Proving directly that ($a+b)^3 \equiv a^3 + b^3 \mod 3$

Question

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.

You already have your proof with what you did and lab's comment. Kudos! +1 — Timbuc, Oct 28 '14 at 18:15
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. — dustin, Oct 28 '14 at 19:07

score 2 · Accepted Answer · answered Oct 28 '14 at 18:18

2

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

answered Oct 28 '14 at 18:18

Dr. Sonnhard Graubner

95,283

The OP already did this. – Timbuc Oct 28 '14 at 18:20

score 1 · Answer 2 · answered Oct 28 '14 at 18:53

1

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?

answered Oct 28 '14 at 18:53

Gregory J. Puleo

6,443

Proving directly that ($a+b)^3 \equiv a^3 + b^3 \mod 3$

2 Answers2