>

Remember that Fermat numbers are of the form 2^(2^a) + 1 .

>

> m integer >1

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> 2^n+1 mod 2^m-1

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> the only solution is : 2^3+1 Mod 2^2-1

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> 9 Mod 3

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> 9 = 0 Mod 3

> thanks

>

But if you want any positive exponent n (not of the form 2^a), then

(2^1 + 1) mod (2^2 - 1) = 0

(2^3 + 1) mod (2^2 - 1) = 0 (yours)

(2^5 + 1) mod (2^2 - 1) = 0

etc.

Mark