Multiplication in elliptical curve cryptography

Question

I'm practising for a cryptography test and came across this question:

Let us assume that for a specific elliptic curve the formulas for the computation of $(x_3, y_3)=R+S$ are given by:

$X_3=\sigma^2-x_1-x_2\ (\textrm{mod}\ p)$

$Y_3=\sigma(x_1-x_3)-y_1 \ (\textrm{mod}\ p)$

$\sigma = \frac{y_2-y_1}{x_2-x_1}\qquad \textrm{if R}\neq\textrm{S}$

$\sigma = \frac{3x_1^2+5}{2y_1}\qquad \textrm{if R = S}$

The following holds: $p=17$, $P=(3, 5)$, $a=3$, $k=2$, $M=(2, 7)$. Compute Q.

I understand the provided solution up until step 5.

1.$\ Q=a\cdot P = 3\cdot P$

2.$\ P= (3, 5) \Rightarrow 3\cdot P=2\cdot P+P$

3.$\ 2\cdot P\Rightarrow(3,5)+(3,5)$

4.$\ \sigma=\frac{3x_1^2+5}{2y_1}\qquad\ \ \ (R=S)$

5.$\ \sigma =\frac{3\cdot 3^2+5}{2\cdot 5}=\frac{32}{10}=\frac{16}{5}=\frac{16\cdot7}{5\cdot 7}\textrm{ mod } 17=112\textrm{ mod }17=10\qquad\ \ \ (5^{-1}=7\textrm{ mod } 17)$

I understand that $16\cdot 7\textrm{ mod }17=10$, but where does the $7$ come from? Also, why is $5^{-1}=7\textrm{ mod } 17$ shouldn't that equal zero..?

Answer 1

I understand that $16\cdot 7\equiv 10 \bmod 17$, but where does the $7$ come from? Also, why is $5^{−1}\equiv 7 \bmod 17$ shouldn't that equal zero..?

When working with modular arithmetic, $a^{-1}$ (if it exists) is a number $b$ such that $a\cdot b\equiv 1$. So, because $5\cdot7\equiv 1\bmod{17}$, we say that $5^{-1}\equiv 7\bmod{17}$. Not sure where $0$ came from in your mind as $0$ times anything is $0$.

In step 5, where you are multiplying by $\frac{7}{7}$, the reason they are doing this is to cancel out the bottom since $5\cdot 7\equiv 1$. This step is not necessary, but it does make things simpler, I guess. You can play around with this modular arithmetic calculator to verify that $\frac{16}{5} = 16\cdot 7\equiv 10\bmod{17}$.