# A generalization of Lander, Parkin, and Selfridge conjecture

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 $$m + n \ge d$$

• 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 $m=1$, $n=2$, $d=4$. Aug 4 '19 at 0:00
• Yes, You are right. Thanks you. I need including $gcd(A_1,...,A_n, B_1,...B_m)=1$ Aug 4 '19 at 1:08

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 $$n, and $$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.

• 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. Aug 9 '19 at 16:45
• You are very famous. I hope that You like the improvement: mathoverflow.net/questions/338117 Aug 14 '19 at 11:30
• @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 Sep 8 '19 at 13:16