the sum comes out to be bm-1-am-1b-a
since a and b are less than m and m is a prime
they are naturally coprime to it
bm-1≡1(modm)
am-1≡1(modm)
If m is a prime number,and a,b ,two numbers less than m prove that am-2+am-3b+am-4b2+...bm-2 is a multiple of m
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1 Answers
akash93
·2009-08-23 23:29:26