prime number proof help

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.

Look at it the "other way around": $\displaystyle p^2= 54K- 81= 9(6- k)$

Quote:

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Originally Posted by HallsofIvy

Look at it the "other way around": $\displaystyle p^2= 54K- 81= 9(6- k)$

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Not sure what you mean....

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

hence 9 divides p^2 ?

so 9 divides p ?

if p=9(mod54), p²=81(mod54)

p-9=kx54
p=9(1+6k)
p can't be prime.

Quote:

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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.

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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.

Quote:

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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 ?

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No, either 9 divides p (if 9 also divides 6-k) or 3 divides p. Either way, p is not prime.