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First, I'm using the settings of https://en.wikipedia.org/wiki/NTRUEncrypt, with $L_f$ set of polynomials with $d_f+1$ coefficients equal to 1, $d_f$ equal to $-1$ and the remaining $N-2d_f-1$ equal to 0; and $L_g$ the set of polynomials with $d_g$ coefficients equal to 1, $d_g$ equal to $-1$ and the remaining $N-2d_g$ equal to 0. The natural numbers $d_f$ and $d_g$ are just fixed parameters of the scheme.

Suppose one receives a polynomial $h$ in the ring $R_{N,q}=\mathbb{Z}_q[X]/\langle X^N-1 \rangle$.

Question: Is it possible to determine if $h$ is a public-key, that is, is it possible to determine if $h$ is of the form $pf_q \cdot g \pmod{q}$?

My attempt: The NTRU hardness assumption says that from $h$ one cannot determine $f$ or $g$, otherwise the scheme would be useless. Although I couldn't answer my question, I came up with a test. Since $g(1)=0$, we must have $h(1)=0$. Hence if $h(1) \neq 0$ then $h$ is not a public-key. What can we test more?

PS: No zero-knowledge proof or similar things from the source of $h$ are given.

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You can't under a standard assumption known as the "Decisional NTRU Assumption". This is essentially the statement that NTRU public keys are pseudorandom. The following is definition 4.4.4 of A Decade of Lattice Cryptography.

NTRU Learning Problem: For an invertible $s\in R_q^*$, and distribution $\chi$ on $R$, define $N_{s, \chi}$ to be the distribution that outputs $e/s\in R_q$ where $e\gets\chi$. The NTRU Learning Problem is: Given independent samples $a_i\in R_q$, where each sample is distributed according to either

  1. $N_{s,\chi}$, for some randomly chosen $s\in R_q^*$ (fixed for all samples), or
  2. the uniform distribution, distinguish which is the case (with non-negligible advantage).

Note that this problem essentially states that you cannot do what you are asking, i.e. states that NTRU keys are computationally indistinguishable from random.

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  • $\begingroup$ Thanks a lot! I understand more now, but here in this definition the uniform distribution is defined over all $R_{N,q}$, which is against the test I provided. Moreover, $\chi$ is a distribution over $L(d_g,d_g)$. Can I define point 2 as uniform distribution over $R_{N,q}$ except for all polynomials which evaluated at 1 give 0? Will this definition / hardness assumption be enough? $\endgroup$
    – Leafar
    Oct 4 '21 at 23:27
  • $\begingroup$ You can probably just define the ring $R$ to have polynomials that evaluate to 0 at 1, e.g. set $R = \mathbb{Z}[x] / (x^n-1)$. You can't just change point 2 unilaterally though, as then the distinguishing problem becomes a different distinguishing problem. $\endgroup$
    – Mark
    Oct 5 '21 at 1:51
  • $\begingroup$ I can't because such polynomials are never invertible and this will mess with $f$, in your answer $s$, which needs to be invertible. $\endgroup$
    – Leafar
    Oct 5 '21 at 20:46

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