Help with a proof on divisibility

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    Divisibility Proof
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Homework Help Overview

The discussion revolves around a proof concerning divisibility, specifically addressing the statement that if 5 divides the sum of squares of three integers \(a\), \(b\), and \(c\), then 5 must divide each of those integers individually.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to prove the statement by assuming the contrary and exploring the implications of different remainders when \(a\), \(b\), and \(c\) are divided by 5. Some participants question the correctness of the theorem itself and suggest looking for counterexamples.

Discussion Status

There is an ongoing exploration of the problem, with some participants providing guidance on considering the remainders of squares when divided by 5. The original poster acknowledges a misunderstanding of the theorem, indicating a shift in focus to a simpler version of the problem.

Contextual Notes

The original statement of the problem was misinterpreted, leading to confusion about the requirements of the proof. This highlights the importance of clarity in problem statements.

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Homework Statement



Prove the following:
If 5 divides [tex]a^2 + b^2 + c^2[/tex] then 5 divides a and 5 divides b and 5 divides c.

Homework Equations



[tex]5 \mid a \implies a=5k , k \in Z[/tex]

The Attempt at a Solution



My idea is to assume 5 divides [tex]a^2 + b^2 +c^2[/tex]

also assume that "5 does not divide a" or "5 does not divide b" or "5 does not divide c"

then by the division algorithm, for some integers k,s,t

[tex]a=5k+1[/tex] or [tex]a=5k+2[/tex] or [tex]a=5k+3[/tex] or [tex]a=5k+4[/tex]

[tex]b=5s+1[/tex] or [tex]b=5s+2[/tex] or [tex]b=5s+3[/tex] or [tex]b=5s+4[/tex]

[tex]c=5t+1[/tex] or [tex]c=5t+2[/tex] or [tex]c=5t+3[/tex] or [tex]c=5t+4[/tex]

Now if we square each possibility for a, b, and c and add them together in each possible way, we would find that there is no possible combination that will give that 5 divides a^2 + b^2 + c^2. Then we would see that this is a contradiction so we must have that 5 divides a,b, and c. However, there must be an easier way than to have to manipulate 3^4 (i think) equations. Does anybody see a better way to go about this? or a quicker way to check all of the equations?
Thanks!
 
Last edited:
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Are you certain that your theorem is even correct?? Have you tried finding counterexample?
 
You are absolutely right! I misread the problem. It says prove that 5 divides a or 5 divides b or 5 divides c. Thanks! This is much easier! Hah i feel stupid
 
You probably want to try this on your own. But it may be worth to see what kind of remainders a2 can have when divided by 5...
 
Yes that is exactly right. I have it figured out now. Thank you
 

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