# prime number proof help

• Nov 19th 2013, 05:58 AM
Tweety
prime number proof help
prove tha there is no prime number p such that $p^{2} -81$ is divisible by 54.

suppose $54/ p^{2} - 81$

than

$p^{2} -81 = 54K$ 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.
• Nov 19th 2013, 06:15 AM
HallsofIvy
Re: prime number proof help
Look at it the "other way around": $p^2= 54K- 81= 9(6- k)$
• Nov 19th 2013, 06:25 AM
Tweety
Re: prime number proof help
Quote:

Originally Posted by HallsofIvy
Look at it the "other way around": $p^2= 54K- 81= 9(6- k)$

Not sure what you mean....

so $p^{2} = 9(6-k)$

hence 9 divides p^2 ?

so 9 divides p ?
• Nov 19th 2013, 10:38 AM
Hartlw
Re: prime number proof help
if p=9(mod54), p2=81(mod54)

p-9=kx54
p=9(1+6k)
p can't be prime.
• Nov 19th 2013, 12:27 PM
Hartlw
Re: prime number proof help
Quote:

Originally Posted by Tweety
prove tha there is no prime number p such that $p^{2} -81$ is divisible by 54.

suppose $54/ p^{2} - 81$

than

$p^{2} -81 = 54K$ 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.
• Nov 19th 2013, 01:39 PM
HallsofIvy
Re: prime number proof help
Quote:

Originally Posted by Tweety
Not sure what you mean....

so $p^{2} = 9(6-k)$

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.