but then it won't satisfy for all condition if I restrict it
BTW can I use this counterexample
say that the number is 3 x 5 x 7 x 11 x 13
and the first 3 prime number is 3,5,and 7... if we use multiply these three number we get 105 but then we add 1 to this number it becomes 106 which is not a prime number.. therefore the conjecture is false
I don't see that to be a disagreeable issue. For instance there are many formulas that are only applicable to certain circumstances. And there are numerous proofs I've seen that use Mathematical Induction where the theorem isn't true for the first few values of n but are true for, say, n = 3 and above.
-Dan
That is false.
$\displaystyle 2+1 = \mbox{ prime}$
$\displaystyle 2\cdot 3 + 1 = \mbox{ prime}$
$\displaystyle 2\cdot 3\cdot 5 + 1 = \mbox{ prime}$
$\displaystyle 2\cdot 3\cdot 5\cdot 7 + 1 = \mbox{ prime}$
$\displaystyle 2\cdot 3\cdot 5\cdot 7\cdot 11+1 = \mbox{ prime}$
$\displaystyle 2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13 + 1 = 59\cdot 509$
I am now totally confused! Why did you mention that there is a number outside 2 x 3 x 5 x ... that if is multiplied can produce the same number. I am confused why you keep mentioning:
$\displaystyle 2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13 + 1 = 59\cdot 509 $
can you explain
Because you are trying to show that for any sequence of primes, their product plus one is a prime. This is not true because
$\displaystyle 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$
which your proposition states should be a prime number is not a prime number.
-Dan