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Chapter 4: Problem 12

If \(n \in \mathbf{Z}^{+}\), prove that 57 divides \(7^{n+2}+8^{2 n+1}\).

Short Answer

Expert verified
By using the principle of mathematical induction, the statement '57 divides \(7^{n+2}+8^{2n+1}\) for every positive integer \(n\)' can be proved to be true.

Step by step solution

01

Base Case

First, we should verify if the condition holds for the smallest possible positive integer. In this case, we can substitute \(n = 1\) into \(7^{n+2}+8^{2n+1}\) and check whether the resulting number can be divided by 57. If so, we have verified our base case. The resulting number should be \(7^3 + 8^3 = 343 + 512 = 855\). When 855 is divided by 57, the quotient is 15 and the remainder is 0, so our base case holds true.
02

Inductive Step

Now, we need to make an assumption for the inductive step. We assume that the given expression is divisible by 57 for some arbitrary but particular \(n\), thus \(7^{n+2}+8^{2n+1}\) equals \(57k\), for some integer \(k\). This is called our 'inductive hypothesis'.
03

Prove for \(n+1\)

Now, we're going to show that the statement is also true for \(n + 1\). This requires substituting \(n + 1\) into the expression and simplifying the resulting expression using the inductive hypothesis. The inductive step means proving that if \(7^{n+2}+8^{2n+1}\) is divisible by 57, then \(7^{(n+1)+2}+8^{2(n+1)+1}\) also has to be divisible by 57.
04

Simplification of the Expression

After substituting \(n + 1\) into the expression, the resulting expression is complex. It could be simplified by using properties of exponents and factorization. We could represent \(7^{n+3}\) as \(7 \times 7^{n+2}\), and represent \(8^{2n+3}\) as \(8^2 \times 8^{2n+1}\). Since we assumed \(7^{n+2}+8^{2n+1}\) equals \(57k\), we could substitute \(57k\) into the expression instead of \(7^{n+2}+8^{2n+1}\), and simplify the expression further.
05

Use the Inductive Hypothesis

The point of having the inductive hypothesis (the assumption that the statement holds for \(n\)) is to use it for the proof of the successor case (\(n + 1\)). Put it into the equation and show that \(7^{(n+1)+2}+8^{2(n+1)+1}\) also must be divisible by 57.

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Most popular questions from this chapter

For any \(n \in \mathbf{Z}^{*}\), prove that \(n\) is a perfect square if and only if \(n\) has an odd number of positive divisors.

Let \(a, b, c \in \mathbf{Z}^{+}\)with \(\operatorname{gcd}(a, b)=1\). If \(a \mid c\) and \(b \mid c\), prove that \(a b \mid c\). Does the result hold if \(\operatorname{gcd}(a, b) \neq 1\) ?

If \(a, b \in \mathbf{Z}^{+}\), and both are odd, prove that \(2 \mid\left(a^{2}+b^{2}\right)\) but \(4 \backslash\left(a^{2}+b^{2}\right)\).

Let \(a, b \in \mathbf{Z}^{*}\) where \(a \geq b\). Prove that \(\operatorname{gcd}(a, b)=\operatorname{gcd}(a-b, b)\).

Let \(n \in \mathbf{Z}^{+}\)where \(n\) is odd and \(n\) is not divisible by 5 . Prove that there is a power of \(n\) whose units digit is 1 .
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