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Prove that the ${8^{th}}$ power of any number is of the form $17n$ or $17n \pm 1$.

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Last updated date: 20th Jun 2024
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Answer
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Hint: Consider two cases for solving this problem. First will be the case where the number is prime with the prime number $17$ and the other where the number is not a prime with $17$. For case one, use Fermat's little theorem and prove that for case one the number with power $8$ can be written as $17n \pm 1$ . For case two, a number is already multiple of $17$ , so it can be written as the product of $17$ and integer $n$ .

Complete step by step answer:
In this problem, we have to prove that any number having power $8$ can be expressed in the form of $17n$ or $17n \pm 1$ where $'n'$ is some integer number.

For this problem, we can use Fermat’s little theorem. But before using this theorem we must understand it beforehand.

Fermat's little theorem is a fundamental theorem in elementary number theory, which helps compute powers of integers modulo prime numbers. According to this theorem, if $a$ is an integer, $p$ is a prime number and $a$ is not divisible by $p$, then ${a^{p - 1}} \equiv 1\left( {{\text{mod }}p} \right)$ .

A frequently used corollary of Fermat's Little Theorem is ${a^{p - 1}} \equiv 1\left( {{\text{mod }}p} \right)$ . As you can see, it is derived by multiplying both sides of the theorem by $a$ . The restated form is nice because we no longer need to restrict ourselves to integers $a$ not divisible by $p$ .

The expression ${a^{p - 1}} \equiv 1\left( {{\text{mod }}p} \right)$ can also be written as ${a^p} = a\left( {\bmod {\text{ }}p} \right)$ which means that the number ${a^{p - 1}}$ will be completely divisible by a prime number $p$ or is an integral multiple of $p$ .

Therefore, ${a^{p - 1}} \equiv 1\left( {{\text{mod }}p} \right) \Rightarrow {a^{p - 1}} - 1 = n \times p$ (i)

Now let’s take the value of $p = 17$ , this will give us:
$ \Rightarrow {a^{17 - 1}} - 1 = n \times 17 \Rightarrow {a^{16}} - 1 = 17n$
In the above expression we can write $1$ as ${1^2}$ and hence we can use the identity ${a^{2m}} - {b^{2n}} = {\left( {{a^m}} \right)^2} - {\left( {{b^n}} \right)^2} = \left( {{a^m} - {b^n}} \right)\left( {{a^m} + {b^n}} \right)$ in it. This will give us:
$ \Rightarrow {\left( {{a^8}} \right)^2} - {1^2} = 17n \rightarrow \left( {{a^8} - 1} \right)\left( {{a^8} + 1} \right) = 17n$
So now this concluded that the product of $\left( {{a^8} - 1} \right)$ and $\left( {{a^8} + 1} \right)$ is a multiple of a prime number $17$ . So either $\left( {{a^8} - 1} \right)$ or $\left( {{a^8} + 1} \right)$ is a multiple of the prime number $17$ .
Therefore, we can say $ \Rightarrow {a^8} = 17n \pm 1$
Now, if we consider the case where $a$ is not prime to $17$ , i.e. they already have a common factor other than one, which can only be $17$ , since it is a prime.
Therefore, $a$ can be represented as multiple of $17$ as $ \Rightarrow a = 17q$
Hence, ${a^8} = {\left( {17q} \right)^8} = 17\left( {{{17}^7}{q^8}} \right) = 17n$

Thus, we proved that a number with a power of $8$ , i.e. of form ${a^8}$ is of the form $17n{\text{ or }}17n \pm 1$.

Note: Notice that the use of Fermat’s little theorem is the most crucial part of the solution. Congruence modulo $n$ is a congruence relation, meaning that it is an equivalence relation that is compatible with the operations of addition, subtraction, and multiplication. Congruence modulo $n$ is denoted: $a \equiv b\left( {\bmod {\text{ }}n} \right)$.
The parentheses mean that $\left( {\bmod {\text{ }}n} \right)$ applies to the entire equation, not just to the right-hand side (here b). This notation is not to be confused with the notation $b{\text{ mod }}n$ (without parentheses), which refers to the modulo operation. Indeed, $b{\text{ mod }}n$ denotes the unique integer $a$ such that $0 \leqslant a \leqslant n{\text{ and }}a \equiv b\left( {{\text{mod }}n} \right)$ (i.e., the remainder of $b$ when divided by $n$ ).