Congruence in Z(integers) mod n

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SUMMARY

In the discussion on congruence in Z mod 5, it is established that (a + b)⁵ = a⁵ + b⁵ due to the property that 5 ≡ 0 mod 5. This allows for the simplification of the expression, as the term involving 5 vanishes. Participants confirmed the correctness of substituting 0 for 5 in this context, emphasizing the equivalence notation. The discussion highlights the importance of understanding modular arithmetic and its implications in algebraic expressions.

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Homework Statement



In Z mod 5, compute (a + b)5.


Homework Equations





The Attempt at a Solution



Noticing that (a+b)5 = a5+5(a4*b+2*a3*b2+2*a2*b3+a*b4)+b5. Since 5=0 in Z mod 5, it follows that 0(a4*b+2*a3*b2+2*a2*b3+a*b4)=0 and hence (a+b)5=a5+b5.

I am just wondering if it is correct for me to say 5=0 in Z mod 5 and just substitute 0 in for 5?
 
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[tex]\equiv[/tex]Sounds reasonable to me. Why don't you test this with a couple of numbers to see if your results are consistent with what you've found?

BTW, instead of saying 5=0 in Z mod 5, you can say 5 [itex]\equiv[/itex] 0 mod 5. That 3-bar equals sign means "is equivalent to".
 
thanks for the response, much appreciated
 

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