prove that if R is less than or equal to S and S is less than or equal to N then P(N,S) is divisible by P(N,R)
Answers
Theorem.
If for some positive integer n, 2n-1 is prime, then so is n.
Proof.
Let r and s be positive integers, then the polynomial xrs-1 is xs-1 times xs(r-1) + xs(r-2) + ... + xs + 1. So if n is composite (say r.s with 1<s<n), then 2n-1 is also composite (because it is divisible by 2s-1).
Notice that we can say more: suppose n>1. Since x-1 divides xn-1, for the latter to be prime the former must be one. This gives the following.
Corollary.
Let a and n be integers greater than one. If an-1 is prime, then a is 2 and n is prime.
Usually the first step in factoring numbers of the forms an-1 (where a and n are positive integers) is to factor the polynomial xn-1. In this proof we just used the most basic of such factorization rules, see for some others.
Answer:
Theorem.
If for some positive integer n, 2n-1 is prime, then so is n.
Proof.
Let r and s be positive integers, then the polynomial xrs-1 is xs-1 times xs(r-1) + xs(r-2) + ... + xs + 1. So if n is composite (say r.s with 1<s<n), then 2n-1 is also composite (because it is divisible by 2s-1).
Notice that we can say more: suppose n>1. Since x-1 divides xn-1, for the latter to be prime the former must be one. This gives the following.
Corollary.
Let a and n be integers greater than one. If an-1 is prime, then a is 2 and n is prime.
Usually the first step in factoring numbers of the forms an-1 (where a and n are positive integers) is to factor the polynomial xn-1. In this proof we just used the most basic of such factorization rules, see for some others.
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Step-by-step explanation: