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:

  • 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 $m=1$, $n=2$, $d=4$. $\endgroup$ Aug 4, 2019 at 0:00
  • $\begingroup$ Yes, You are right. Thanks you. I need including $gcd(A_1,...,A_n, B_1,...B_m)=1$ $\endgroup$ Aug 4, 2019 at 1:08

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Aug 9, 2019 at 16:45
  • $\begingroup$ You are very famous. I hope that You like the improvement: mathoverflow.net/questions/338117 $\endgroup$ Aug 14, 2019 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 $\endgroup$ Sep 8, 2019 at 13:16

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