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-

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?

ReplyDeletehere is a proof:

ReplyDeletehttp://www.mathematicsgre.com/viewtopic.php?f=1&t=177

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