
#19
May2505, 12:26 PM

P: 79

Let's see if I can apply it. Let's say p = 5, does that mean then it is congruent to: [tex]0 + 2x + 1 (mod~5)[/tex]? How I derived this: [tex]5 \equiv 0 (mod~5)[/tex] therefore [tex]5x^2 \equiv 0x^2 (mod~5) = 0[/tex] [tex]7 \equiv 2 (mod~5)[/tex] therefore [tex]7x \equiv 2x (mod~5)[/tex] [tex]1 \equiv 1 (mod~5)[/tex] 



#20
May2505, 01:58 PM

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#21
May2505, 02:07 PM

P: 79

Do I have to apply mod 5 to every term?
For example, [tex]3^{2n} + 7[/tex], could I apply mod p to the 7 only and still have it being congruent? i.e., p = 8; [tex]3^{2n} + 7 \equiv 3^{2n} 1 (mod~8)[/tex]? or do I have to apply it to the 3 as well, that is, ... oh wait... it is already mod 8... but I could change it to [tex]3^{2n} + 7 \equiv 4^{2n} 1 (mod~8)[/tex] correct? 



#22
May2505, 02:10 PM

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Since 0*p is divisible with p, yes.




#23
May2505, 02:11 PM

P: 79





#24
May2505, 02:12 PM

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#26
May2505, 02:15 PM

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Do as Darth Gokul saith, be sure you understand the definition of congruence. 



#27
May2505, 02:15 PM

P: 79

n divides ab, ahh




#28
May2505, 02:20 PM

P: 79

[tex] 7 \equiv [/tex]1 [tex](mod~8)[/tex] 



#29
May2505, 02:30 PM

P: 79

LOL, which pretty much states that I did not apply mod 8 to just the 7. Sorry, I'm a goof




#30
May3005, 02:24 AM

P: 79

My professor has posted up the proof to this question and I am not understanding one thing: The part where I have to prove that [tex]6k^3 + 5k[/tex]. His proof states that k must take one of six forms, that is 6m, 6m+1, 6m+2, 6m+3, 6m+5, and must be divisible by 6 when substituted in for k.
The case thing is what I don't understand. Why does this prove that [tex]6k^3 + 5k[/tex]? 



#31
May3005, 03:36 AM

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You're missing out 6m+4, but the point is if you were to put any of those numbers into the formula k^3+5k you get a number divisible by 6. He's just doing the "unclever" way of solving the problem, that is to say he's doing modulo arithmetic without telling you he's doing modulo arithmetic.
Every given any number n and another number p we can express n=pq+r where q and r are integers 0<=r<p and q is then uniquely determined. That is just kindergarten mathematics (which is far more mathematically mature than knowing how to write out decimals if you ask me) where r is the remainder after dividing by p. All he is now claiming is that whatever the remainder is after dividing by 6 (ie what class it is in modulo r) we get something divisible by 6, eg if k=6m+1 k^3+5k= (6m+1)^3+5(6m+1) = (6m^3) +3(6m)^2+4(6m)+1+5(6m)+5 obvisouly anyting with a 6m in it is divisible by 6 so that just leaves 1+5 part, which is also divisible by 6. Hopefully you see that if I expand (6m+r)^3 i get something divisible by 6 plus r^3, and the 5k=30m+5r, so I just need to show that r^3+5r is divisible by 6 for r from 0 to 5. 



#32
May3005, 03:48 PM

P: 79

Oooh.. so those forms, 6m, 6m+1, etc., are all possible ways of expressing any number in terms of 6?
That is q=6, the uniquely determined integer? But that doesn't mean that n is divisible by 6? Oh I get it... so for any integer k, k can be expressed using one of the six forms. (Now help me with my understanding of why we do this...) We use q=6 so that we may be able to find 6 as a common term... I'm afraid to say it but I think I have the idea... 



#33
May3005, 04:36 PM

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Note if k=6m+r then k=r mod 6. 



#34
May3105, 01:49 AM

P: 646

Dividend = Divisor*Quotient + Remainder Now he was well aware of this, then i made a connection of this equation with the above proposed "i can express any number as......". Then i had to get back to the original question. To this day, i still hope he understood what i said that day.  AI 


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