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# Help with Congruence of integers! watch

1. Hello! can anyone help me on this? Much appreciated!

Prove that a^2 = k mod 9 , where k is an element of (0,1,4,7)

Thanks!
2. Any interger a is either -4, -3, -2, -1, 0, 1, 2, 3, or 4 mod 9. Should be obvious from there.
3. (Original post by Lord of the Flies)
Any interger a is either -4, -3, -2, -1, 0, 1, 2, 3, or 4 mod 9. Should be obvious from there.

I get it that any of those integers squared would be 0 ,1 ,4 ,9 and 16 mod 9. Which can be reduced to 0,1,4,7 mod 9. Which is the values of k that I'm looking for.

However, why do we limit the integer value of "a" from -4 to 4? That's the part I don't get. What's preventing "a" from being any other integer like say 6? I know 6^2 = 36 = 0 mod 9. But it doesn't make sense to prove this for every number as it's not possible.

Thanks!
4. (Original post by iamlovinit05)

I get it that any of those integers squared would be 0 ,1 ,4 ,9 and 16 mod 9. Which can be reduced to 0,1,4,7 mod 9. Which is the values of k that I'm looking for.

However, why do we limit the integer value of "a" from -4 to 4? That's the part I don't get. What's preventing "a" from being any other integer like say 6? I know 6^2 = 36 = 0 mod 9. But it doesn't make sense to prove this for every number as it's not possible.

Thanks!
It can be 6. But 6 = -3 mod 9

I listed them in that particular way because it does the job twice as fast, but you could equally say "any integer is either 0, 1, 2, 3, 4, 5, 6, 7, or 8 mod 9", square these and deduct an appropriate multiple of 9 and we get 0, 1, 4, 7.

Or is it the fact that we have regrouped infinitely many integers into a finite number of categories which bothers you? The point of a mod is usually to characterise a number by its remainder when divided by another number. Of course, there are only 9 possible remainders when you divide by 9, so we can regroup all the integers in those 9 categories. These categories can range from -4 to 4, from 0 to 8, from 1 to 9, or whatever you like - it doesn't matter, they are all correct. The -4 to 4 is just easiest to work with!
5. (Original post by Lord of the Flies)
It can be 6. But 6 = -3 mod 9

I listed them in that particular way because it does the job twice as fast, but you could equally say "any integer is either 0, 1, 2, 3, 4, 5, 6, 7, or 8 mod 9", square these and deduct an appropriate multiple of 9 and we get 0, 1, 4, 7.

Or is it the fact that we have regrouped infinitely many integers into a finite number of categories which bothers you? The point of a mod is usually to characterise a number by its remainder when divided by another number. Of course, there are only 9 possible remainders when you divide by 9, so we can regroup all the integers in those 9 categories. These categories can range from -4 to 4, from 0 to 8, from 1 to 9, or whatever you like - it doesn't matter, they are all correct. The -4 to 4 is just easiest to work with!
Ahh I get what you mean now! I didn't realize that the other integers were "taken care of" by the negative integers Thanks so much!

But what happens if the 9 was changed to a bigger number, say 30. Do we have to list down a set of 30 integers and do this whole process again?
6. (Original post by iamlovinit05)
But what happens if the 9 was changed to a bigger number, say 30. Do we have to list down a set of 30 integers and do this whole process again?
Exact same problem with 30 instead of 9? Yes you would have to list them, as tedious as that may sound. Again, the best way would be to list them as "part negative part positive", that way you only need to square half of them.
7. (Original post by Lord of the Flies)
Exact same problem with 30 instead of 9? Yes you would have to list them, as tedious as that may sound. Again, the best way would be to list them as "part negative part positive", that way you only need to square half of them.

Alright thanks for clearing my doubt! You're a boss at this stuff. Respect.

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Updated: January 4, 2013
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