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 $$\begin{align}f(x)<n \end{align}$$

  • 2
    $\begingroup$ In $y^3 - f(x) \equiv 0 \pmod{n}$, how are $x$ and $y$ related? Are you looking for any $(x, y)$ pair that satisfies the equation? Is $x$ a known constant? Or, did you mean $y^3 - f(y) \equiv 0 \pmod{n}$ $\endgroup$ – poncho Nov 12 '20 at 14:04
  • $\begingroup$ yeah @poncho I am looking for any valid pair of (x,y) where x and y are natural numbers. I will update the question $\endgroup$ – Saurav Kumar Singh Nov 12 '20 at 14:08

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.

  • $\begingroup$ If we had to find for smallest value possible of x, then can we do it? $\endgroup$ – Saurav Kumar Singh Nov 12 '20 at 14:35
  • $\begingroup$ why do u think that x will not be smallest? when You stated ${f(x)<n}$, and a, b are constant. $\endgroup$ – SSA Dec 13 '20 at 6:42
  • $\begingroup$ @SauravKumarSingh: if $y$ happened to have a cube root modulo $n$ (which it will quite a lot of the time), then a solution with $x=0$ will be minimal. However, finding that solution would imply solving the RSA problem with $e=3$; that's a hard problem. $\endgroup$ – poncho Dec 13 '20 at 17:13

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