This question is half of a larger question: Show that the equation $5x = 3$ modulo $20$ has no integer solution but $3x = 5$ does.
I can show that $3x = 5$ has an integer solution in $U(20)$ by using the fact that the multiplicative inverse of 3 is 7 in $U(20)$ since $(3 * 7)$ mod $20 = 21$ mod $20 = 1$, the identity of $U(20)$.
Multiplying, mod $20$, on the left,
$$\begin{aligned}7*3x &= 7 * 5 \pmod{20}\\ x &= 15 \end{aligned}$$
Now, for the first part of the question, how do I show that there is no integer solution for $5x = 3$ mod $20$?
Does it suffice to say that there is no multiplicative inverse of 5 in $U(20)$, so there is no way to solve the equation?
If you multiply both sides of the equation by 4, you get a contradiction.