The following theorem gives a partial answer to Problem 2 on the structure of almost squared groups.

Theorem. Let $G$ be an almost squared group and $R$ be an almost square root of $G$. Then for every normal subgroup $H\subseteq G$ with abelian quotient group $G/H$ there exists $\epsilon\in\{-1,1\}$ such that:

1. $|G/H|$ divides $|R|+\epsilon$ and $|H|$ is divisible by $|R|-\epsilon$.

2. The set $L=\big\{gH\in H/G:|R\cap gH|=\frac{|R|+\epsilon}{|G/H|}\big\}$ has cardinality $|L|=|G/H|-1$.

3. For the unique element $gH$ of the singleton $(G/H)\setminus L$ we have $|R\cap gH|=\frac{|R|+\epsilon}{|G/H|}-\epsilon$.

Proof. For a subset $S\subseteq G$ let $\sum S$ be the element $\sum_{x\in S}x$ in the group algebra $\mathbb C G$. Taking into account that $R$ is an almost square root of $G$, we conclude that $(\sum R)^2=\sum G+s$ for some element $t\in G$ (having two representations as products of elements of the set $R$).

Consider the Abelian group $A=G/H$ and in its group algebra $\mathbb C A$, consider the element $y=\sum_{gH\in G/H}|R\cap gH|\cdot gH$. The equality $(\sum R)^2=\sum G+s$ implies $y*y=|H|\cdot\sum A+\tau$ where $\tau=tH\in A$. Now consider the elements $z=y-\frac{|R|\pm1}{|A|}\sum A$ of $\mathbb C A$ and observe that $$ \begin{aligned} z*z&=(y-\tfrac{|R|\pm1}{|A|}\sum A)*(y-\tfrac{|R|\pm1}{|A|}\sum A)=\\ &=y*y-\tfrac{|R|\pm1}{|A|}\sum_{b\in A}y*b-\tfrac{|R|\pm1}{|A|}\sum_{b\in A}b*y+\tfrac{(|R|\pm1)^2}{|A|^2}\sum_{a,b\in A}ab=\\ &=|H|\cdot\sum A+\tau-\tfrac{|R|\pm1}{|A|}\sum_{a,b\in A}|R\cap a|\cdot (ab+ba)+\tfrac{(|R|\pm1)^2}{|A|}\sum A=\\ &=\tfrac{|H|\cdot|A|+(|R|\pm1)^2}{|A|}\cdot\sum A+\tau-2\tfrac{|R|\pm1}{|A|}|R|\sum A=\\ &=\tfrac{|G|+|R|^2\pm2|R|+1-2(|R|\pm1)|R|}{|A|}\sum A+\tau=\\ &=\tfrac{|R|^2-1-|R|^2+1}{|A|}\sum A+\tau=\tau. \end{aligned} $$

Let $\hat z$ be the Fourrier transformation of $z$. It is the function assigning to every character $\chi:A\to\mathbb C$ the complex number $$\hat z(\chi)=\sum_{a\in A}z(a)\chi(a)$$where $z=\sum_{a\in A}z(a)\cdot a$. Since the function $\hat \tau$ has absolute value $1$ at each character, $\hat z\cdot \hat z=\widehat{z{*}z}=\hat \tau$, the function $\hat z:\hat A\to\mathbb C$ also has its values in the unit circle and hence $\hat z$ has norm $1$ in the Hilbert space $\mathbb C\hat A$ endowed with the inner product $$\langle f,g\rangle=\frac1{|A|}\sum_{a\in \hat A}f(a)\cdot\overline{g(a)}.$$ Endow the group algebra $\mathbb C A$ with the inner product $$\langle f,g\rangle=\sum_{a\in A}f(a)\cdot\overline{g(a)}.$$ Since the Fourrier transformation is an isometry isomorphism of the Hilbert spaces $\mathbb C A$ and $\mathbb C\hat A$, we obtain that $$ (\star)\quad\|z\|^2=\sum_{a\in A}\big||R\cap a|-\tfrac{|R|\pm 1}{|A|}\big|^2=\|\hat z\|=1. $$ Since $|A|$ divides $|G|=|R|^2-1$, the number $\frac{|R|}{|A|}$ is not integer. Let $\varepsilon>0$ be the smallest natural number such that $\frac{|R|+\varepsilon}{|A|}$ is integer.

The equation ($\star$) implies that for every $a\in A=G/H$ the number $|R\cap a|$ is equal either to $\frac{|R|+\varepsilon}{|A|}$ or to $\frac{|R|+\varepsilon}{|A|}-1$. Let $$P=\{a\in A:|R\cap a|=\tfrac{|R|+\varepsilon}{|A|}\}\quad\mbox{and}\quad Q=\{a\in A:|R\cap a|=\tfrac{|R|+\varepsilon}{|A|}-1\}.$$ Then the equation ($\star$)reduces to the equation $$(\star\star)\quad \begin{aligned} |A|^2&=|P|\cdot(\varepsilon\mp1)^2+|Q|\cdot(\varepsilon\mp1-|A|)^2=\\ &=|P|\cdot(\varepsilon\mp 1)^2+|Q|\cdot(\varepsilon\mp 1)^2-2|Q|\cdot|A|\cdot (\varepsilon\mp1)+|Q|\cdot|A|^2=\\ &=|A|\cdot(\varepsilon\mp 1)^2-2|Q|\cdot|A|\cdot(\varepsilon\mp1)+|Q|\cdot|A|^2 \end{aligned} $$ as $|P|+|Q|=|A|$.

Subtracting the equations ($\star$) for different signs after $\varepsilon$, we obtain obtain that $\varepsilon=|Q|$. After substitution of $\varepsilon=|Q|$ into the equation ($(\star\star$) we obtain that $\varepsilon=|Q|$ is equal to $1$ or $|A|-1$.

If $\varepsilon=|Q|=1$, then $|A|$ divides $|R|+\varepsilon=|R|+1$. If $\varepsilon=|Q|=|A|-1$, then $|A|$ divides $|R|+\varepsilon=|R|+|A|-1$ and hence divides $|R|-1$. Put $\epsilon=1$ if $\varepsilon=1$ and $\epsilon=-1$ if $\varepsilon=|A|-1$. In both cases we obtain that $|A|$ divides $|R|+\epsilon$.

Observe that $$|H|=\frac{|G|}{|A|}=\frac{|R|^2-1}{|A|}=\frac{|R|+\epsilon}{|A|}(|R|-\epsilon),$$ which means that $|H|$ is divisible by $|R|-\epsilon$.

If $|Q|=1$, then $\epsilon=1$, the set $$L=\{gH\in A:|R\cap gH|=\tfrac{|R|+\epsilon}{|A|}\}=|P|$$ has cardinality $|P|=|A|-|Q|=1$, and $$\{gH:|R\cap gH|=\tfrac{|R|+\epsilon}{|A|}-\epsilon\}$$ is a singleton.

If $|Q|=|A|-1$, then $\epsilon=-1$, the set $$\{gH:|R\cap gH|=\tfrac{|R|+|A|-1}{|A|}-1\}= \{gH:|R\cap gH|=\tfrac{|R|+\epsilon}{|A|}\}=L$$ has cardinality $|A|-1$.

On the other hand, the set $$P=\{gH\in A:|R\cap gH|=\tfrac{|R|+|A|-1}{|A|}\}=\{gH\in A:|R\cap gH|=\tfrac{|R|+\epsilon}{|A|}-\epsilon\}$$ is a singleton.

The following theorem gives a partial answer to Problem 2 on the structure of almost squared groups. Let us recall that the Abelianization of a group $G$ is the quotient group $G/[G,G]$ of $G$ by its commutator subgroup.

Theorem. Let $G$ be an almost squared group and $R$ be an almost square root of $G$. Let $H=[G,G]$ be the commutator subgroup of $G$ and $A=G/H$ be the Abelianization of $G$. Then there exists $\epsilon\in\{-1,1\}$ such that:

1. $|A|$ divides $|R|+\epsilon$.

2. The set $L=\big\{gH\in A:|R\cap gH|=\frac{|R|+\epsilon}{|A|}\big\}$ has cardinality $|L|=|A|-1$.

3. For the unique element $gH$ of the singleton $A\setminus L$ we have $|R\cap gH|=\frac{|R|+\epsilon}{|A|}-\epsilon$.

Proof. For a subset $S\subseteq G$ let $\sum S$ be the element $\sum_{x\in S}x$ in the group algebra $\mathbb C G$. Taking into account that $R$ is an almost square root of $G$, we conclude that $(\sum R)^2=\sum G+s$ for some element $t\in G$ (having two representations as products of elements of the set $R$).

In the group algebra $\mathbb C A$ of the Abelianization $A=G/H$, consider the element $y=\sum_{gH\in A}|R\cap gH|\cdot gH$. Since $R$ is an almost square root of $G$, $(\sum R)^2=\sum G+t$ for some elememt $t\in G$ (that has a double representation as a product of two elements in $R$). The equality $(\sum R)^2=\sum G+t$ implies $y*y=|H|\cdot\sum A+\tau$ where $\tau=tH\in A$. Now consider the elements $z=y-\frac{|R|\pm1}{|A|}\sum A$ of $\mathbb C A$ and observe that $$ \begin{aligned} z*z&=(y-\tfrac{|R|\pm1}{|A|}\sum A)*(y-\tfrac{|R|\pm1}{|A|}\sum A)=\\ &=y*y-\tfrac{|R|\pm1}{|A|}\sum_{b\in A}y*b-\tfrac{|R|\pm1}{|A|}\sum_{b\in A}b*y+\tfrac{(|R|\pm1)^2}{|A|^2}\sum_{a,b\in A}ab=\\ &=|H|\cdot\sum A+\tau-\tfrac{|R|\pm1}{|A|}\sum_{a,b\in A}|R\cap a|\cdot (ab+ba)+\tfrac{(|R|\pm1)^2}{|A|}\sum A=\\ &=\tfrac{|H|\cdot|A|+(|R|\pm1)^2}{|A|}\cdot\sum A+\tau-2\tfrac{|R|\pm1}{|A|}|R|\sum A=\\ &=\tfrac{|G|+|R|^2\pm2|R|+1-2(|R|\pm1)|R|}{|A|}\sum A+\tau=\\ &=\tfrac{|R|^2-1-|R|^2+1}{|A|}\sum A+\tau=\tau. \end{aligned} $$

Let $\hat z$ be the Fourrier transformation of $z$. It is the function assigning to every character $\chi:A\to\mathbb C$ the complex number $$\hat z(\chi)=\sum_{a\in A}z(a)\chi(a)$$where $z=\sum_{a\in A}z(a)\cdot a$. Since the function $\hat \tau$ has absolute value $1$ at each character, $\hat z\cdot \hat z=\widehat{z{*}z}=\hat \tau$, the function $\hat z:\hat A\to\mathbb C$ also has its values in the unit circle and hence $\hat z$ has norm $1$ in the Hilbert space $\mathbb C\hat A$ endowed with the inner product $$\langle f,g\rangle=\frac1{|A|}\sum_{a\in \hat A}f(a)\cdot\overline{g(a)}.$$ Endow the group algebra $\mathbb C A$ with the inner product $$\langle f,g\rangle=\sum_{a\in A}f(a)\cdot\overline{g(a)}.$$ Since the Fourrier transformation is an isometry isomorphism of the Hilbert spaces $\mathbb C A$ and $\mathbb C\hat A$, we obtain that $$ (\star)\quad\|z\|^2=\sum_{a\in A}\big||R\cap a|-\tfrac{|R|\pm 1}{|A|}\big|^2=\|\hat z\|=1. $$ Since $|A|$ divides $|G|=|R|^2-1$, the number $\frac{|R|}{|A|}$ is not integer. Let $\varepsilon>0$ be the smallest natural number such that $\frac{|R|+\varepsilon}{|A|}$ is integer.

The equation ($\star$) implies that for every $a\in A=G/H$ the number $|R\cap a|$ is equal either to $\frac{|R|+\varepsilon}{|A|}$ or to $\frac{|R|+\varepsilon}{|A|}-1$. Let $$P=\{a\in A:|R\cap a|=\tfrac{|R|+\varepsilon}{|A|}\}\quad\mbox{and}\quad Q=\{a\in A:|R\cap a|=\tfrac{|R|+\varepsilon}{|A|}-1\}.$$ Then the equation ($\star$)reduces to the equation $$(\star\star)\quad \begin{aligned} |A|^2&=|P|\cdot(\varepsilon\mp1)^2+|Q|\cdot(\varepsilon\mp1-|A|)^2=\\ &=|P|\cdot(\varepsilon\mp 1)^2+|Q|\cdot(\varepsilon\mp 1)^2-2|Q|\cdot|A|\cdot (\varepsilon\mp1)+|Q|\cdot|A|^2=\\ &=|A|\cdot(\varepsilon\mp 1)^2-2|Q|\cdot|A|\cdot(\varepsilon\mp1)+|Q|\cdot|A|^2 \end{aligned} $$ as $|P|+|Q|=|A|$.

Subtracting the equations ($\star$) for different signs after $\varepsilon$, we obtain obtain that $\varepsilon=|Q|$. After substitution of $\varepsilon=|Q|$ into the equation ($(\star\star$) we obtain that $\varepsilon=|Q|$ is equal to $1$ or $|A|-1$.

If $\varepsilon=|Q|=1$, then $|A|$ divides $|R|+\varepsilon=|R|+1$. If $\varepsilon=|Q|=|A|-1$, then $|A|$ divides $|R|+\varepsilon=|R|+|A|-1$ and hence divides $|R|-1$. Put $\epsilon=1$ if $\varepsilon=1$ and $\epsilon=-1$ if $\varepsilon=|A|-1$. In both cases we obtain that $|A|$ divides $|R|+\epsilon$.

Observe that $$|H|=\frac{|G|}{|A|}=\frac{|R|^2-1}{|A|}=\frac{|R|+\epsilon}{|A|}(|R|-\epsilon),$$ which means that $|H|$ is divisible by $|R|-\epsilon$.

If $|Q|=1$, then $\epsilon=1$, the set $$L=\{gH\in A:|R\cap gH|=\tfrac{|R|+\epsilon}{|A|}\}=|P|$$ has cardinality $|P|=|A|-|Q|=1$, and $$\{gH:|R\cap gH|=\tfrac{|R|+\epsilon}{|A|}-\epsilon\}$$ is a singleton.

If $|Q|=|A|-1$, then $\epsilon=-1$, the set $$\{gH:|R\cap gH|=\tfrac{|R|+|A|-1}{|A|}-1\}= \{gH:|R\cap gH|=\tfrac{|R|+\epsilon}{|A|}\}=L$$ has cardinality $|A|-1$.

On the other hand, the set $$P=\{gH\in A:|R\cap gH|=\tfrac{|R|+|A|-1}{|A|}\}=\{gH\in A:|R\cap gH|=\tfrac{|R|+\epsilon}{|A|}-\epsilon\}$$ is a singleton.