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No. The procedure exposes a square-root oracle for the Rabin cryptosystem. The verifier chooses $m$, sends $r=m^2\mathrm{mod}N$, and receives another square root $m_1$. Ifthen the factorization from distinct square roots modulo a semiprime givesFor a product of two odd primes, a square normally has four roots, and a returned root lies outside the pair $\{\pm m\}$ with probability about $1/2$. Repeating chosen-square challenges therefore factors $N$ with overwhelming probability. The attacker can then compute square roots and impersonate Alice. This is the chosen-square attack on Rabin authentication, so the proposed protocol is insecure.

For Rabin cryptosystem, choose primes $p,q\equiv 3(\mathrm{mod}4)$, publish $N=pq$, and encrypt $x$ as $x^2\mathrm{mod}N$. The factors let the receiver take square roots modulo $p$ and $q$ and combine them using the Chinese remainder theorem; redundancy identifies the intended one of four roots.

For RSA cryptosystem, choose $N=pq$, select $e$ coprime to $\phi (N)$ and $d$ with $ed\equiv 1(\mathrm{mod}\phi (N))$. Encrypt $x$ as $x^e$ and decrypt by raising to $d$ modulo $N$. Rabin inversion is provably equivalent to factoring, while ordinary RSA lacks that reduction; Rabin's disadvantage is its fourfold decryption ambiguity.

In the Rabin cryptosystem, the public key is $N=pq$ and encryption sends an encoded message $m$ toKnowing $p,q$, the receiver finds the two square roots modulo each prime and combines them with the Chinese remainder theorem to obtain four roots modulo $N$; prescribed redundancy identifies the intended one.

Omicron hasThe Chinese remainder theorem determines $m^2$ uniquely modulo $N{N}^{\prime }$ (assuming the independently generated moduli are coprime; a nontrivial gcd would itself factor them). Since $1\le m<N<N^{\prime }$, one has $m^2<N{N}^{\prime }$, so this residue is the ordinary integer $m^2$. Taking its positive integer square root recovers $m$.

This does not normally decrypt another ciphertext sent under only one modulus. The recovered pair reveals no nontrivial square root collision modulo that modulus and hence supplies no factorization; breaking a single Rabin instance remains equivalent to factoring its modulus.

In the Rabin cryptosystem, the public key is $N=pq$, usually with distinct secret primes $p,q\equiv 3(\mathrm{mod}4)$. A message $m$ is encrypted asKnowing $p$ and $q$, the receiver finds the two square roots of $c$ modulo each prime and combines them by the Chinese remainder theorem, obtaining the four square roots modulo $N$; redundancy identifies the intended message.

Factoring $N$ plainly enables this decryption. Conversely, suppose an algorithm returns a square root of a chosen quadratic residue. Choose random invertible $x$, submit $x^2\mathrm{mod}N$, and receive a root $y$. With probability at least $1/2$, $y\not\equiv \pm x(\mathrm{mod}N)$; thenis a nontrivial factor. Thus decryption and factoring are equivalent up to a randomized polynomial-time reduction.

A Rabin cryptosystem chooses Blum primes $p,q$, publishes $N=pq$, and encrypts $m$ as $c=m^2\mathrm{mod}N$. Decryption takes square roots modulo $p$ and $q$, combines them with CRT, and uses redundancy to select the intended one of four roots. Here $2355^2\equiv 25(\mathrm{mod}2773)$, while the obvious root is $5$. ThusSo $2773=47\cdot 59$ is factored, allowing all future square roots and hence all ciphertexts to be decrypted.