Euler's sum of powers conjecture: Difference between revisions

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1=m^5-x^5-y^5-z^5 are the obvious ones (that aren't distinct\all >0). Nonetheless, 1 (as the zeroeth 'prime' Po) can be the first of 3 factors (one of
the others being a power of the first 4 primes) for specifying a LHS argument. Given 4 LHS arguments
each raised to a 5th power as is m on the RHS of the equation for which m<=ΣΠPi<2^(3+5), i=0 to 4, the LHS solution (if one exists) will consist of 4 elements of a factorial domain that are each a triplet
(a prime\Po * a prime^1st\2nd power * a prime^1st\4th power) multiple having a distinct pair of the first 4 primes present &
absent. Such potential solutions are generated by aiming for simplicity rather than