proving this modular problem?

bonfire09

I have to prove "For every integer n, n^2 is congruent to exactly one 0,2,or 4mod 7? I don't even know where to start? Apparently its a problem that has 6 other problems to it too meaning a-f and my professor assigned us the last one.

SteveL27

Hint: You don't have to do it for every n; just for n = 0, 1, 2, 3, 4, 5, 6. Why? And if you think about it a little, you really only need to test n = 0, 1, 2, and 3. Why?

sweet springs

Hi
Any integer is written as 7m+k where k=0,1,2,3,4,5,6. mod((7m+k)^2) = mod(k^2) for modulus 7.
You can get possible cases in seven calculations. As SteveL27 says you can reduce number of calculations half with attention that mod((7m+k)^2)=mod((7m-k)^2)=mod({7(m-1)+(7-k)}^2)=mod((7-k)^2)=mod(k^2)

In similar ways
For modulus 2, Ans {0,1}
For modulus 3, Ans {0,1}
For modulus 4, Ans {0,1}
For modulus 5, Ans {0,1,4}
For modulus 6, Ans {0,1,3,4}
For modulus 7, Ans {0,1,2,4}
For modulus 8, Ans {0,1,2,3,4}
For modulus 10, Ans {0,1,4,5,6,9}

Regards.

bonfire09

thanks for the explanation! I think I get what you guys say. I'm going to figure it out and see how far I can get.

bonfire09

Oh that's because 4 and 6 are multiples of 2 and 3 and 5 is basically 2(2)+1.

SteveL27

What I had in mind was that 4 = -3 (mod 7) so you dont have to test 4 since you're squaring.

4^2 is already the same as (-3)^2 = 3^2. So you only need to check 3 but not 4. Likewise you don't have to check 5 or 6.

mathwonk

the whole point of modular arithmetic is that every specific problem can be done in a finite amount of time by anyone no matter how clueless, just by trying all the cases.

your predicament suggests that you lack the basic idea. hopefully it is getting clearer with this example.

bonfire09

I proved it by using 4 cases where n=0,n=1,n=2, and n=3. And showed that each case was congruent to exactly one thing. As you use 4,5,6... it repeats so I chose not to show those in my proof.

ramsey2879

I don't see the work where you showed or gave an explanation why it repeats for 4,5,6?

Eval

@bonfire09:
Depending on the situation, you may want to show that the cases repeat. For example, showing these 3 computations should be satisfactory (though if this is for an introductory class, it might be prudent to be more rigorous than this):

0^2 mod 7 = 0
1^2 mod 7 = 1 = 6^2 mod 7
2^2 mod 7 = 4 = 5^2 mod 7
3^2 mod 7 = 2 = 4^2 mod 7

Also, in response to your very first post, 1^2 is not in {0,2,4} mod 7 :P

(I think you made a typo)

ramsey2879

No, Bonfire wrote 0, 1, 4, 2 as "one, 0, 4, 2". In my post, I tried to say that any proof must be complete,i.e. you can't merely ignore the 4,5,6.

Eval

I know, that was what I was saying. I was responding to him, not you, sorry :/

Also, I read that as a different syntax. I thought he was saying "congruent to exactly one 0,2,or 4mod 7" meaning that it was one of the values in the list {0,2,4}

EDIT: This is one of those cases where you could get away with "WLOG" (Without Loss of Generality), depending on the level at which you are writing the proof.

ramsey2879

Yeah, it could be read either way, which is how some professors try to confuse things to make the student think. As for your edit, I disagree.

bonfire09

Oh yeah here is my proof:
Case:1 let n=0
Since n=0 then n^2=0. Hence 0 is congruent to 0mod 7.
Case 2:let n=1
Since n=1 then n^2=1. Hence 1 is congruent to 1mod7.
Case 3 let n=2
Since n=2 then n^2=4. Hence 4 is congruent to 4mod7.
Case 4 let n=3
Since n=3 then n^2=9. Hence 9 is congruent to 2mod7.
This is the rundown of my proof. There is not much to assume here except that n is an integer. And prove that each integer n^2 is congruent to exactly one of the four cases.

SteveL27

What happens if n = 5793459834895479437?

bonfire09

Dang! Oh i forgot to represent the integer n as an arbitrary value meaning that n=7m, 7m+1,7m+2,etc
Where m is an integer.

ramsey2879

How does your proof address the question of whether n = 7m + 4 , 7m +5 and 7m +6 give one of the same values for n = 0 to 3?