6

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.

Rolandb
  • 445
  • are the numbers positive? – Will Jagy Jan 16 '14 at 02:55
  • What is "rad"???? – Igor Rivin Jan 16 '14 at 03:46
  • 1
    @IgorRivin, it is the product of all primes that divide the thing; it is squarefree. And it is multiplicative, as in $rad(ab) = rad(a) ; rad(b)$ when $\gcd(a,b) = 1.$ Which applies here, $x,y,x+y,x^2 + xy + y^2$ are pairwise coprime – Will Jagy Jan 16 '14 at 03:49
  • @IgorRivin http://en.wikipedia.org/wiki/Radical_of_an_integer – Will Jagy Jan 16 '14 at 03:58
  • @WillJagy I had actually guessed that, but I did not realize that $\mathop{rad}$ was standard notation... – Igor Rivin Jan 16 '14 at 03:59
  • user, recommend you give much more background; where did you see this in the first place. Also, i am doing a huge computer search myself. You should do a computer run with, say, $1 \leq x < y \leq 100$ with $\gcd(x,y) = 1,$ confirm my calculation that the four $(x,y,x+y,x^2 + xy+y^2)$ are then pairwise relatively prime, and that your inequality holds. And, again, where did you get this? – Will Jagy Jan 16 '14 at 18:51

2 Answers2

3

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 $(x,y) = (250000, 1)$.

1

Certainly seems plausible, now that I've had a chance to test it. Other than $x=1,y=0$ the right hand side is generally much larger than the left hand side. I decided to simply take the ratio of the logarithms, for a long time $x=18,y=1$ is best, and then $x=3^7, y=5^4$ does just a bit better, largely because $6540469 = 229 \cdot 13^4.$ Here are the best (smallest) log ratios I have found so far. Well above 2.0 in all cases.

log ratio     x     y   x+y    x^2+xy+y^2  RADICAL(stuff)
2.26089      2187  625  2812     6540469   62784930 = 2 3 5 13 19 37 229
2.2694        18    1    19         343   798 = 2 3 7 19
2.27367      3267   10  3277    10706059   98408310 = 2 3 5 7 11 13 29 113
2.27453      3277  -10  3267    10706059   98408310 = 2 3 5 7 11 13 29 113
2.31185        19   -1    18         343   798 = 2 3 7 19
2.33479      2812 -625  2187     6540469   62784930 = 2 3 5 13 19 37 229
2.35811       324    1   325      105301   838110 = 2 3 5 7 13 307
2.35937       325   -1   324      105301   838110 = 2 3 5 7 13 307
2.39648      2917   27  2944     8588377   205701006 = 2 3 7 23 73 2917
2.39925      2944  -27  2917     8588377   205701006 = 2 3 7 23 73 2917
2.4078       123    2   125       15379   111930 = 2 3 5 7 13 41
2.41308      2891  565  3456    10310521   345817290 = 2 3 5 7 13 19 59 113
2.41587       125   -2   123       15379   111930 = 2 3 5 7 13 41
2.42585       243   13   256       62377   695058 = 2 3 7 13 19 67
2.44886       256  -13   243       62377   695058 = 2 3 7 13 19 67
2.45693      4624    1  4625    21386001   1011413130 = 2 3 5 7 13 17 19 31 37
2.45699      4625   -1  4624    21386001   1011413130 = 2 3 5 7 13 17 19 31 37
2.46713      3456 -565  2891    10310521   345817290 = 2 3 5 7 13 19 59 113
2.4783      3125   34  3159     9873031   470922270 = 2 3 5 7 13 17 73 139
2.48164      3159  -34  3125     9873031   470922270 = 2 3 5 7 13 17 73 139
2.48222      4293 2500  6793    35412349   3251061870 = 2 3 5 7 43 53 6793
2.48649       361  179   540      226981   6223830 = 2 3 5 19 61 179
2.49388       983   41  1024     1008273   32161794 = 2 3 7 19 41 983
2.50061       729  112   841      625633   20595162 = 2 3 7 29 37 457
2.50347       323   37   360      117649   2509710 = 2 3 5 7 17 19 37
2.50867      1024  -41   983     1008273   32161794 = 2 3 7 19 41 983
2.51652       379  128   507      208537   6414954 = 2 3 7 13 31 379
2.55046       360  -37   323      117649   2509710 = 2 3 5 7 17 19 37
2.55285      2084  125  2209     4619181   344532090 = 2 3 5 7 47 67 521
2.554      3843  125  3968    15264649   1551508770 = 2 3 5 7 31 61 3907
2.55482       841 -112   729      625633   20595162 = 2 3 7 29 37 457
2.5639      3968 -125  3843    15264649   1551508770 = 2 3 5 7 31 61 3907
2.56399       121    4   125       15141   237930 = 2 3 5 7 11 103
2.57008       800   33   833      667489   32083590 = 2 3 5 7 11 17 19 43
2.57142         5    3     8          49   210 = 2 3 5 7
2.57231      2209 -125  2084     4619181   344532090 = 2 3 5 7 47 67 521
2.57277        39   16    55        2401   30030 = 2 3 5 7 11 13
2.57615       729  640  1369     1407601   120187470 = 2 3 5 13 37 8329
2.57754       175   32   207       37249   932190 = 2 3 5 7 23 193
2.58137       125   -4   121       15141   237930 = 2 3 5 7 11 103
2.58562       833  -33   800      667489   32083590 = 2 3 5 7 11 17 19 43
2.58717      4563 4312  8875    59089969   16389683310 = 2 3 5 7 11 13 71 7687
2.5937       459  245   704      383161   24308130 = 2 3 5 7 11 17 619
2.59497       513   16   529      271633   11675766 = 2 3 19 23 61 73
2.60005      4880   81  4961    24216241   4061448930 = 2 3 5 7 11 19 37 41 61
2.60042       161   15   176       28561   690690 = 2 3 5 7 11 13 23
2.60047       640  243   883      624169   45801210 = 2 3 5 7 13 19 883
2.60509      4961  -81  4880    24216241   4061448930 = 2 3 5 7 11 19 37 41 61
2.60774       529  -16   513      271633   11675766 = 2 3 19 23 61 73
2.60842      2816 2325  5141    19882681   4785911130 = 2 3 5 7 11 13 31 53 97
2.61086      2080   53  2133     4439449   491511930 = 2 3 5 7 13 43 53 79
2.61418       225   64   289       69121   2711670 = 2 3 5 13 17 409
2.61498       499    1   500      249501   11422110 = 2 3 5 7 109 499
2.61582       500   -1   499      249501   11422110 = 2 3 5 7 109 499
2.61798       208   17   225       47089   1438710 = 2 3 5 7 13 17 31
2.61946      2133  -53  2080     4439449   491511930 = 2 3 5 7 13 43 53 79
2.62002      4875  328  5203    25472209   5453117670 = 2 3 5 7 11 13 41 43 103
2.62174       800  531  1331     1346761   155156430 = 2 3 5 11 13 59 613
2.62297        81    2    83        6727   108066 = 2 3 7 31 83
2.62706      3055 1089  4144    13845841   3185552370 = 2 3 5 7 11 13 37 47 61
2.63682      2048  227  2275     4710729   710807370 = 2 3 5 7 13 31 37 227
2.63753        83   -2    81        6727   108066 = 2 3 7 31 83
log ratio     x     y   x+y    x^2+xy+y^2  RADICAL(stuff)
Will Jagy
  • 146,052