Life Satisfaction Scale By Singh And Joseph Pdf Freegolkes

Life Satisfaction Scale By Singh And Joseph Pdf Freegolkes

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Life Satisfaction Scale By Singh And Joseph Pdf Freegolkes

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Number of natural numbers $N \leq \log_2 2$ such that $N^{14} \equiv N^{13} \mod 16$

How many natural numbers $N \leq \log_2 2$ are there that satisfy:
$$N^{14} \equiv N^{13} \mod 16?$$
I know that when $N = 5$ we would have:
$$5^{14} \equiv 5^{13} \mod 16 \iff 16 = 5^{13} \iff 5^{12} = 5^2 \iff 5^2 \equiv 1 \mod 16$$
Thus, $N = 5$ is the only solution. But how do I show this in general?

A:

If $a,b$ are relatively prime and $a \mid b^n-b^{n-1}$, $n \ge 1$, then $a \mid b^k-b^{k-1}$ for all $k \ge 1$. So $N^{14} \equiv N^{13} \mod 16$ if and only if $N \mid 16$ or $N$ is a square modulo $16$. So let’s assume $N
mid 16$.

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