Let $N=pq$ where $p$ and $q$ are primes of the form $4k+1$. Let $\mathbb{Z}_N$ be the set of integers modulo $N$ and $\mathbb{Z}_N^*$ be the units in $\mathbb{Z}_N$. Let $QR$ be the quadratic residues in $\mathbb{Z}_N^*$. If none of $p$ and $q$ is $5$, then show that $\mathbb{Z}_N=\{a-b: a, b \in QR\}$. That is, we need to show that $\mathbb{Z}_N$ can be expressed as difference of quadratic residues.

Actually I observed that for N being product of two primes, apart from 3 and 5, Z_N can always be represented as difference of its set of quadratic residues. I tried to prove it but I failed. That's why I asked it here.

Initially I started with primes of the form 4k+1 as -1 is a quadratic residue there. However, nothing seems to work out.

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Also posted to m.se, without any notice at either site. How rude. math.stackexchange.com/questions/832608/… – Gerry Myerson Jun 14 '14 at 4:54

For any prime $p$, and any nonzero residue $x\in(\mathbb{Z}/p\mathbb{Z})^\times$, the equation $a^2-b^2=x$ has $p-1$ solutions in $\mathbb{Z}/p\mathbb{Z}$, hence at least $p-5$ solutions in $(\mathbb{Z}/p\mathbb{Z})^\times$. It follows that, for $p>5$, any residue modulo $p$ is the difference of two quadratic residues modulo $p$. By the Chinese Remainder Theorem, it follows that if $N$ is the product of distinct primes exceeding $5$, any residue modulo $N$ is the difference of two quadratic residues modulo $N$ (i.e. two squares of units modulo $N$).
In any ring where $2$ is invertible, every element is the difference of two squares, namely $a=(a+4^{-1})^2-(a-4^{-1})^2$.
EDIT: I missed that the squares are supposed to be units. If $N$ is odd, the Chinese remainder theorem and Hensel’s lifting show that the property holds in $\mathbb Z/N\mathbb Z$ iff it holds in $\mathbb Z/p\mathbb Z$ for all prime divisors $p\mid N$. The identity above shows that an element $a$ of $\mathbb Z/p\mathbb Z$ is a difference of nonzero squares if we can find $b\ne0$ such that $ab^2\ne\pm4^{-1}$; the latter is always possible unless the only quadratic residues are $\pm1$, i.e., $p=3,5$. Since the property also fails for $2$, we see that every element of $\mathbb Z/N\mathbb Z$ is a difference of two square units iff $N$ is not divisible by 2, 3, or 5.