Re: prime number proof help
Look at it the "other way around": $\displaystyle p^2= 54K- 81= 9(6- k)$
Re: prime number proof help
Quote:
Originally Posted by
HallsofIvy
Look at it the "other way around": $\displaystyle p^2= 54K- 81= 9(6- k)$
Not sure what you mean....
so $\displaystyle p^{2} = 9(6-k) $
hence 9 divides p^2 ?
so 9 divides p ?
Re: prime number proof help
if p=9(mod54), p2=81(mod54)
p-9=kx54
p=9(1+6k)
p can't be prime.
Re: prime number proof help
Quote:
Originally Posted by
Tweety
prove tha there is no prime number p such that $\displaystyle p^{2} -81 $ is divisible by 54.
suppose $\displaystyle 54/ p^{2} - 81 $
than
$\displaystyle p^{2} -81 = 54K $ for some K in the naturals.
how would I proceed from here? very confused, I am suppose to come up with some sort of contradiction.
any help appreciated.
p2=81+54k
p2=9(9+6k)
p=3(9+6k)1/2
so p can't be prime even if (9+6k)1/2 is an integer.
Re: prime number proof help
Quote:
Originally Posted by
Tweety
Not sure what you mean....
so $\displaystyle p^{2} = 9(6-k) $
hence 9 divides p^2 ?
so 9 divides p ?
No, either 9 divides p (if 9 also divides 6-k) or 3 divides p. Either way, p is not prime.