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Question: If a and b are distinct integers, prove that a-b is a factor of \( {a^n} - {b^n} \) , whenever n is ...

If a and b are distinct integers, prove that a-b is a factor of an−bn{a^n} - {b^n} , whenever n is a positive integer.

Explanation

Solution

This question will be solved by the principal of mathematical induction. First of all, we will see the definition of mathematical induction and then its principal. The principal basically consists of two steps: one is a base step which checks stamen for initial value and the second step is an inductive step which checks for n=k and n= k+1.

Complete step-by-step answer:
Let us first see the definition of mathematical induction:
Mathematical induction is a method which is used to prove a theorem or formula which is true for every natural number.
Now, we will see the principle of mathematical induction.
Suppose the given statement is p(n) , involving the natural number n such that:
(i) The statement is true for n= 1i.e. P(1) is true, and
(ii) If the statement is true for n=k (where k is some positive integer), then the statement is also true for n= k+1, truth of P(k) implies the truth of P(k+1).
Following the procedure, let us take the initial value of n=1.
The given expression is an−bn{a^n} - {b^n} .
For n = 1, the expression an−bn{a^n} - {b^n} = a – b which is factor of an−bn{a^n} - {b^n}
Now let us take n = k.
The expression that we get = ak−bk.{a^k} - {b^{k.}} .
Suppose ak−bk.{a^k} - {b^{k.}} = λ\lambda (a-b). (1)
Again take the value of n = k+1.
Putting in the given expression, we get:
ak+1−bk+1.{a^{k + 1}} - {b^{k + 1.}} .
On further solving the above expression, we have:
ak+1−bk+1.{a^{k + 1}} - {b^{k + 1.}} = a.ak−b.bk.a.{a^k} - b.{b^{k.}}

Putting the value from equation 1, we get:
a.ak−b.bk.=a[bk+λ(a−b)]−bk+1=abk+bk+1+λa(a−b)a.{a^k} - b.{b^{k.}} = a\left[ {{b^k} + \lambda (a - b)} \right] - {b^{k + 1}} = a{b^k} + {b^{k + 1}} + \lambda a(a - b) .
= bk(a−b)+λa(a−b){b^k}(a - b) + \lambda a(a - b)
= (a−b)(bk+λa)(a - b)({b^k} + \lambda a) = λ’(a−b)\lambda^{’} (a - b)
So for every iteration we get the same value.
Therefore, from the principle of mathematical induction, an−bn{a^n} - {b^n} is divisible by (a-b) for every natural number n.

Note: Important thing to be noted is that you have to follow the two steps to prove this kind of question. First step is the base step where you have to take the initial value. There may be situations when a statement is true for all n⩾4n \geqslant 4 . In this case, step1 will start from n=4 and we shall verify the result for n=4 i.e.P(4).