Question

Show that p-1 is a factor of p¹⁰ -1 and also of ¹¹ -1.​

Answer 1

how to prove that (p^-1) is a factor of both p^10-1 and p^11-1

Lets Go

Assuming here that  g(p) = p^10-1.

  H (p) = p^11-1, we try and plug in values for p -1 in equation g(p) = p^10-1, we get

  g(1) = 1^10-1 = 1 – 1 (as 1^10 = 1)

  g(1) = 1-1 = 0. So, p-1 is a factor of g(p).

Now again, we plug in p = 1 in second equation, and we get

  h(1) = (1) 11-1 = (1)^10-1 = 1-1 = 0.

Hence p-1 is a factor of h(p).

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Answer 2

Answer: Hope this helps.

Step-by-step explanation:

Let , p - 1 = 0

p = 1

For p−1 to be factor, the equation on substituting p=1 should equate to 0

i) f(p) =  $p^{10}$  − 1

f(1) = 1 − 1=0

ii) f(p )=   $p^{11}$ − 1

f(1) = 1 − 1 = 0

∴ Hence p−1 is a factor of both $p^{10}$  − 1  and  $p^{11}$ − 1