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Let a and b be distinct integers such that a^2 + ab + b^2|ab(a+b). Show that

     $$ |a - b| > \sqrt[3]{ab} $$

1. Show that |a - b| > \sqrt[3]{ab}

This is a tricky problem. The key here is that we can re-express the problem via coprime terms to gain some insight.

Express a,b as multiples of coprime naturals x,y

Let g = (a,b), then it must be the case that there exists a coprime pair x,y such that a = gx, b = gy. We know that a^2 + ab + b^2|ab(a+b), let’s see what it divides to

(1)   \begin{align*} \frac{ab(a+b)}{a^2 + ab + b^2} &= \frac{g^3xy(x+y)}{g^2(x^2 + xy + y^2)} \\ &= \frac{xy(x+y)g}{x^2 + xy + y^2} \in \mathbb{Z}^+ \end{align*}

Here, we will employ a natural property of \gcd where f|xyg \wedge (f,x)=(f,y)=1 \implies f|g to show that it must be the case that x^2 + xy + y^2|g. In order to claim this, we only need to show that (x^2 + xy + y^2, x) = (x^2 + xy + y^2, y) = (x^2 + xy + y^2, x+y) = 1

Show that (x^2 + xy + y^2,x)=1

Using the simple euclidean algorithm, we reduce the problem to

    \[(x^2 + xy + y^2, x) = (y^2, x)\]

Because by definition (y,x) = 1, then naturally (y^2,x) = 1. \blacksquare

Show that (x^2 + xy + y^2, y)=1

In the same vein, this reduces down to (x^2,y)=1. \blacksquare

Show that (x^2 + xy + y^2, x+y) =1

    \begin{align*} (x^2 + xy + y^2, x+y) &= (x(x + y) + y^2, x+y) \\ &= (y^2,x+y) \end{align*}

Now, because (x+y,y)=(x,y)=1, it must also be the case that (y^2,x+y)=1. Together with the lemma stated above, this concludes our proof that x^2 + xy + y^2|g. \blacksquare

To the finish line!

Because x^2 + xy + y^2|g, it must be the case that g = n(x^2 + xy + y^2) : n \in \bf Z^+. Then, we can derive the following expression from the problem. Ordering a,b such that a>b so that |a-b| = a-b

    \begin{align*} (a-b)^3 &= g^3(x-y)^3 \\ &= g^2(x-y)^3\times g \\ &= g^2 n(x-y)^3 (x^2 + xy + y^2) \\ &> g^2 xy \\ &= ab \\ &\implies \\ a-b &> \sqrt[3]{ab} \end{align*}

This concludes our proof that |a-b| > \sqrt[3]{ab}. \blacksquare

Posted By Lee

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