RSA polynomial cubic root Let's suppose we are given a linear polyonmial
$$\begin{align}f(x) = ax + b\end{align}$$
where a and b is known which satisfies this equation
$$\begin{align}y^3 - f(x) \equiv 0\mod(n) \end{align}$$ where n is RSA modulus.
Is there any way to solve for pair(x,y)?
Follow-up question: If there is restriction where only those values of x are allowed for which $$\begin{align}f(x)&lt;n \end{align}$$
Is there any way to solve for pair(x,y)?

If you're looking for an arbitrary pair, it's easy (assuming $a \ne 0$).

Pick an arbitrary $y$

Solve for $f(x) = y^3$; that'd be $x = a^{-1}(y^3 - b) \pmod{n}$


You're done.