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 Divisibility (Posted on 2007-01-08)
Let A be an integer, P an odd prime and n=3 be the smallest integer for which A^n - 1 is divisible by P.
Determine the smallest integer m for which (A+1)^m - 1 is divisible by P.

 No Solution Yet Submitted by atheron No Rating

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 re: Some more thoughts | Comment 3 of 6 |
(In reply to Some more thoughts by Dej Mar)

obviously m can't be negative or else (A+1)^m would not be an integer.

Also m can not be 1 because we would then have that A is a multiple of P and if that was true we would have that A=Pk and from the first part we would get A^3-1=Pt or (Pk)^3-1=Pt or

t=[(Pk)^3-1]/P=P^2k^3-(1/P) and thus t would not be an integer thus giving us a contradiction therefore m can not be 1

now where it gets tricky is if we allow m=0 because in that case the second part breaks down to saying that 0 is divisble by P and since technically 0 is divisible by any number that would hold true and so it would seem that 0 is the smallest integer m that is being looked for.

So in conlcusion if you allow m=0 then 0 is the answer otherwise m=2 is the answer as shown by Dej Mar

 Posted by Daniel on 2007-01-09 11:02:29

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