4
$\begingroup$
FullSimplify[ Divisible[p^2 - 1, 24] , Element[p, Primes] && p > 3]

Should evaluate to True, but I get

Divisible[-1 + p^2, 24]

Compare with

FullSimplify[Divisible[(6 k - 1)^2 - 1, 24], k \[Element] Integers] 

Which does evaluate to

True
$\endgroup$
  • 2
    $\begingroup$ Actually you can show a much simpler case for Divisible: Try this: Assuming[Element[p, Primes] && p > 3, Divisible[p, 2]] I think Divisible works only for actual numbers. From help it says : works for integers or rational number but this is my educated guess ;) $\endgroup$ – Nasser Aug 21 '13 at 6:17

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