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), p^{2}=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.

p^{2}=81+54k

p^{2}=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.