I have the following trigonometric equation of $3$ variables:

$$f(\theta,\lambda,\phi)=3 \cos (\theta ) \cos (\lambda ) \cos (\phi )-(\cos (\theta )+3) \sin (\lambda ) \sin (\phi )$$

$$-\sin (\theta ) \cos (\lambda )+\sin (\theta ) \cos (\phi )+3 \cos (\theta )+\cos (\lambda ) \cos (\phi )-7$$

I want to prove that the solution set to the equation $f(\theta,\lambda,\phi)=0$ is

$$S_f=\{(2\pi k,\lambda+2\pi m,-\lambda+2\pi n):k,m,n\in\mathbb{Z},\lambda\in\mathbb{R}\}$$

Graphically, it is easy to see that this is the case but everything I have tried so far has failed to prove the conjecture. Perhaps my best attempt was extrapolating this equation into an unwieldy polynomial of $3$ variables

$$P(x,y,z)=9 x^4 y^2+16 x^4-18 x^3 y^2-192 x^3+9 x^2 y^4+z^4 \left(\left(x^2-1\right) y^2+1\right) \left(16 \left(x^2-1\right) y^2+(3 x+5)^2\right)-64 x^2 y^2+z^2 \left((x-1) (x+1) \left(9 x^2+30 x+41\right) y^4+2 x (x (48 (x-5) x+187)+234) y^2+x (x (9 (x-2) x-64)-78)-74 y^2+151\right)+736 x^2+2 y z^3 \left(9 x^4+12 x^3-20 x^2+(x-1) (x+1) (x (31 x-78)-33) y^2-96 x-33\right)+2 y z \left(31 x^4-270 x^3+560 x^2+(x (x (3 x (3 x+4)-20)-96)-33) y^2-210 x-111\right)+30 x y^4-78 x y^2-960 x+25 y^4+151 y^2+400$$

over the domain $(x,y,z)\in [-1,1]^3$. If I can prove that the solution set to $P(x,y,z)=0$ over this domain is


then the original conjecture would be solved. The motivation behind this is actually proving a certain type of quantum error detection encoding exists. It's a little difficult to explain (although if anyone wants details I am more than happy to provide them) but suffice to say that after a lot of work I managed to whittle my existence proof down to the conjecture above.

  • $\begingroup$ There's no way the solution set is what you claim it is. Why can't you have $(2\pi k,\lambda+2\pi, -\lambda)$? $\endgroup$ Jun 27, 2021 at 1:26
  • $\begingroup$ Fair enough, originally I was only considering the solutions over the space $[0,2\pi)^3$. When I expanded to $\mathbb{R}^3$ I forgot to update correctly $\endgroup$
    – QC_QAOA
    Jun 27, 2021 at 1:46
  • 1
    $\begingroup$ it's cool that your question comes from "proving a certain type of quantum error detection encoding exists"!! $\endgroup$ Jun 27, 2021 at 1:50
  • $\begingroup$ "Graphically, it is easy to see...": how can you display a function of $3$ variables ?? $\endgroup$
    – user65203
    Jun 28, 2021 at 16:37
  • $\begingroup$ I make a plot of $\lambda$ and $\phi$ and then I vary $\theta$ over $[0,2\pi]$ $\endgroup$
    – QC_QAOA
    Jun 28, 2021 at 16:39

4 Answers 4


You can get a more tractable, hand-bashable quadratic equation if you do the famous half tangent substitution: $$\tan\frac\theta 2 = x,\,\, \tan\frac\phi 2 = y,\,\,\tan\frac\lambda 2 = z.$$ Then for example, one has: $$\cos\theta = \dfrac{1-x^2}{1+x^2}\text{ and } \sin\theta = \dfrac{2x}{1+x^2}.$$ This saves us the trouble of repeated squaring and once you multiply everything out, you will get a quadratic in $x$ whose discriminant is: $$\Delta = -16(y+z)^2(21y^2z^2+13y^2+13z^2+16yz+21)<0$$ unless $y+z = 0.$

The whole thing took me about 20 mins of hand checking, but once you know the plan of attack with the right substitution, it's nothing but guaranteed to be completed.


We prove $f(\theta,\lambda,\phi) \le 0$ for each $\theta,\lambda,\phi$ and that equality holds exactly when you conjecture it does.

Write $$f(\theta,\lambda,\phi) = \cos(\theta)a+\sin(\theta)b+\cos(\lambda)\cos(\phi)-3\sin(\lambda)\sin(\phi)-7$$ for $a = 3\cos(\lambda)\cos(\phi)-\sin(\lambda)\sin(\phi)+3$ and $b = \cos(\phi)-\cos(\lambda)$. To maximize $f$, we of course to choose $\theta$ so that $\cos(\theta)a \ge 0$ and $\sin(\theta)b \ge 0$. So, for ease, let's just pretend $a,b \ge 0$ so that $\cos(\theta),\sin(\theta) \ge 0$. To maximize $xa+\sqrt{1-x^2}b$ for $x \in [0,1]$, one takes $x = \frac{a}{\sqrt{a^2+b^2}}$ (easy to prove by basic calculus), yielding a maximum of $\sqrt{a^2+b^2}$; and note that any other $x$ yields a strictly smaller value. Therefore, we wish to show $$\sqrt{(3\cos(\lambda)\cos(\phi)-\sin(\lambda)\sin(\phi)+3)^2+(\cos(\phi)-\cos(\lambda))^2}+\cos(\lambda)\cos(\phi)-3\sin(\lambda)\sin(\phi)-7 \le 0$$ with equality if and only if $\lambda = -\phi+2\pi m$ for some $m \in \mathbb{Z}$ (since then $\cos(\theta)$ will be $1$). For ease, let $x = \cos(\lambda)\cos(\phi)$ and $y = \sin(\lambda)\sin(\phi)$. We can rewrite the above as $$\sqrt{(3x-y+3)^2+(\cos(\phi)-\cos(\lambda))^2}+x-3y-7 \le 0.$$ We will show $$\sqrt{(3x-y+3)^2+(\cos(\phi)-\cos(\lambda))^2+(\sin(\phi)+\sin(\lambda))^2}+x-3y-7 \le 0$$ with equality if and only if $\lambda = -\phi+2\pi m$, which clearly suffices. The point of introducing the $(\sin(\phi)+\sin(\lambda))^2$ is that $$(\cos(\phi)-\cos(\lambda))^2+(\sin(\phi)+\sin(\lambda))^2 = 2-2x+2y.$$ So, we just wish to show that $$\sqrt{(3x-y+3)^2+2-2x+2y}+x-3y-7 \le 0$$ for $\{(x,y) \in [-1,1]^2 : |x-y| \le 1\}$ with equality if and only if $x-y = 1$. Note that $|3x-y+3| = |2x+(x-y)+3| \le 6$ and $|2-2x+2y| = 2|1-(x-y)| \le 4$, so $\sqrt{(3x-y+3)^2+2-2x+2y} \le \sqrt{40}$. Therefore, we cannot have $y \ge 0$ and $x \le 0$.

Let's first start with $y \ge 0$. Then, as just explained, $x \ge 0$. We will show that $$\sqrt{(3x-y+3)^2+2-2x+2y}+x-7 \le 0$$ with equality if and only if $x=1,y=0$. Note the derivative of the term inside the square root with respect to $y$ is $2y+2-6(x+1)$, which is negative, so we choose $y=0$ to maximize. We then wish to show $$\sqrt{(3x+3)^2+2-2x} -7 \le 0$$ with equality if and only if $x = 1$. But this is equivalent to $(3x+3)^2+2-2x \le 49$, which is true for $x \le 1$ with equality at and only at $x=1$.

Now let's deal with the case $x \le 0$. As explained before, this means $y \le 0$. We show $$\sqrt{(3x-y+3)^2+2-2x+2y}+x-3y-7 \le 0$$ for any $x,y \in [-1,0]^2$ with equality if and only if $y=-1,x=0$. The derivative of the term inside the square root with respect to $y$ is, as before, $2y+2-6(x+1)$, which is negative, so we take $y=-1$ to maximize. We then obtain $$\sqrt{9x^2+22x+16}+x-4,$$ which is maximized at $x=0$, yielding the value $0$. We're done.

  • 1
    $\begingroup$ It's not immediately clear where the other cases such as $a>0>b$ follows identically because their domains are technically different. For example if $a < 0<b,$ then your function to maximize is $-xa + \sqrt{1-x^2}b$ But I do think it will just be similar length bashing. $\endgroup$
    – dezdichado
    Jun 27, 2021 at 3:25
  • $\begingroup$ @dezdichado It does follow essentially identically. If $a < 0 < b$, we choose $\theta$ so that $\cos(\theta) < 0$ and $\sin(\theta) > 0$; or, we can just replace $x$ with $-x$. $\endgroup$ Jun 27, 2021 at 3:36

Use Lagrange multipliers and Grobner bases to find the minimum and maximum of the function.

This will show that $f(\theta,\lambda,\phi)= 0$ is the maximum when $S_f=\{(2\pi k,\lambda+2\pi m,-\lambda+2\pi n):k,m,n\in\mathbb{Z},\lambda\in\mathbb{R}\} \tag{1}$

Let $x_1 = \cos(\theta)$, $y_1 = \sin(\theta)$. A constraint is $x_1^2 + y_1^2 - 1 = 0$.

Similarly $x_2= \cos(\lambda)$, $y_2= \sin(\lambda)$, $x_3= \cos(\phi)$, $y_3= \sin(\phi)$.

$$F(x_1,y_1,x_2,y_2,x_3,y_3) = 3x_{1}x_{2}x_{3} -\left(x_{1}+3\right)y_{2}y_{3}-x_{2}y_{1}+x_{3}y_{1}+3x_{1} +x_{2}x_{3}-7 \tag{2}$$

The constraints are:

$$g_1 = x_1^2 + y_1^2 - 1 \,, g_2 = x_2^2 + y_2^2 - 1 \,, g_3 = x_3^2 + y_3^2 - 1 \tag{3}$$

The Lagrangian is: $$\mathcal{L} = F - L_1g_1 - L_2g_2 -L_3g_3 \tag{4}$$

Solve: $$\nabla \mathcal{L} = 0 \tag{5}$$


G(x1,y1,x2,y2,x3,y3,L1,L2,L3) :=  3*x1*x2*x3 - (x1+3)*y2*y3
 -y1*x2 + y1*x3 +3*x1 + x2*x3 - 7 - L1*(x1^2 + y1^2 - 1) 
- L2*(x2^2 + y2^2 - 1) -L3*(x3^2 + y3^2 - 1);

J : jacobian([G(x1,y1,x2,y2,x3,y3,L1,L2,L3)],[x1,y1,x2,y2,x3,y3,L1,L2,L3]);
J : transpose(J);

$$\nabla \mathcal{L} = \begin{pmatrix}-\mathit{y_2}\, \mathit{y_3}+3 \mathit{x_2}\, \mathit{x_3}-2 \mathit{L_1}\, \mathit{x_1}+3\\ -2 \mathit{L_1}\, \mathit{y_1}+\mathit{x_3}-\mathit{x_2}\\ -\mathit{y_1}+3 \mathit{x_1}\, \mathit{x_3}+\mathit{x_3}-2 \mathit{L_2}\, \mathit{x_2}\\ -\left( \mathit{x_1}+3\right) \, \mathit{y_3}-2 \mathit{L_2}\, \mathit{y_2}\\ \mathit{y_1}-2 \mathit{L_3}\, \mathit{x_3}+3 \mathit{x_1}\, \mathit{x_2}+\mathit{x_2}\\ -2 \mathit{L_3}\, \mathit{y_3}-\left( \mathit{x_1}+3\right) \, \mathit{y_2}\\ -{{\mathit{y_1}}^{2}}-{{\mathit{x_1}}^{2}}+1\\ -{{\mathit{y_2}}^{2}}-{{\mathit{x_2}}^{2}}+1\\ -{{\mathit{y_3}}^{2}}-{{\mathit{x_3}}^{2}}+1\end{pmatrix} \tag{6}$$

Find a Grobner basis:


vars : [x1,y1,x2,y2,x3,y3,L1,L2,L3];
eqns : transpose(J)[1];
gb : poly_reduced_grobner(eqns,reverse(vars));

The Grobner basis:

$$\begin{pmatrix} -y_{1}^2- x_{1}^2+1\cr -y_{2}^2-x_{2}^2+1\cr x_{1}-x_{1}^3\cr x_{1}^2y_{2}- y_{2}\cr 3y_{3}^2+4x_{1}x_{2}y_{1}+3x_{2}^2-3\cr x_{1} y_{2}y_{3}-4x_{1}^2x_{2}y_{1}+2x_{2}y_{1}-3x_{1} x_{2}^2-3x_{1}+2L_{1}\cr 3x_{1}y_{2}y_{3}+9y_{2}y_{3}- 6x_{1}^2x_{2}y_{1}-2x_{1}x_{2}y_{1}+3x_{2}y_{1}-9 x_{1}x_{2}^2-3x_{2}^2-6x_{1}^2+6L_{2}+6\cr -24x_{2}y_{3} -3x_{1}y_{1}y_{2}-7y_{1}y_{2}-24x_{1}x_{2}y_{2}\cr 3 x_{1}y_{2}y_{3}+9y_{2}y_{3}-12x_{1}^2x_{2}y_{1}-2 x_{1}x_{2}y_{1}+3x_{2}y_{1}-9x_{1}x_{2}^2-3x_{2}^2-6 x_{1}^2+6L_{3}+6\cr y_{1}y_{3}+x_{1}y_{1}y_{2}\cr x_{1}^2 y_{3}-y_{3}\cr 2x_{1}y_{1}-3x_{3}+6x_{1}^2x_{2}-3x_{2} \cr -3x_{2}^2y_{1}-6x_{1}^2y_{1}-2x_{1}y_{1}+3y_{1} \cr -x_{1}^2x_{2}^2+x_{2}^2+x_{1}^2-1\cr \end{pmatrix} \tag{7}$$

The third equation gives $x_1(x_1^2-1) = 0$. So the maximum and minimum occur at $x_1 = [-1,0,1]\tag{8}$

Consequently $x_1y_1 = 0$ due to constraint $g_1$ equation $(3)$.

The fifth equation gives $3y_{3}^2+4x_{1}x_{2}y_{1}+3x_{2}^2-3 = 0$. Substituting $x_1y_1 = 0$ gives $y_{3}^2+x_{2}^2-1 = 0$. This is a unity constraint use ($g_2,g_3$) which give $x_2^2 = x_3^2 \tag{9}$ $y_2^2 = y_3^2\tag{10}$

So we can add equations to the basis: $[x_1y_1,x_2^2 - x_3^2, y_2^2 - y_3^2]$.

Recalculate the basis:


eqns2 : append(eqns,[x1*y1,x2^2-x3^2,y2^2-y3^2]);   
gb2 : poly_reduced_grobner(eqns2,reverse(vars));
gb2 : transpose(gb2);

$$\begin{pmatrix} -y_{1}^2- x_{1}^2+1\cr -y_{2}^2-x_{2}^2+1\cr x_{1}\,y_{1}\cr x_{1}-x_{1}^3\cr x_{3}-2\,x_{1}^2\,x_{2}+x_{2}\cr y_{1}\,y_{2}\cr y_{1}-x_{2}^2\, y_{1}\cr x_{1}^2\,x_{2}^2-x_{2}^2-x_{1}^2+1\cr x_{1}^2\,y_{2}-y_{2} \cr y_{3}^2+x_{2}^2-1\cr -x_{1}\,y_{2}\,y_{3}-3\,y_{2}\,y_{3}-x_{2} \,y_{1}+3\,x_{1}\,x_{2}^2+x_{2}^2+2\,x_{1}^2-2\,L_{3}-2\cr -x_{1}\, y_{2}\,y_{3}-2\,x_{2}\,y_{1}+3\,x_{1}\,x_{2}^2+3\,x_{1}-2\,L_{1}\cr x_{2}\,y_{3}+x_{1}\,x_{2}\,y_{2}\cr -x_{1}\,y_{2}\,y_{3}-3\,y_{2}\, y_{3}-x_{2}\,y_{1}+3\,x_{1}\,x_{2}^2+x_{2}^2+2\,x_{1}^2-2\,L_{2}-2 \cr y_{1}\,y_{3}\cr x_{1}^2\,y_{3}-y_{3}\cr \end{pmatrix} \tag{11}$$

Equations $6$ and $15$ give: $y_{1}\,y_{2} = 0 \tag{12}$ $y_{1}\,y_{3} = 0 \tag{13}$ These may be useful later but I cannot see any more useful reductions.

At this point brute force is an option. $(x_1,y_1):(-1,0),(0,1),(0,-1),(1,0)$. $x_2 = x_3$, $x_2 = -x_3$. $y_2 = y_3$, $y_2 = -y_3$. Which gives $4\times 2 \times 2 = 16$ test cases. The number of cases will be reduced when $(x_1,y_1)$ are added to the Grobner basis and recalculated.

The function to evaluate:


F(x1,y1,x2,y2,x3,y3) :=  3*x1*x2*x3 - (x1+3)*y2*y3 -y1*x2 + y1*x3 +3*x1 + x2*x3 - 7;

$ ------------------------------------$

case $(x_1,y_1) = (-1,0)$.

There should be no solutions for this case.

Add $x_1$ and $y_1$ to the basis $(11)$.


eqns_minus1_0 : append(eqns2,[x1+1,y1]);    
gb_minus1_0 : poly_reduced_grobner(eqns_minus1_0,reverse(vars));
gb_minus1_0 : transpose(gb_minus1_0);

$$\begin{pmatrix} -y_{2}^2- x_{2}^2+1\cr x_{1}+1\cr y_{1}\cr x_{3}-x_{2}\cr y_{2}\,y_{3}+x_{2}^2 +L_{3}\cr x_{2}\,y_{3}-x_{2}\,y_{2}\cr y_{2}\,y_{3}+x_{2}^2+L_{2} \cr y_{3}^2+x_{2}^2-1\cr -y_{2}\,y_{3}+3\,x_{2}^2+2\,L_{1}+3\cr \end{pmatrix} \tag{14}$$

The fourth equation gives $x_2 = x_3$. So the only test cases required are $y_3 = y_2$ and $y_3 = -y_2$ (see equation $(10)$).

case $y_3 = y_2$

$F(-1,0,x2,y2,x2,y2) = -2\,y_{2}^2-2\,x_{2}^2-10 = -2\,(y_{2}^2+\,x_{2}^2)-10 = -12$

By $g_2$ $(3)$.

This is a local minima and not a solution to $F = 0$.

case $y_3 = -y_2$:

$F(-1,0,x2,y2,x2,-y2) = 2\,y_{2}^2-2\,x_{2}^2-10 = -4\,x_{2}^2-8$

This is an inverted parabola with a maximum of $-8$ at $x_2 = 0$.

There are no solutions to $F = 0$ for the case $(x_1,y_1) = (-1,0)$.

$ ------------------------------------$

case $(x_1,y_1) = (0,-1)$.

There should be no solutions for this case.

Add $x_1$ and $y_1$ to the basis $(11)$.


eqns_0_minus1 : append(eqns2,[x1,y1+1]);    
gb_0_minus1 : poly_reduced_grobner(eqns_0_minus1,reverse(vars));
gb_0_minus1 : transpose(gb_0_minus1);

$$\begin{pmatrix} x_{1}\cr y_{1}+1\cr x_{2}-2\,L_{3}-1\cr x_{3}+x_{2}\cr 1-x_{2}^2\cr y_{2}\cr L_{1}-x_{2}\cr y_{3}\cr -x_{2}+2\,L_{2}+1\cr \end{pmatrix} \tag{15}$$

From the fourth equation $x_3 = -x_2$.

From the sixth equation $y_2 = 0$.

From the eighth equation $y_3 = 0$.

$F(0,-1,x2,0,-x2,0) = -x_{2}^2+2\,x_{2}-7$

This is an inverted parabola with a minimum of $F = -6$ at $x_2 = 1$.

There is no solution to $F = 0$ for the case $(x_1,y_1) = (0,-1)$.

$ ------------------------------------$

case $(x_1,y_1) = (0,1)$.

There should be no solutions for this case.

Add $x_1$ and $y_1$ to the basis $(11)$.


eqns_0_1 : append(eqns2,[x1,y1-1]); 
gb_0_1 : poly_reduced_grobner(eqns_0_1,reverse(vars));
gb_0_1 : transpose(gb_0_1);

$$\begin{pmatrix} x_{1}\cr y_{1}-1\cr -x_{2}-2\,L_{3}-1\cr x_{3}+x_{2}\cr x_{2}^2-1\cr y_{2} \cr x_{2}+L_{1}\cr y_{3}\cr x_{2}+2\,L_{2}+1\cr \end{pmatrix} \tag{16}$$

From the fourth equation $x_3 = -x_2$.

From the sixth equation $y_2 = 0$.

From the eighth equation $y_3 = 0$.

$F(0,1,x2,0,-x2,0) = -x_{2}^2-2\,x_{2}-7$

This is an inverted parabola with a maximum of $F = -6$ at $x_2 = -1$.

There is no solution to $F = 0$ for the case $(x_1,y_1) = (0,1)$.

$ ------------------------------------$

case $(x_1,y_1) = (1,0)$.

There should be solutions for this case.

Add $x_1$ and $y_1$ to the basis $(11)$.


eqns_1_0 : append(eqns2,[x1-1,y1]); 
gb_1_0 : poly_reduced_grobner(eqns_1_0,reverse(vars));
gb_1_0 : transpose(gb_1_0);

$$\begin{pmatrix} -y_{2}^2- x_{2}^2+1\cr x_{1}-1\cr y_{1}\cr x_{3}-x_{2}\cr 2\,y_{2}\,y_{3}-2\, x_{2}^2+L_{3}\cr x_{2}\,y_{3}+x_{2}\,y_{2}\cr 2\,y_{2}\,y_{3}-2\, x_{2}^2+L_{2}\cr y_{3}^2+x_{2}^2-1\cr -y_{2}\,y_{3}+3\,x_{2}^2-2\, L_{1}+3\cr \end{pmatrix} \tag{17}$$

From the fourth equation $x_3 = x_2$. So the only test cases required are $y_3 = y_2$ and $y_3 = -y_2$.

case $y_3 = y_2$.

$F(1,0,x2,y2,x2,y2) = -4\,y_{2}^2+4\,x_{2}^2-4 = 8\,x_{2}^2-8$

The maximum occurs when $x_2 = \pm 1$ then $F = 0$. Note $y_2 = 0$ by $g_2$ equation $(3)$. Note $0 = -0$. Note $y_3 = y_2 = 0 = -y_3$. The only solution in this case still obeys $(x_1,y_1,x_2,y_2,x_3,y_3) = (1,0,x_2,y_2,x_2,-y_2)$.

case $y_3 = -y_2$.

$F(1,0,x2,y2,x2,-y2) = 4\,y_{2}^2+4\,x_{2}^2-4 = 0$.

By $g_2$ equation $(3)$.

$F = 0$ whenever $(x_1,y_1,x_2,y_2,x_3,y_3) = (1,0,x_2,y_2,x_2,-y_2)$.

$ ------------------------------------$

The only solutions to $F = 0$ occur at $(x_1,y_1,x_2,y_2,x_3,y_3) = (1,0,x_2,y_2,x_2,-y_2)$.

This translates to $x_1 = \cos(\theta) = 1$ so $\theta = 2\pi k$.

$x_2 = x_3$ translates to $\cos(\lambda) = \cos(\phi)$.

$y_2 = - y_3$ translates to $\sin(\lambda) = -\sin(\phi)$.

From the identities $\cos(\alpha) = \cos(-\alpha)$ and $\sin(\alpha) = -\sin(-\alpha)$ with periodicity of $2\pi$.

These conditions are represented by $S_f=\{(2\pi k,\lambda+2\pi m,-\lambda+2\pi n):k,m,n\in\mathbb{Z},\lambda\in\mathbb{R}\}$.


Let $\;x=\tan\dfrac\theta2,\;y=\dfrac{\lambda+\varphi}2,\;z=\dfrac{\lambda-\varphi}2,\;$ then $$g(x,y,z) = (1-x^2)(2\cos2y+\cos 2z+3)+4x\sin y\sin z\\ +(1+x^2)(-\cos2y+2\cos2z-7)\\ = (-3\cos2y+\cos2z-10)x^2+4x\sin y\sin z+\cos2y+3\cos2z-4\\ = (-6\cos^2y-2\sin^2z-6)x^2+4x\sin y\sin z - 2\sin^2y-6\sin^2z=0,$$ $$-6(1+\cos^2y)x^2-2(x\sin z-\sin y)^2-6\sin^2z=0.\tag1$$ From $(1)$ should $$x=0,\quad y=\pi j,\quad z=\pi l,\tag2$$ and finally $$\color{green}{\mathbf{\theta=2\pi k,\quad \lambda=\pi m,\quad \varphi=2\pi n-\lambda,\quad k,m,n\in\mathbb Z.}}$$


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