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The greatest common divisor of all numbers of the form $p^2 − 1$, where $p \geq 7$ is a prime, is

  1. $6$
  2. $12$
  3. $24$
  4. $48$
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4 Comments

Option B
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Why not A??
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Any prime number is of the form 6k+1 or 6k +5

Now put 6k+1 or 6k+5 in place of p in the expression (p2 -1) you will see it is always divisible by 12.

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2 Answers

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$p^2-1$=$(p-1)$$\times$$(p+1)$.

Since p is a prime number, one of p-1 or p+1 must be divisible by 3, and since p-1 is even and p+1 is also even. then one of them is divisible by 2 and the other by 4 . Therefore it is divisible by 4$\times$3$\times$2=24.

$\therefore$ G.C.D would be 24.
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Answer is (C)
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