# Is there an easy way to calculate “d” in the RSA algorithm?

Quite frankly, it is a pain to use the Extended Euclidean Algorithm to calculate d (the private exponent) in RSA. The equation used to find d is:

$$e d \equiv1~(\mathrm{mod}~ \varphi(n)).$$

Does anyone have a way to solve for d using basic algebra or something simpler? If not, can someone explain how to use the Extended Euclidean Algorithm to find d?

• if $\frac{1+k\cdot \varphi}{e}, k\in\mathbb N$ is an integral number, you have found $d$. This doesn't scale as nicely as the EEA does though. – SEJPM Jul 30 '16 at 20:14
• On the other hand, $0 < k < e$, and so there are only $e-1$ possible values of $k$ to try; if $e$ is small, this would be practical (if less efficient than EE) – poncho Jul 30 '16 at 20:20
• @SEJPM That's an answer, right? Or should I brush up on my RSA wrt "doesn't scale as nicely"? Connor C, $d$ is the private exponent. Together with the modulus it is the private key (although Chinese Remainder Theorem parameters could also be used). – Maarten Bodewes Jul 30 '16 at 20:32
• – e-sushi Jul 31 '16 at 10:43

## 1 Answer

The easiest way which is widely known to calculate a modular inverse is finding the smallest $k\in\mathbb N$ such that the following expression is an integral integer:

$$\frac{1+k\cdot \varphi}{e}$$

If this is an integral integer, it is the inverse of $e$ modulo $\varphi$ and thus $d$ and you'll find it with at most $e$ tries.

The issue obviously know is that the run-time of this will grow linearly with the size of $e$, as opposed to the extended euclidean algorithm (EEA) which will run in time proportional to (a power of) the logarithm of $e$, which will especially be much faster than the above method when $e$ is chosen as large as the modulus. If you want to see examples and how-tos on the algorithm, Wikipedia is your friend as usual.