Sunday, May 23, 2010

GR9768.61: Greatest Integer, Prime

61.  What is the greatest integer that divides $p^4 -1$ for every prime number p greater than 5?
       (A) $12$
       (B) $30$
       (C) $48$
       (D) $120$
       (E) $240$

Solution:
$p^4 -1 = (p-1)(p^3 + p^2 + p + 1)$.
And we see that $(p-1)$ and $(p^3 + p^2 + p + 1)$ are the divisors of $p^4 -1$.
Guess and check:
$p = 13$ works so A is a divisor.
$p = 31$ works so B is a divisor.
$\vdots$
$p = 241$ works so E is a divisor.
And $p-1 = 240$ is the greatest integer provided.
The answer is E.
-sg-






3 comments:

  1. I'm not sure that I understand your solution. Are you saying that since 12 divides 13^4 - 1, 12 divides p^4 - 1 for all prime p > 5?

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  2. here is a proof:
    http://www.mathematicsgre.com/viewtopic.php?f=1&t=177

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