If $a \mid b^2$, $b \mid c^2$, $c \mid a^2$ then prove $abc \mid (a+b+c)^7$

Question

I tried using the binomial theorem but doesn't makes sense. Tried multiplying and dividing but couldn't get to something. Can someone tell how to approach these questions?

Answer 1

Notice that by definition of divisiblity we have $b^2=ra$, $c^2=sb$ and $a^2=tc$ for some $r,s,t \in \mathbb{Z}$, and hence $$r^2s^2t^2a^2 = s^2t^2b^2 = t^2c^2 = a^2$$ Assuming wlog$^{\dagger}$ that $a \ne 0$, we have $r^2s^2t^2 = 1$, and hence each of $r,s,t$ are equal to $1$ or $-1$. But then we have $$c^2 = (\pm b)^2 = b^2 = (\pm a)^2 = a^2$$ and so $|a|=|b|=|c| = k$, say. But then $$abc = {\pm k^3} \quad \text{and} \quad (a+b+c)^7 = {\pm 3^7 k^7} \text{ or } {\pm k^7}$$

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$^{\dagger}$If $a=b=c=0$ then the result is trivial. Otherwise we can assume $a^2$ without loss of generality because the problem is symmetric in $a,b,c$.

Answer 2

$$(a+b+c)^7 \equiv (a+b)^7 + (b+c)^7 + (a+c)^7 - a^7 - b^7 - c^7 \pmod {abc}$$

The above computation wasn't really necessary, but it verifies that we are left with terms that only have two variables. So we only have 4 things to check.

Assume $b^2=ra$, $c^2=sb$, $a^2=tc$.

1. $a^7 = aa^6 = a(tc)^3 = a(tc)(t^2sb) = abc(st^3)$
2. $a^6 b = abc(a^3t)$
3. $a^5 b^2 = abc(a^2bt)$
4. $a^4 b^3 = abc(ab^2t)$

So each term in $(a+b+c)^7$ is divisible by $abc$ (which is a stronger result).

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This also shows that it is true in the universal ring $\Bbb Z[a,b,c,r,s,t]/(b^2-ra,c^2-sb,a^2-tc)$ (Grothendieck's point of view), hence in every commutative ring.