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- jhonyy9

- for example :
(2-1) (2-1)
----- + ------ +1 = 2
2 2
(3-1) (3-1)
----- + ----- +1 =3
2 2
(5-1) (3-1)
----- + ----- +1 = 2 +1 +1 =4
2 2
(5-1) (5-1)
----- + ----- +1 = 2+2+1=5
2 2
can be this proven that is true for every natural numbers ?

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- jhonyy9

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- jhonyy9

can be do it by induction ?

- EscherichiaRinku

The difficult part here is probably getting a general formula. Then proving it by induction shouldn't be hard.

- hoblos

(a-1) (a-1) 2a -2
----- + ------ +1 = ------- +1 = a-1+1 =a
2 2 2

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- hoblos

this is true for any natural number

- jhonyy9

,,hoblos" yes i see it but this need to be proven

- jhonyy9

-so if we know that p and k are primes

- jhonyy9

- 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

- jhonyy9

this was my previosly question

- jhonyy9

so than p=2a+1
and k=2b+1

- hoblos

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

- jhonyy9

so but do you can prove it in one proning style ,method that is true for every natural numbers ?

- jhonyy9

sorry proving stayle ,method

- jhonyy9

style ,method

- hoblos

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

- jhonyy9

not right because if you see the first sentence of this exercise there is that p and k are prims

- hoblos

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!!

- jhonyy9

but is indifferent from this exercise p and k are prims

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