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I am currently studying for my next semester's course in mathematics and I have this problem. I know that I have to use Fermat's Little Theorem, but I am lost as to how to start. Also, what property of modulo can I use for exponentiating $a$ in $a \equiv b (\mod p)$

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    Please explain your notations. what is D? E? Are they exponentiations in d,e? is there a connection between d,e or do you need to find such pairs so that the equation holds? Also, you should choose one way to write "mod N" and be consistent. I suggest that you use $\equiv$ instead of =, and write (mod N) in the end of the line. – 2017-02-27
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    @Ofir Assuming this is the definition of RSA Encryption, then $D(x)=x^d \mod N$ and $E(x) = x^e \mod N$ for $ed \equiv 1 \mod \varphi(N)$ – 2017-02-27
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    @lioness99a I assumed that this is the case, but it doesn't change the fact that he should be able to properly formulate the question before trying to answer it (or in this case, ask for help with it). – 2017-02-28

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By the looks of things, this is the definition of RSA Encryption. I suggest you watch this 4 part series on Khan Academy which might clear things up for you

Put simply though, when encrypting $m$, we compute $$c \equiv m^e \mod N$$ for $N=pq$ where $p$ and $q$ are prime

We then want to find a value of $d$ such that $$c^d\mod N \equiv m$$

That is to say, we want $$\begin{align}\left(m^e\mod N\right)^d \mod N &\equiv m\\ m^{e^d}\mod N &\equiv m \\ m^{e\times d} \mod N &\equiv m \end{align}$$

For any choice of $e$, we want a way of finding $d$ such that this equivalence is true

We use Euler's Phi Function to find $\varphi(N)=(p-1)(q-1)$.

This is true as $p$ and $q$ are coprime, so $$\varphi(N)\equiv\varphi(p)\varphi(q)$$

As $p$ and $q$ are prime, then $$\varphi(p)\varphi(q)=(p-1)(q-1)$$

We then use Euler's Theorem to say that $$\begin{align}m^{\varphi(N)} & \equiv 1\mod N \\ m^{k\times \varphi(N)} &\equiv 1\mod N \tag{as $1^k=1$} \\ m\times m^{k\times\varphi(N)} & \equiv m\mod N\tag{as $1\times m = m$} \\ m^{k\times \varphi(N) + 1} &\equiv m\mod N\end{align}$$

We now have a function for finding $e\times d$ which depends on $\varphi(N)$:

$$\begin{align}m^{e\times d}\mod N & \equiv m^{k\times \varphi(N) + 1}\mod N \\ e\times d&\equiv k\times \varphi(N)+1 \\ e\times d &\equiv 1 \mod \varphi(N) \\ d &\equiv e^{-1}\mod \varphi(N) \end{align}$$

Hopefully this gives you somewhere to start