Squares in the Set of Gauss Integers and the Pythagorean Triple Equation // กำลังสองในเซตของจำนวนเต็มเกาส์และสมการสามเท่าของพีทาโกรัส

 

          Complex numbers of the form  $a+\imath b$  with  $a,b \in \mathbb{Z}$  are called Gaussian integers.

          Here we will consider the problem of finding Gaussian integers that are "perfect squares". For example, we are interested in when the result  $\sqrt{a+\imath b}$ is still a Gaussian integer.

           The problem is solved for example in W. Sierpinski (chapter XIII is dedicated to Gauss integers), page 450, Problem 4. We see that this problem is related to Pythagorean triples. This problem is not completed, unlike Problem 2 about representing as sums of two squares.

          What is said here is only a necessary and sufficient condition that solves our problem :

      "The complex integer  $a+\imath b$  is the square of a complex integer if and only if 

$a^2+b^2=c^2,\;\;\;c+a=2x^2,\;\;\;c-a=2y^2$

   where  $c$ is a natural number and  $x,\;y$  are rational integers. Then

$a+\imath b=(\pm x\pm y\imath)^2$

   the sign should be identical if  $b>0$  and opposite if  $b<0$."

       So the perfect squares in the set of Gauss integers are related to the solutions of the equation of Pythagorean numbers. As shown in another post, for the numbers  $a$  and  $b$  we have the possibilities :

$a=d\cdot 2st,\;\;b=d\cdot (s^2-t^2) \tag{1}$

$a=d\cdot (s^2-t^2),\;\;b=d\cdot 2st \tag{2}$

where  $d\in \mathbb{Z}$  and  $s,\;t$  are coprime integers of opposite parity.

          In the first case  $c+a=d\cdot (s^2+t^2)+d\cdot 2st=d\cdot (s+t)^2$, and $c-a=d\cdot (s^2+t^2)-d\cdot 2st=d\cdot (s-t)^2$. Hence

$\frac{c\pm a}{2}=\frac{d}{2}\cdot (s \pm t)^2 \tag{3}$

          In the second case  $c+a=d\cdot (s^2+t^2)+d\cdot (s^2-t^2)=2ds^2$, and  $c-a=d\cdot (s^2+t^2)-d\cdot (s^2-t^2)=2dt^2$,  hence

$\frac{c+a}{2}=d\cdot s^2,\;\; \frac{c-a}{2}=d\cdot t^2 \tag{4}$

So for  $\frac{c\pm a}{2}$  to be perfect squares,  $d$  must be  $2\alpha^2$ in (3) and  $\alpha^2$ in (4).

Answer CiP: The perfect squares in the set of Gaussian integers are

$4\alpha^2st+2\alpha^2(s^2-t^2)\imath=\left ( \pm\alpha[(s+t)+(s-t)\imath]\right )^2$

$\alpha^2(s^2-t^2)+2\alpha^2st \imath=\left (\pm \alpha [s+t\imath] \right )^2$

where  $\alpha \in \mathbb{Z}$, and  $s,t$ are coprime integers of opposite parity. 

          Examples CiP

$3+4\imath=(2+\imath)^2=(-2-\imath )^2$ :   $3^2+4^2=5^2,\;\frac{5\pm 3}{2}\in \{4,\;1\}$

$2\imath=(1+\imath)^2=(-1-\imath)^2$  :   $0^2+2^2=2^2,\;\frac{2\pm 0}{2}=1$

$5\pm 12 \imath=(3\pm 2\imath)^2$   :   $5^2+12^2=13^2,\;\frac{13\pm 5}{2}\in \{9,\;4\}$

$12\pm 5\imath \neq perfect\;squares$   :    $12^2+5^2=13^2$  although  $\frac{13\pm 12}{2}\in \left \{\frac{25}{2},\;\frac{1}{2}\right \}$

$24\pm 10\imath=(5\pm \imath)^2$   :   $24^2+10^2=26^2,\;\frac{26\pm 24}{2}\in\{25,\;1\}$

Problem E : 17346

 From GMB 11/2025, page 554. In translation :

  "E:17346.  A natural number  $n$ gives when divided by 7 the remainder 4

                       and when divided by 9 the remainder 5. Find out what remainder is

                       obtained  when dividing  $n$  by 63.

 [Author] Ștefan GOBLEJ, Curtea de Argeș"

ANSWER CiP

32

                            Solution CiP

                    Let  $a$ and  $b$  be the quotients of the divisions of  $n$  by 7 and 9 respectively. Then

$n=7\cdot a+4,\;\;\;n=9\cdot b+5 \;\;\;\;\;a,\;b\in \mathbb{N}\tag{1}$

The equation  $7\cdot a+4=9\cdot b+5$   is equivalent to

$7\cdot a-9\cdot b=1\;\;\;\;\;\;a,b \in \mathbb{N} \tag{2}$

Noting that  $7\cdot 4-9\cdot 3=28-27=1$  , equation  (2)  has a particular solution  $a_0=4,\;b_0=3$.

So the general solution of this [linear Diophantine] equation is

$a=4+9\cdot k,\;\;b=3+7\cdot k,\;\;\;k \in \mathbb{N} \tag{3}$

Substituting the number  $a$  from (3) into (1) we obtain  $n=7\cdot (4+9\cdot k)+4$  so  

$$n=63 \cdot k+32$$

and from this it follows that the remainder of dividing the number  $n$  by 63 is 32.

$\blacksquare$

             Remark CiP  We could have thought like that too, without the theory of Diophantine equations :

$7\cdot a\overset{(2)}{=}1+9\cdot b=1+2\cdot b+7\cdot b\;\Rightarrow\; 7\mid 1+2\cdot b\;\Rightarrow\; 2\cdot b+1=7\cdot k_1,\;k_1 \in \mathbb{N}$. But in the equality  $2\cdot b=7\cdot k_1-1$  we must have  $k_1=odd\;number,\;\; k_1=2\cdot k+1$  so  $2\cdot b=7(2\cdot k+1)-1=14 \cdot k+6$  and from here we obtain  $b=7\cdot k+3$  that is precisely  (3).

          COPILOT's solution : 

We write the numbers that give remainder of 4 when divided by 7 :

4, 11, 18, 25, 32, 39, 46, 53, 60, 67, 74, ...

We check which of them gives a remainder of 5 when divided by 9 :

4  mod 9  = 4

11  mod 9  = 2

18  mod 9  = 0

25  mod 9  =7

32  mod  9  =5  :  BINGO

(CiP note : After at most nine (=9) attempts, the result is definitely found, whether it exists or not.)

So the number that satisfies both conditions is  $n \equiv 32\;\;(mod\;63)$  , hence the remainder when dividing  $n$  by 63 is 32.

LUDO // oder Mensch ärgere dich nich // Another calculation problem with Vectors

           The game known in Romania as "Don't Be Angry, Brother!" it's basically LUDO. There is also a German version, mentioned in the title. I would have liked to say "Don't Be Upset, Sister!", because alongside Problem S:L25.296 in Post here, there is also its sister, Problem S:L25.295, which I initially neglected.

In translation :

"Let the triangle  $ABC$  and the points  $M\in (AB),\; N\in (AC), \;P\in BC$

  be such that  $\overrightarrow{BC}=2\overrightarrow{CP},\;\frac{MA}{MB}=\frac{1}{2},\;\;and\;\;\frac{NC}{AC}=\frac{2}{5}.$

     a) Express the vectors  $\overrightarrow{AM},\;\overrightarrow{AN}$  and  $\overrightarrow{AP}$ in terms of the vectors  $\overrightarrow{AB}\;and\;\overrightarrow{AC}$.

     b) Express the vectors  $\overrightarrow{MN}$  and  $\overrightarrow{MP}$  in terms of the vectors  $\overrightarrow{AB}\;and\;\overrightarrow{AC}$.

 c) Show that the points  $M,\;N,\;P$  are collinear and determine the ratio  $\frac{MN}{MP}.$

No author"

ANSWER CiP

a)  $\overrightarrow{AM}=\frac{1}{3}\overrightarrow{AB}$

$\overrightarrow{AN}=\frac{3}{5}\overrightarrow{AC}$

$\overrightarrow{AP}=-\frac{1}{2}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC}$

b)  $\overrightarrow{MN}=-\frac{1}{3}\overrightarrow{AB}+\frac{3}{5}\overrightarrow{AC}$

$\overrightarrow{MP}=-\frac{5}{6}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC}$

c)  $\overrightarrow{MN}=\frac{2}{5}\overrightarrow{MP}$

$\frac{MN}{MP}=\frac{2}{5}$

                         Solution CiP

          a) From  $\frac{MA}{MB}=\frac{1}{2}$  and  $M\in (AB)$  it follow  $\frac{\overline{MA}}{\overline{MB}}=\frac{-1}{2}$ and so  

$\frac{\overline{AM}}{\overline{AB}}=-\frac{\overline{MA}}{\overline{MB}-\overline{MA}}=-\frac{-1}{2-(-1)}=\frac{1}{3}$, hence  $\overrightarrow{AM}=\frac{1}{3}\overrightarrow{AB}.$

From  $\frac{NC}{AC}=\frac{2}{5}$  and  $N\in (AC)$  it follow  $\frac{\overline{NC}}{\overline{AC}}=\frac{2}{5}$  so

  $\frac{\overline{AN}}{\overline{AC}}=\frac{\overline{AC}+\overline{CN}}{\overline{AC}}=\frac{\overline{AC}-\overline{NC}}{\overline{AC}}=\frac{5-2}{5}=\frac{3}{5}$,  hence  $\overrightarrow{AN}=\frac{3}{5}\overrightarrow{AC}$

Finally  $\overrightarrow{AP}=\overrightarrow{AC}+\overrightarrow{CP}=\overrightarrow{AC}+\frac{1}{2}\overrightarrow{BC}=\overrightarrow{AC}+\frac{1}{2}(\overrightarrow{AC}-\overrightarrow{AB})=-\frac{1}{2}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC}$.

The same answer can be obtained by observing that  $\frac{\overline{PB}}{\overline{PC}}=\frac{3}{1}$  and we apply the formulas (1.1) and (3.2)

  $\overrightarrow{AP}=\frac{1}{1-3}\overrightarrow{AB}-\frac{3}{1-3}\overrightarrow{AC}.$

          b)  $\overrightarrow{MN}=\overrightarrow{AN}-\overrightarrow{AM}=\frac{3}{5}\overrightarrow{AC}-\frac{1}{3}\overrightarrow{AB}$  hence the answer.

     $\overrightarrow{MP}=\overrightarrow{AP}-\overrightarrow{AM}=\left (-\frac{1}{2}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC} \right )-\frac{1}{3}\overrightarrow{AB}=-\frac{5}{6}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC}$

          c)  With the results from  b)  we have

$\overrightarrow{MN}=-\frac{1}{3}\overrightarrow{AB}+\frac{3}{5}\overrightarrow{AC}=\frac{2}{5} \cdot \left (-\frac{5}{6}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC} \right )=\frac{2}{5}\overrightarrow{MP}$

hence the points  $M,\;N,\;P$  are collinear and  $\frac{\overline{MN}}{\overline{MP}}=\frac{2}{5}.$

     The collinearity of points  $M,\;N,\;P$  also results from observing the  $\overrightarrow{AP}=-\frac{1}{2}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC}=-\frac{1}{2}\cdot 3\overrightarrow{AM}+\frac{3}{2}\cdot \frac{5}{3}\overrightarrow{AN}$  or,

$$\overrightarrow{AP}=\frac{5}{2}\overrightarrow{AN}-\frac{3}{2}\overrightarrow{AM}=\frac{1}{\frac{2}{5}}\overrightarrow{AN}+\left ( 1-\frac{1}{\frac{2}{5}}\right )\overrightarrow{AM}$$

to which we apply formulas  (1.1)  and  (2.2).

$\blacksquare$

Comments on BARICENTRIC RELATIONS between THREE COLLINEAR POINTS // ҮС КОЛЛИНЕАРНАЙ ТОЧКА икки ардыгар БАРИЦЕНТРИЧЕСКАЙ СЫҺЫАННАР тустарынан комментарийдар

 We will further specify the relationships between the barycentric coordinates of three points on a line, which appear in the Post here. The notations are slightly modified.

          

          Let  $X,\;Y,\;Z$  be three collinear points. The following relations hold :

$\frac{\overline{XY}}{\overline{XZ}}=r\;\tag{1.1}$

$\overrightarrow{OY}=(1-r)\overrightarrow{OX}+r\overrightarrow{OZ}\;\tag{1.2}$

$Y=(1-r)X+rZ\;\tag{1.3}$

and the writing of (1.2)  $\overrightarrow{OY}=r\overrightarrow{OZ}+(1-r)\overrightarrow{OX}$  also says that

$\frac{\overline{ZY}}{\overline{ZX}}=1-r\;\tag{1.4}$

$\frac{\overline{XZ}}{\overline{XY}}=\frac{1}{r}\;\tag{2.1}$

$\overrightarrow{OZ}=\frac{1}{r}\overrightarrow{OY}+(1-\frac{1}{r})\overrightarrow{OX}\;\tag{2.2}$

$Z=\frac{1}{r}Y+(1-\frac{1}{r})X\;\tag{2.3}$

$\frac{\overline{YZ}}{\overline{YX}}=1-\frac{1}{r}\;\tag{2.4}$

$\frac{\overline{YX}}{\overline{YZ}}=-\frac{r}{1-r}\;\tag{3.1}$

$\overrightarrow{OX}=\frac{1}{1-r}\overrightarrow{OY}-\frac{r}{1-r}\overrightarrow{OZ}\;\tag{3.2}$

$X=\frac{1}{1-r}Y-\frac{r}{1-r}Z\;\tag{3.3}$

$\frac{\overline{ZX}}{\overline{ZY}}=\frac{1}{1-r}\;\tag{3.4}$

Problem S:L25.296 of Vector Calculus in Parallelogram // Parallelogrammda vektorlı esaplavnıñ 296-nci meselesi

 "Gazeta Matematika. Meşğuliyetlernen qoşma" 10/2025 sanlı mecmuasından alındı, 10 saife.

          To further practice vector calculus, consider the problem (in translation):

 "Consider the parallelogram  $ABCD$  and the points  $M\in AB,\; N \in AC$

              such that  $\overrightarrow{AM}=\frac{1}{4}\overrightarrow{AB},\;\; \overrightarrow{AN}=\frac{1}{5}\overrightarrow{AC}$. Express the vectors  $\overrightarrow{DN},\; \overrightarrow{MN}$

  in terms of the vectors  $\overrightarrow{AB}$  and $\overrightarrow{AD}$ , and then show that

                                           the points  $D,\; N,\; M$ are collinear.

 No author."

We will use formulas from the February 20, 2024 Post "Barycentric coordinate on the Straight Line".

Figure according to the statement. The green line  $k$  is a suggestion of the conclusion.

ANSWER  CiP

See  (1)  and  (3)

$$\overrightarrow{DM}=\frac{5}{4}\overrightarrow{DN}=5\overrightarrow{NM}$$

                    Solution CiP

          Let us first consider the point  $M$.  From  $\overrightarrow{AM}=\frac{1}{4}\overrightarrow{AB}\;\;\Leftrightarrow\;\;\frac{\overline{AM}}{\overline{AB}}=\frac{1}{4}$, it 

follows by the Corollary to the LEMMA  $\overrightarrow{DM}=\frac{3}{4}\overrightarrow{DA}+\frac{1}{4}\overrightarrow{DB}$  which is further equal to

$\frac{3}{4}(-\overrightarrow{AD})+\frac{1}{4}(\overrightarrow{DA}+\overrightarrow{AB})=-\frac{3}{4}\overrightarrow{AD}-\frac{1}{4}\overrightarrow{AD}+\frac{1}{4}\overrightarrow{AB}$  so

$$\overrightarrow{DM}=\frac{1}{4}\overrightarrow{AB}-\overrightarrow{AD} \tag{1}$$

From  $\overrightarrow{AN}=\frac{1}{5}\overrightarrow{AC}\;\;\Leftrightarrow\;\;\frac{\overline{AN}}{\overline{AC}}=\frac{1}{5}$  it also results  $\overrightarrow{DN}=\frac{4}{5}\overrightarrow{DA}+\frac{1}{5}\overrightarrow{DC}$  so

$$\overrightarrow{DN}=\frac{1}{5}\overrightarrow{AB}-\frac{4}{5}\overrightarrow{AD} \tag{2}$$

Finally

$\overrightarrow{MN}=\overrightarrow{DN}-\overrightarrow{DM}\;\underset{(1),(2)}{=}\;\left ( \frac{1}{5}\overrightarrow{AB}-\frac{4}{5}\overrightarrow{AD}\right )-\left (\frac{1}{4}\overrightarrow{AB}-\overrightarrow{AD}\right )$  so

$$\overrightarrow{MN}=-\frac{1}{20}\overrightarrow{AB}+\frac{1}{5}\overrightarrow{AD}. \tag{3}$$

          From (1), (2) and (3) we observe the equations  $\overrightarrow{DM}=\frac{5}{4}\overrightarrow{DN}=5\overrightarrow{NM}$,  and it results that the points  $D,\;N,\;M$  are collinear.

$\blacksquare$

How many MISTAKES are allowed in a book ? // Hoeveel fouten zijn toegestaan ​​in een boek ?

 Answer: NOT TOO MUCH

Let's take the first problem from a book.

Page 21, Chapter 1 EQUALITIES, Section 1.1 Equalities of one complex variable, Paragraph 1.A, Problem  #1 (in translation)

               "Let  $z\in\mathbb{C}$ such that  $z^2=-3+4\imath$. Calculate:

                       a)  $A=z^2+\bar z^2;$

                       b)  $B=\left | z-\frac{1}{\bar z } \right |;$

                       c)  $C=z+\frac{1}{\bar z};$

                       d)  $D=Re\;z+Im\;z.$"

Do you see the mistakes ?

ANSWER CiP

A. $-6$  ; B. $\frac{4\sqrt{5}}{5}$  ; C. $\pm\frac{6}{5}(1+2\imath)$  ;  D. $\pm 3$.

In reality  $z=\pm(1+2\imath)$

                         Solution CiP

               Let  $z=x+\imath y$. From  $-3+4\imath=z^2=(x+\imath y)^2=(x^2-y^2)+2xy \cdot \imath$

it follow  $x^2-y^2=-3,\;\;2xy=4$  so  $y=\frac{2}{x}\;\;and\;\;x^2-\frac{4}{x^2}=-3$.  Hence

$x^2=1$  and therefore  $z_{1,2}=\pm(1+2\imath).$

          Regardless of the possible values ​​for  $z$, we have the calculations :

$A=z^2+\overline {z^2}=(-3+4\imath)+\overline {(-3+4\imath)}=(-3+4\imath)+(-3-4\imath)=-6;$

     We then have  $|z^2|=|-3+4\imath|=\sqrt{(-3)^2+4^2}=5=|z|^2=|\bar z|^2$  so on one side  $|\bar z|=\sqrt{5}$, and

$B=\left | \frac{z\cdot \bar z-1}{\bar z} \right |=\left | \frac{|z|^2-1}{\bar z} \right |=\left | \frac{|z^2|-1}{\bar z} \right |=\left | \frac{5-1}{\bar z}\right |=\frac{4}{|\bar z|}=\frac{4}{\sqrt{5}};$

$C=\frac{z \cdot \bar z+1}{\bar z}=\frac{|z|^2+1}{\bar z}=\frac{5+1}{\bar z}=\frac{6\cdot z}{          \bar z \cdot z}=\frac{6z}{|z|^2}=\frac{6}{5}\cdot z\;\tag{1}$

     Finally, we still need

$\bar z=\frac{5}{z}$  and  $\frac{1}{z}=\frac{z}{-3+4\imath}$

to calculate  $D=Re\;z+Im\;z=\frac{z+\bar z}{2}+\frac{z-\bar z}{2\imath}=\frac{z+\frac{5}{z}}{2}+\frac{z-\frac{5}{z}}{2\imath}=\frac{z^2+5}{2z}+\frac{z^2-5}{2\imath z}=$

$\overset{z^2=-3+4\imath}{=}\;\;\frac{2+4\imath}{2z}+\frac{-8+4\imath}{2\imath z}=\frac{1}{z}+\frac{2\imath}{z}+\frac{8\imath ^2}{2\imath z}+\frac{2}{z}=\frac{3+6\imath}{z}=$

$=3(1+2\imath) \cdot \frac{1}{z}=3(1+2\imath) \cdot \frac{z}{-3+4\imath}=3(1+2\imath)\cdot \frac{(-3-4\imath)z}{25}=\frac{3(5-10\imath)}{25}\cdot z=\frac{3}{5}(1-2\imath)\cdot z \tag{2}$

Substituting the values  $z_{1,2}$  ​​into (1) and (2) we obtain the answer.

$\blacksquare$

          Remark CiP  Compare with the official solution :

An equation in integers // Een vergelijking in gehele getallen

 Problem  S:E25.261  from the Exercise Supplement  9/2025. In translation :

                  "Find the pairs of integers  $(x,y)$ for which the relation

$x(x+1)=y(y+7) \tag{E}$

                     is verified.

 Liliana NICULESCU, Târgu Mureș"

ANSWER CiP

$(x,y)\in \{(-6,-10),\;(-6,3),\;(-1,-7),\;(-1,0),\;(0,-7),\;(0,0),\;(5,-10),\;(5,3)\}$

Solution CiP

 Equation (E) is written equivalently, successively

$x^2+x=y^2+7y\Leftrightarrow 4x^2+4x+1=4y^2+28y+1\Leftrightarrow (2x+1)^2=(2y+7)^2-48 \Leftrightarrow$

 $\Leftrightarrow (2y-7)^2-(2x+1)^2=48 \Leftrightarrow (2y+7-2x-1)(2y+7+2x+1)=48 \Leftrightarrow$

$\Leftrightarrow 2(y-x+3)\cdot 2(y+x+4)=48 \Leftrightarrow$

$\Leftrightarrow \;\;\;(y-x+3)(y+x+4)=12 \tag{1}$

In integers, for  (1)  we have the possibilities

\begin{array}{c|c|}A=y-x+3&-12&-6&-4&-3&-2&-1&1&2&3&4&6&12\\ \hline B=y+x+4&\underset{[1]}{-1}&\underset{\color{Red}{impossible}}{-2}&\underset{[2]}{-3}&\underset{[3]}{-4}&\underset{\color{Red}{impossible}}{-6}&\underset{[4]}{-12}&\underset{[5]}{12}&\underset{\color{Red}{impossible}}{6}&\underset{[6]}{4}&\underset{[7]}{3}&\underset{\color{Red}{impossible}}{2}&\underset{[8]}{1} \end{array}

But  $A+B=2y+7$  is odd so some cases in the Table are impossible in integers. Only cases [1],...[8] remain.

          [1] \begin{cases}y-x+3=-12\\y+x+4=-1\end{cases} $\Leftrightarrow$ \begin{cases}y-x=-15\\y+x=-5\end{cases}

from where, adding  $2y=-15-5=-20$  means  $y=-10$  and  $x=-5-y=-5+10=5$. We check  (E)  :  $5 \cdot 6=-10 \cdot (-3)$.

[2] \begin{cases}y-x=-7\\y+x=-7\end{cases}

from where  $y=-7,\;x=0$  and  (E) is verified :  $0\cdot 1=-7 \cdot 0$.

[3] \begin{cases}y-x=-6\\y+x=-8\end{cases}

from where  $y=-7,\;x=-1$  and  (E)  is verified :  $-1\cdot 0=-7\cdot 0$.

[4]\begin{cases}y-x=-4\\y+x=-16\end{cases}

from where  $y=-10,\;x=-6$  and  (E)  is verified :  $-6\cdot (-5)=-10\cdot (-3)$.

[5]\begin{cases}y-x=-2\\y+x=8\end{cases}

from where  $y=3,\;x=5$  and  (E)  is verified :  $5\cdot 6=3\cdot 10$.

[6]\begin{cases}y-x=0\\y+x=0\end{cases}

from where  $y=0,\;x=0$  and  (E)  is obvious.

[7]\begin{cases}y-x=1\\y+x=-1\end{cases}

from where  $y=0,\;x=-1$  and  (E)  is obvious.

[8]\begin{cases}y-x=9\\y+x=-3\end{cases}

from where  $y=3,\;x=-6$  and  (E)  is verified : $-6\cdot (-5)=3\cdot 10$.

      I got the answer.

$\blacksquare$