TITLE: Proof of $(x+y)^2<\operatorname{rad}\bigl[xy(x+y)(x^2+xy+y^2)\bigr]$ for coprime integers $x,y$
QUESTION [6 upvotes]: An ancient statement, and I once read an elementary proof but I don't recall. Who can help me? The origin of the inequality is a regular magazine for high school teachers issued by P. Noordhoff since 1914 (Nieuw Tijdschrift voor Wiskunde).
I considered various methods to prove the statement above. For instance using the fact that integers $n$ which are of the form $x^2+xy+y^2$ for two relatively prime integers $x,y$, are precisely those positive integers occurring as divisors of $m^2+m+1$ for some integer $m$.
Another approach I tried is to consider the general solutions $x,y$ of $n=x^2+xy+y^2.$ An integer $n$ can be uniquely written as $n=sz^2$ with $s$ squarefree. Let
$$\begin{align*}
s=f^2+fg+g^2,&\quad z=m^2+mn+n^2;\\
x=(f+g)n^2+2gmn-fm^2,&\quad y=-gn^2+2fnm+(f+g)m^2\,.
\end{align*}$$
Then we have $x^2+xy+y^2=(f^2 + fg + g^2)(m^2 + mn + n^2)^2$. The radical of $xy(x+y)(x^2+xy+y^2)$ has the form $xy(x+y)sz$, which leads to the (interesting) identity
$$\bigl(xy(x+y)\bigr)sz\bigl(fg(f+g)\bigr)=sz\prod_{i=1}^3{(u_i^2-sn^2)}\,,$$
where $u_1=gn-fm$, $u_2=(f+g)m+fn$, and $u_3=gm+(f+g)n$.
I also tried to use the following summation identity: let  $e>0$  be an integer and  $w=(-1)^e$. Choose $r=\frac{1}{3}(2^e-w)$ which is an integer. For $f,g,m,n$ define
$$x=\frac14\sum_{i=0}^r\sum_{j=0}^2\binom{2^e}{3i+j}n^jm^{-j}(-1)^in^{3i}m^{2^e-3i}\,\times\\
\bigr[(f+(f + 2g)w)j^2+(-f + 2g +(-5f - 4g)w)j -2f - 2g +(2f - 2g)w\bigl]\,,$$
$$y=\frac14\sum_{i=0}^r\sum_{j=0}^2\binom{2^e}{3i+j}n^jm^{-j}(-1)^in^{3i}m^{2^e-3i}\,\times\\
\bigr[(-f-g-(-f+g)w)j^2+(3f+g -(-f - 5g)w)j + 2g -(4f+2g)w\bigl]\,.$$
Then the equality $x^2+xy+y^2=(f^2+fg+g^2)(m^2+mn+n^2)^{2^e}$ holds, which enable us to find all solutions with $s_i$ square-free of the form $s_0\prod_k{s_k^{2k}}$.
N.B.: Please note that the transformation $x\mapsto x+y,y\mapsto-y$ yields the same value for $x^2+xy+y^2$.
Why of interest? A corollary of $(x+y)^2<\operatorname{rad}\bigl[xy(x+y)(x^2+xy+y^2)\bigr]$ is that for $ABC$ triples with $A+B=C$ and $\operatorname{rad}(ABC)<C$, we have $A^2+AB+B^2=C^2-AB$. Thus,
$$\operatorname{rad}(C^2-AB)<\leq C^2-AB\operatorname{rad}(ABC)\operatorname{rad}(C^2-AB)-AB<C\operatorname{rad}(C^2-AB)-AB\,,$$
which implies $AB<(C-1)\operatorname{rad}(C^2-AB)\,.$ We can also conclude that $C=A+B<\operatorname{rad}(C^2-AB)$, and thus for small values of $\operatorname{rad}(C^2-AB)=3,7,13,19,21\dots$ (OEIS sequence A034017) there are only a limited number of ABC triples.

REPLY [3 votes]: Assuming the ABC conjecture, for each $\theta < 3$ there exist
only finitely many solutions of
$$
(|x|+|y|)^\theta < \text{rad}\left( xy(x+y)(x^2+xy+y^2) \right)
$$
in relatively prime integers $x,y$.  It may be that $\theta = 2$
is small enough that there are no solutions at all,
though it is conceivable that a counterexample exists
(see below for some further examples of $\theta < 2.5$).
I doubt that one can prove such a result by elementary means
for any $\theta > 0$; it seems to be comparable in difficulty
with the ABC conjecture itself.
The reduction to ABC can be obtained from the polynomial identity
$$
\left( (x-y)(2x+y)(x+2y) \right)^2 + 27 (xy(x+y))^2 = 4 (x^2+xy+y^2)^3.
$$
If $\gcd(x,y) = 1$ then $\gcd(x^2+xy+y^2, xy(x+y)) = 1$,
so $\gcd(4(x^2+xy+y^2)^3, 27(xy(x+y))^2)$ is bounded
(it must be a factor of $4 \cdot 27$).  Divide through
by the common factor to get an identity $A+B=C$ in coprime integers
with $C \gg H^6$ and $\text{rad}(A)$ no larger than
$$
\text{rad}\left( \bigl( (x-y)(2x+y)(x+2y) \bigr)^2 \right)
 = \text{rad}\bigl( (x-y)(2x+y)(x+2y) \bigr) \ll H^3,
$$
where $H = |x| + |y|$.  So the ABC conjecture implies
$\text{rad}(B) \, \text{rad}(C) \gg_\epsilon H^{6(1-\epsilon) - 3}$
for all $\epsilon>0$.  Since $\text{rad}(B)$ and $\text{rad}(C)$
are within bounded factors of the radicals of $xy(x+y)$ and $x^2+xy+y^2$,
we deduce that the radical of $xy(x+y)(x^2+xy+y^2)$ is $\gg_\theta H^\theta$
for each $\theta<3$, as claimed.
In general, suppose $P(x,y)$ is a homogeneous polynomial of degree
$d \geq 3$ without repeated factors.  Then one expects that
$\text{rad}(P(x,y)) \gg_\theta H^\theta$ for all $\theta < d-2$.
The ABC conjecture is the special case $P(x,y) = xy(x+y)$ (with $d=3$).
The general case can be deduced from this via a suitable polynomial identity,
such as the one used above for the $d=5$ case $P(x,y) = xy(x+y)(x^2+xy+y^2)$;
such identities are provided by Belyi's theorem, as in my paper

Noam D. Elkies: ABC implies Mordell, International Math. Research Notices 1991 #7, 99-109 [bound with Duke Math. J. 64 (1991)].

P.S. Here are some more examples with small $\theta$ that fall outside
the range of Will Jagy's search.  None of these $\theta$ is
smaller than his record of $2.261$ for $(x,y) = 3^7, 5^4$.
I have not tried an exhaustive search, just things like
$x,y$ with small radicals or for which $x^2+xy+y^2$ is divisible
by cube or higher power of a prime.
 2.28587:  77824, 17
 2.30953:  15125, 496
 2.37769:  26973, 169
 2.37809:  1092663, 64
 2.39904:  5375, 2401
 2.40001:  6889472, 74925

with a couple dozen further examples of $\theta < 2.5$, including
$\theta \approx 2.48742$ for $(x,y) = (18428663, 5371345)$
and $\theta \approx 2.48838$ for $(x,y) = (250000, 1)$.