jhonyy9
  • 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 ?
Mathematics
katieb
  • katieb
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jhonyy9
  • jhonyy9
can be do it by induction ?
EscherichiaRinku
  • EscherichiaRinku
The difficult part here is probably getting a general formula. Then proving it by induction shouldn't be hard.
hoblos
  • hoblos
(a-1) (a-1) 2a -2 ----- + ------ +1 = ------- +1 = a-1+1 =a 2 2 2

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hoblos
  • hoblos
this is true for any natural number
jhonyy9
  • jhonyy9
,,hoblos" yes i see it but this need to be proven
jhonyy9
  • jhonyy9
-so if we know that p and k are primes
jhonyy9
  • 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
  • jhonyy9
this was my previosly question
jhonyy9
  • jhonyy9
so than p=2a+1 and k=2b+1
hoblos
  • 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
  • jhonyy9
so but do you can prove it in one proning style ,method that is true for every natural numbers ?
jhonyy9
  • jhonyy9
sorry proving stayle ,method
jhonyy9
  • jhonyy9
style ,method
hoblos
  • 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
  • jhonyy9
not right because if you see the first sentence of this exercise there is that p and k are prims
hoblos
  • 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
  • jhonyy9
but is indifferent from this exercise p and k are prims

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