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> > > **My question:** Are the conjectures as follows correct?
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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 $d=min(h(A\_1), h(A\_2), ...., h(A\_n), h(B\_1),...,h(B\_n))$.



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> > **Conjecture:** if $\sum\_{i=1}^{n} A\_i = \sum\_{j=1}^{n} B\_j$ then $2n \ge d$
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**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 $d=min(h(A\_1), h(A\_2), ...., h(A\_n), h(B\_1),...,h(B\_m))$.



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> > **Conjecture:** if $\sum\_{i=1}^{n} A\_i = \sum\_{j=1}^{m} B\_j$ then $m + n \ge d$
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**See also:**


* [Lander, Parkin, and Selfridge conjecture](https://en.wikipedia.org/wiki/Lander,_Parkin,_and_Selfridge_conjecture)
* [Minimal exponent in prime factorization of n A051904](https://oeis.org/A051904)
* [Nivens Constant](http://mathworld.wolfram.com/NivensConstant.html)



A generalization of Lander, Parkin, and Selfridge conjecture
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.
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> 


(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.


