Quadratic residuosity problem reduction to integer factorization How can one show how to reduce the quadratic residuosity problem to an integer factorization?
Factoring $\longrightarrow$ square roots.
Computing square roots modulo a prime $p$ is easy: if $p \equiv 3 \pmod 4$ and $a$ is a quadratic residue modulo $p$, then $a^{(p + 1)/4}$ is a square root of $a$ modulo $p$; for $p \equiv 1 \pmod 4$, you can use Cipolla or Tonelli–Shanks.

Computing square roots modulo a composite of known prime factorization is easy: compute square roots modulo each factor separately and combine with the Chinese remainder theorem.
Square roots $\longrightarrow$ factoring.
It is well-known that an oracle $S$ for finding square roots of random elements of $\mathbb Z/n\mathbb Z$ can be used to make an efficient algorithm to factor $n$: pick $x$ at random, and if $y := S(x^2) \not\equiv \pm x \pmod n$, then $\operatorname{gcd}(x \pm y, n)$ is a nontrivial factor of $n$ since $y^2 \equiv x^2 \pmod n$ so $n \mid x^2 - y^2 = (x + y) (x - y)$, and by hypothesis neither $x + y$ nor $x - y$ is congruent to zero modulo $n$.
Factoring $\longrightarrow$ quadratic residuosity.
Deciding quadratic residuosity for a prime modulus $p$ is easy: just compute the Legendre symbol $(a|p) \equiv a^{(p - 1)/2} \pmod p$ and see whether it is 0 or 1.

Deciding quadratic residuosity for a composite modulus of known prime factorization $n = pq$ is easy: $a$ is a quadratic residue modulo $n$ iff it is a quadratic residue modulo $p$ and a quadratic residue modulo $q$, which can be computed by the respective Legendre symbols.
Quadratic residuosity $\stackrel?\longrightarrow$ factoring.
We don't know!

If you could find a way to factor $n$ using a decision oracle $D$ for deciding whether random elements of $\mathbb Z/n\mathbb Z$ are quadratic residues, you could publish it in a top-tier conference, because the state of the art to my knowledge is that such a reduction is known only in the generic ring model.