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Question: If a and b distinct integers, prove that a-b is a factor of \[a^{n}-b^{n}\], whenever n is a positiv...

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

Explanation

Solution

Hint: In this question it is given that if a and b distinct integers, then we have to prove that a-b is a factor of an−bna^{n}-b^{n}, whenever n is a positive integer. So to prove this we need to know the binomial expansion, which is-
(a+b)n= nC0 an+ nC1 an−1b+ nC2 an−2b2+⋯+ nCn bn\left( a+b\right)^{n} =\ ^{n} C_{0}\ a^{n}+\ ^{n} C_{1}\ a^{n-1}b+\ ^{n} C_{2}\ a^{n-2}b^{2}+\cdots +\ ^{n} C_{n}\ b^{n}......(1)
And after that if we are able to show that an−bn=(a−b)⋅ka^{n}-b^{n}=\left( a-b\right) \cdot k (where ‘k’ is a natural number) then we can easily say that a-b is a factor of an−bna^{n}-b^{n}.
Complete step-by-step solution:
So in order to find the solution we can write ana^{n} as,
an=(a)na^{n}=\left( a\right)^{n}
⇒an=(a−b+b)n\Rightarrow a^{n}=\left( a-b+b\right)^{n}
\Rightarrow a^{n}=\left\\{ b+\left( a-b\right) \right\\}^{n}
Now applying the formula (1) by taking a=a and b=a-b, we get,
⇒an= nC0 bn+ nC1 bn−1(a−b)+ nC2 bn−2(a−b)2+⋯ + nCn(a−b)n\Rightarrow a^{n}=\ ^{n} C_{0}\ b^{n}+\ ^{n} C_{1}\ b^{n-1}\left( a-b\right) +\ ^{n} C_{2}\ b^{n-2}\left( a-b\right)^{2} +\cdots \ +\ ^{n} C_{n}\left( a-b\right)^{n}
⇒an=bn+ nC1 bn−1(a−b)+ nC2 bn−2(a−b)2+⋯ + nCn(a−b)n\Rightarrow a^{n}=b^{n}+\ ^{n} C_{1}\ b^{n-1}\left( a-b\right) +\ ^{n} C_{2}\ b^{n-2}\left( a-b\right)^{2} +\cdots \ +\ ^{n} C_{n}\left( a-b\right)^{n} [since, nC0=1{}^{n}C_{0}=1]
⇒an−bn= nC1 bn−1(a−b)+ nC2 bn−2(a−b)2+⋯ + nCn(a−b)n\Rightarrow a^{n}-b^{n}=\ ^{n} C_{1}\ b^{n-1}\left( a-b\right) +\ ^{n} C_{2}\ b^{n-2}\left( a-b\right)^{2} +\cdots \ +\ ^{n} C_{n}\left( a-b\right)^{n}
⇒an−bn= nC1 bn−1(a−b)+ nC2 bn−2(a−b)(a−b)+⋯ + nCn(a−b)(a−b)n−1\Rightarrow a^{n}-b^{n}=\ ^{n} C_{1}\ b^{n-1}\left( a-b\right) +\ ^{n} C_{2}\ b^{n-2}\left( a-b\right) \left( a-b\right) +\cdots \ +\ ^{n} C_{n}\left( a-b\right) \left( a-b\right)^{n-1}
⇒an−bn=(a−b)[ nC1 bn−1+ nC2 bn−2(a−b)+⋯ + nCn(a−b)n−1]\Rightarrow a^{n}-b^{n}=\left( a-b\right) [\ ^{n} C_{1}\ b^{n-1}+\ ^{n} C_{2}\ b^{n-2}\left( a-b\right) +\cdots \ +\ ^{n} C_{n}\left( a-b\right)^{n-1} ]
⇒an−bn=(a−b)⋅k\Rightarrow a^{n}-b^{n}=\left( a-b\right) \cdot k
Where, k=[ nC1 bn−1+ nC2 bn−2(a−b)+⋯ + nCn(a−b)n−1]k=[\ ^{n} C_{1}\ b^{n-1}+\ ^{n} C_{2}\ b^{n-2}\left( a-b\right) +\cdots \ +\ ^{n} C_{n}\left( a-b\right)^{n-1} ] is a real numbers.
Therefore, we can say that a-b is a factor of an−bna^{n}-b^{n}.
Hence proved.
Note: You can also solve this problem by using the Mathematical induction, i.e, firstly, consider an initial value for which the statement is true. It is to be shown that the statement is true for n = initial value=1.
Secondly, assume the statement is true for any value of n = k. Then prove the statement is true for n = k+1. We actually break n = k+1 into two parts, one part is n = k (which is already proved) and try to prove the other part.