Beal's conjecture is a generalization of Fermat's Last Theorem. It states: If $A^x + B^y = C^z$, where $A, B, C, x, y$ and $z$ are positive integers and $x, y$ and $z$ are all greater than $2$, then $A$, $B$ and $C$ must have a common prime factor.
According to Fermat's Little Theorem $(A^x-A)/x$, where $A$ and $x$ are positive odd integers > 3 will always lead to a whole number, but so does $(A^x-A)/6$.
It opens the possibility to compare different $A^x$'s by feeding any value to $A$ of $x$ and determining how many $6's$ there are and what is the remaining such as in the following:
$3 ^ 3 = 27 = 4 * 6 + 3 , 3 ^ 5 = 243 = 40 * 6 + 3 , 3 ^ 7 = 2187 = 364 * 6 + 3 , 3 ^ 9 = 19683 = 3280 * 6 + 3$
$(4,40,364,3280...)$ grows by: $* 9 + 4 = * 3 ^ 2 + 4$
$5 ^ 3 = 125 = 20 * 6 + 5 , 5 ^ 5 = 3125 = 520 * 6 + 5 , 5 ^ 7 = 78125 = 13020 * 6 + 5 , 5 ^9 = 1953125 = 325520 * 6 + 5$
$(20,520,13020,325520...)$ grows by: $* 25 + 20 = * 5 ^ 2 + 20$
$7 ^ 3 = 343 = 57 * 6 + 1 , 7 ^ 5 = 16807 = 2801 * 6 + 1 , 7 ^ 7 = 823543 = 137257 * 6 + 1 , 7 ^ 9 = 40353607 = 6725601 * 6 + 1$
$(57,2801,137257,6725601...)$ grows by:$* 49 + 8 = * 7 ^ 2 + 8$
$9 ^ 3 = 729 = 121 * 6 + 3 , 9 ^ 5 = 59049 = 9841 * 6 + 3 , 9 ^ 7 = 4782969 = 797161 * 6 + 3 , 9 ^ 9 = 387420489 = 64570081 * 6 + 3$
$(121,9841,797161,64570081...)$ grows by:$* 81 + 40 = 9 ^ 2 + 40$
$11 ^ 3 = 1331 = 221 * 6 + 5 11 ^ 5 = 161051 = 26841 * 6 + 5 11 ^ 7 = 19487171 = 3247861 * 6 + 5 11 ^ 9 = 2357947691 = 392991281 * 6 + 5$
$(221,26841,3247861,392991281...)$ grows by:$* 121 + 100 = 11^ 2 + 100$
$13 ^ 3 = 2197 = 366 * 6 +1 , 13 ^ 5 = 371293 = 61882 * 6 + 1 , 13 ^ 7 = 62748517 = 10458086 * 6 + 1 , 13 ^ 9 = 10604499373 = 1767416562 * 6 + 1$
$(366,61882,10458086,1767416562...)$ grows by:$* 169 + 28 = 13 ^ 2 + 28$
$15 ^ 3 = 3375 = 562 * 6 + 3 , 15 ^ 5 = 759375 = 126562 * 6 + 3 , 15 ^ 7 = 170859375 = 28476562 * 6 + 3 , 15 ^ 9 = 38443359375 = 6407226562 * 6 + 3$
$(562,126562,28476562,6407226562 ...)$ grows by:$* 225 + 112 = 15 ^ 2 + 112$
Going back to the Beal's conjecture, heuristics obviously will not prove the theory for infinity, however for $A^x + B^y = C^z$, where $A$ and $B$ are different constants, can we use the above info to know if they can add to a $C^z$, for example if $A = 5$ and $B = 3$?