Prove that $5 \nmid (a+1)^3 - a^3$

Question

Prove that difference between two consecutive cubes cannot be divided by $5$.

Here's what I've done, but I'm not sure about one step:

Let two cubes be $(a+1)^3$, and $a^3$. $$(a+1)^3 - a^3 = 3a^2+3a+1$$. Suppose for a moment that it is actually divisible by $5$, then we must have $3a^2 +3a+1=5k$ for some $k$.

This implies that $3a^2+3a\equiv-1\pmod 5$. This in turn translates to equation: $$[3a^3 + 3a]=[-1]=[4]$$ since we are working in $\text{Z}_5$. (Not sure about this part)

Since there are only 5 elements in $\text{Z}_5$, we can just test them all. And indeed there's no solution in $\text{Z}_5$. So $5$ cannot divide two consecutive cubes.

Can someone explain in detail why we can translate an equation into a "modulo equation"? Thanks.

Answer 1

The function $\eta : i \mapsto [i]$ that maps $\mathbb{Z}$ to $\mathbb{Z}_5$ is a homomorphism, i.e., it satisfies $\eta(x + y) = \eta(x) + \eta(y)$ and $\eta(xy) = \eta(x)\eta(y)$. If you take a polynomial in $a$, like $f(a) = (a + 1)^3 - a^3 = 3a^2 + 3a + 1$, then $\eta(f(a)) = f_5(\eta(a))$, where I am writing $f_5$ for the "same" polynomial as $f$ but with the coefficients taken over $\mathbb{Z}_5$ (i.e., $[3]a^2 + [3]a + [1]$). Because $\eta$ is a homomorphism it will translate valid equations between polynomials over $\mathbb{Z}$ into valid equations between polynomials over $\mathbb{Z}_5$. (The "converse" of this is false, e.g., $(x^2 - 1)(x^2 - 4)x = 0 $ is valid over $\mathbb{Z}_5$, but not over $\mathbb{Z}$.)

Answer 2

Notice that $[a] \in \mathbb Z_5$ is an residue class whose representant is $a$.

We define the set $$[a] = a + I = \{a + x; x \in I\}$$

where $I$ is an ideal in $\mathbb Z$.

The following are true:

1) $ [ a] = [b] \iff a \equiv b \mod I(5) $, where $I(5)$ is an ideal in $\mathbb Z$.

2) $ [a] \cap [b] \neq \emptyset \implies [a] = [b] $

3) $ \displaystyle {\bigcup_{a \in \mathbb Z}} [a] = \mathbb Z$

In other words a partition. You may define

Addition: $[a + b] = [a] + [b]$

Multiplication: $[a \dot \, b] = [a]\dot\,[b]$

And then you may "translate" to your equation.

Answer 3

Modulo 5, $$0^3\rightarrow0$$ $$1^3\rightarrow1$$ $$2^3\rightarrow3$$ $$3^3\rightarrow2$$ $$4^3\rightarrow4$$ What do you notice about the differences between consecutive integers in the right hand column? How do you think that might answer your question?

Answer 4

For any pair of real numbers $\;x,y\;$ , we have that $\;x^3-y^3=(x-y)(x^2+xy+y^2)\;$ , and thus in this case:

$$(a+i)^3-a^3=(a+1-a)((a+1)^2+(a+1)a+a^2)=3a^2+3a+1$$

The squares modulo $\;5\;$ are resp. $\;0,1,4,4,1\;$ , so in the above we get $\;1\,,\,2\,,\,4\,,\,2\,,\,\,1\;$