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can be do it by induction ?
The difficult part here is probably getting a general formula. Then proving it by induction shouldn't be hard.
(a-1) (a-1) 2a -2 ----- + ------ +1 = ------- +1 = a-1+1 =a 2 2 2
this is true for any natural number
,,hoblos" yes i see it but this need to be proven
-so if we know that p and k are primes
- prove that for every n natural numbers grater or equal 2 exist one p and k numbers prim grater or equal 2 such that the below equation is true : (p-1) (k-1) ----- + ----- +1 = n 2 2
this was my previosly question
so than p=2a+1 and k=2b+1
actually there exist more than one p & k such that the equation is true!! for example: if n=4 you may take (p=5 k=3) or (p=7 k=1) but there is one condition that must be satisfied which is n=(p+k)/2
so but do you can prove it in one proning style ,method that is true for every natural numbers ?
sorry proving stayle ,method
well.. i have only this method for now for all natural numbers p & k , there exist n such that (p-1) (k-1) ----- + ----- +1 = n => (p+k)/2 -1+1=n => n=(p+k)/2 2 2
not right because if you see the first sentence of this exercise there is that p and k are prims
they dont have to be primes try p=6 k=4 and you will get n=5 the exercise is also stating that we have only one p&k while we have more than one!!
but is indifferent from this exercise p and k are prims