Re: Math Problem

Gene Wirchenko <genew@xxxxxxxx> wrote: (*edited*)

a = b^c, a,b,c,d,e integers.

b mod 8 is neither 1 nor prime.

b is even, not 1,3,5,7 mod 8

c = e 2^d, d prime, e = d^2 - 3

4 | c > 0 via d >= 2

not 5|a(b + c)(d + e)

not 5|b, else 5 | b^c = a contra prior

In simplest terms, what is the remainder when a is divided by 10?

a = b^c, not 5|b, 4|c -> a = 1 (mod 5) by Fermat's Little Theorem

b even -> a even -> a = 6 (mod 10)

--Bill Dubuque
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