-3?

Yes, modular arithmetic and the concept of divisibility have applications in fields such as cryptography, computer science, and engineering. By understanding the properties of numbers and how they interact with each other, we can develop more efficient algorithms and systems for various applications.

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