2
$\begingroup$

My question: Are the conjectures as follows correct?

Given a positive integer $P>1$, let its prime factorization be written $$P=p_1^{a_1}p_2^{a_2}p_3^{a_3}...p_k^{a_k}$$.
Define the functions $h(P)$ by $h(1)=1$ and $h(P)=min(a_1,a_2,...,a_k)$

Case 1: Let $n \ge 1 $ be positive integers, and $A_i \ne B_j$ are positive integers for all $1 \le i \le n$ and $1 \le j \le n$ with $\gcd(A_1,...,A_n, B_1,...B_n) = 1$

Let $d=min(h(A_1), h(A_2), ...., h(A_n), h(B_1),...,h(B_n))$.

Conjecture: if $\sum_{i=1}^{n} A_i = \sum_{j=1}^{n} B_j$ then $2n \ge d$

Case 2: Let $n \ne m$ and $n, m \ge 1 $ be positive integers, and $A_i, B_j$ are positive integers for all $1 \le i \le n$ and $1 \le j \le m$ with $\gcd(A_1,...,A_n, B_1,...B_m) = 1$

Let $d=min(h(A_1), h(A_2), ...., h(A_n), h(B_1),...,h(B_m))$.

Conjecture: if $\sum_{i=1}^{n} A_i = \sum_{j=1}^{m} B_j$ then $m + n \ge d$

See also:

$\endgroup$
  • 1
    $\begingroup$ I take it you want the $A_i$ to be distinct, likewise the $B_i$, else $A_1=A_2=16$, $B_1=32$ would seem to be a counterexample with $m=1$, $n=2$, $d=4$. $\endgroup$ – Gerry Myerson Aug 4 '19 at 0:00
  • $\begingroup$ Yes, You are right. Thanks you. I need including $gcd(A_1,...,A_n, B_1,...B_m)=1$ $\endgroup$ – Đào Thanh Oai Aug 4 '19 at 1:08
6
$\begingroup$

The conjectures could not be true as stated, due to simple counterexamples such as $3^8+3^8+3^8+2^9=2^8+2^8+3^9$.

One could exclude such constructions by conjecturing, in the spirit of Schmidt's Subspace Theorem, that:

if $n<d$, and $A_i$ ($1 \leq i \leq n$) are nonzero integers with $\gcd(A_1,\ldots,A_n)$ such that $h(|A_i|) \geq d$ for each $i$ and $\sum_{i=1}^n A_i = 0$, then some proper subsum of the $A_i$ vanishes.

(This accounts for the above "simple counterexample": $A_i = 3^8, 3^8, 3^8, 2^9, -2^8, -2^8, -3^9$ has $(n,d)=(7,8)$ but $3^8+3^8+3^8+(-3^9)=0$.)

However, even this refined conjecture is false: there are has counterexamples with $(n,d) = (5,6)$. One is $p^6 + q^6 + q^6 + 61^9 r^6 = 2 s^6$ where $$ \begin{gather} p \; = \!\! & 37471640786194861459344702995419531,\cr q \; = \!\! & 20793522547111333210520476761092295,\cr r \; = \!\! & 3391542261700904858222899444621,\phantom{0000}\cr s \; = \!\! & 33700711308284627431803214879783946, \end{gather} $$ and each of $p^6, q^6, 61^9 r^6, 2 s^6$ has $h=6$ (the last because $s$ is even -- were it not for the single factor of $2$ in $2q^6$, this identity would have given a counterexample with $(n,d)=(4,6)$. A similar counterexample, this one with three positive and two negative $A_i$, is $p^6 + q^6 + q^6 = 11^9 r^6 + 2 s^6$ where $$ {\small \begin{gather} p \; = \!\! & 122143812902307972831486996789219854509652892482229598069 \phantom{0} \cr q \; = \!\! & 1754343120851725061884697722096469904639987931170348892227 \cr r \; = \!\! & 53451023851036429085688858950495539530964060758748930439 \phantom{00} \cr s \; = \!\! & 1088043146197825196095684124547610617079707688400198829578. \end{gather} } $$

Both of these solutions were obtained using the identity $$ (q^2+qs-s^2)^3 + (q^2-qs-s^2)^3 = 2(q^6-s^6). $$ (This identity is not new; Dickson's History of the Theory of Numbers, Vol. II attributes an equivalent identity to Gérardin in 1910, see page 562 note 107.) We cannot nontrivially make both of $|q^2 \pm qs - s^2|$ squares, because that yields elliptic curves of rank zero. But we can make one of them $p^2$ and the other $\delta r_1^2$ for some choices of $\delta$ that yield elliptic curves $E$ of positive rank, and then search the group of rational points for examples with $\delta | r_1$ (so we can use $r = r_1 / \delta$ and obtain solutions of $p^6 \pm \delta^9 r^6 = 2(q^6-s^6)$). The first such $\delta$ is $11$, with $(q,s) = (3,-2)$ making $q^2+qs-s^2 = -1$ and $q^2-qs-s^2 = 11$. One must multiply the generator by $11$ to get $11|r_1$; that's how I found the second example. The first has $\delta = 61$, using an elliptic curve of rank $2$ with independent solutions $(q,s) = (10,3)$ and $(26,15)$; while these are more complicated than the $\delta = 11$ generator, and $61 | r_1$ is harder to get than $11 | r_1$, we still end up with a smaller example thanks to the freedom to choose two multipliers $-$ the one above uses multipliers $4$ and $5$ respectively.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks you very much for the answer. The answer is true. But the equations $3^8+3^8+3^8+2^9=2^8+2^8+3^9$ is sum of two trivial equations: First $3^8+3^8+3^8=3^9$ and second $2^9=2^8+2^8$ but these sub-equations the $gcd(3,3,3,3)=3 \ne 1$, $gcd(2,2,2)=2 \ne 1$. I my mind, I want remove the trivial solution in my conjecture. You are OK if I replace $gcd(A_1,...,A_n,B_1,..,B_m)=1$ by $gcd(A_i,A_j)=1$ and $gcd(A_i,B_j)=1$ ? Do you agree that? If You ok I will state question in the next answer. $\endgroup$ – Đào Thanh Oai Aug 9 '19 at 16:45
  • $\begingroup$ You are very famous. I hope that You like the improvement: mathoverflow.net/questions/338117 $\endgroup$ – Đào Thanh Oai Aug 14 '19 at 11:30
  • $\begingroup$ @NoamD.Ekies, Can You help me to publish one paper in arxiv.org ? mathoverflow.net/questions/339813 I have been already deleted the false conjectures $\endgroup$ – Đào Thanh Oai Sep 8 '19 at 13:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.