If $a^3-b^3\equiv 0\bmod3$ prove that $a^3-b^3\equiv 0\bmod 9$

Question

If $a^3-b^3\equiv 0\bmod3$ prove that $a^3-b^3\equiv 0\bmod 9$.

Guys, this question appeared in a preparation leaflet for a national exam, which I recently did. However, I was incapable of doing it. Can you guys please help me?

Thanking you in advance

Michael

Answer 1

Hints:

(1) If $a^3-b^3 \equiv 0 \pmod 3$, then $a-b\equiv 0 \pmod 3$

(2) If $a-b\equiv0\pmod3$ then $0\equiv(a-b)^2\equiv a^2+ab+b^2\pmod 3$

Answer 2

Observe that for any $\;a\in\Bbb Z\;$, we have that $\;a=a^3\pmod 3\;$ , because $\;a^3-a=a(a-1)(a+1)\;$ ...

From here that $\;a=b\pmod 3\implies a=b+3k\,,\,\,k\in\Bbb Z\;$

and now evaluate $\;a^3=(b+3k)^3;$ ....

Answer 3

Given the initial goal (orig)

$$\forall a b \mathop. \big(\;a^3 - b^3 \equiv_3 0 \implies a^3 -b^3 \equiv_9 0\;\big) \tag{orig}$$

let's first remark that $(-b)^3 = -(b^3)$. Since negation is a bijection, we can replace $b$ with $-b$ in our original problem, yielding (101).

$$ \forall a b . \big(\; a^3 + b^3 \equiv_3 0 \implies a^3+b^3 \equiv_9 0 \;\big) \tag{101} $$

Next, let's look at our condition $a^3+b^3 \equiv_3 0$ and change it into a fact mod 9 instead (102).

$$ \forall a b \mathop. \big(\; \exists w\mathop.(a^3 + b^3 \equiv_9 3w) \implies a^3+b^3 \equiv_9 0 \big) \tag{102}$$

If $w \equiv_3 0$, we're done.

I'm now going to show that $w \equiv_{3} 1$ and $w \equiv_3 2$ are impossible.

Consider the table of cubes mod 9, given below (103)

    n    n^3    n^3%9        (103)
    0      0      0
    1      1      1
    2      2      8
    3     27      0
    4     64      0
    5    125      8
    6    216      0
    7    343      1
    8    512      8

Note that the only congruence classes in (103) are $\{0, 1, 8\}$ . Let's consider what congruence classes we can hit with a sum of two of these (104):

    0   1   8              (104)
   +---------
 0 |0   1   8
 1 |1   2   0
 8 |8   0   7

$3$ and $6$ are not expressible as the sum of two cubes mod $9$, therefore $w \not\equiv_3 1$ and $w \not\equiv_3 2$ .

Therefore $w \equiv_3 0$ .

That exhausts the cases.

Therefore:

$$ \forall ab \mathop. \big(\; a^3 - b^3 \equiv_3 0 \implies a^3 - b^3 \equiv_9 0 \;\big)$$