Prove that 7^51-1 is a multiple of 103.
According to the rule Ferm 7^102 is a multiple of 103.
I think if 7^51 when divided by 103 gives remainder -1,or say 42, then 7^102 = (7^51)^2 should also give remainder (-1)^2 ,ie, 1 ... so i think you got it wrong somewhere. Correct me if i'm wrong.
but I did it in the calculator and it didn't work!
Learn 'BASIC' and program it. "103" is an odd number, you cannot 'halve' the multiplier and get the same answer.