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Assuming a and b are integers, I must prove directly that: $$ (a + b)^3 \equiv (a^3 + b^3) \mod 3 $$ First, my peers and I made the mistake of assuming what we are trying to prove and thus failed. I've tried expanding $(a + b)^3$ into $a^3 + b^3 + 3(a^2)b + 3a(b^2)$ but I'm not sure where to go from there.

I keep wanting to use the definition of congruence (a≡b(modn) means $a - b = nk$ for some integer $k$) but I believe that is restricted since it is the conclusion I'm trying to prove. I'm not really sure how to get this started.

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    $\begingroup$ $$3n\equiv0\pmod3$$ for any integer $n$ $\endgroup$
    – lab bhattacharjee
    Oct 28, 2014 at 18:14
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    $\begingroup$ You already have your proof with what you did and lab's comment. Kudos! +1 $\endgroup$
    – Timbuc
    Oct 28, 2014 at 18:15
  • $\begingroup$ I think I understand, it seems too simple to me, however. $\endgroup$
    – Blue Wizard
    Oct 28, 2014 at 18:34
  • $\begingroup$ We know anything that is a multiple of 3 is 0 mod 3; that is, 3, 6 = 3(2), 9 = (3)(3), etc. You have 3(a^2)b = 3N and 3(a)b^2 = 3M where M and N are arbitrary numbers. Therefore, 3M and 3N are congruent to 0 mod 3. $\endgroup$
    – dustin
    Oct 28, 2014 at 19:07

2 Answers 2

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we have $(a+b)^3=a^3+b^3+3(a^2b+ab^2)\equiv a^3+b^3 \mod 3$

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  • $\begingroup$ The OP already did this. $\endgroup$
    – Timbuc
    Oct 28, 2014 at 18:20
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You already got $(a+b)^3 = a^3 + b^3 + 3(a^2b + ab^2)$. What can you now say -- just as an equation, forgetting about congruence for the moment -- about the difference $[(a+b)^3 - (a^3+b^3)]$? Now how does that compare to the definition of congruence?

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