Convergence of a family of series











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I want to show that for $D>0, Dequiv 0,1 pmod4$, $z in mathbb{C}$ with $ operatorname{Im}(z)>0$ and $k>1$ the sum



$$sum_{a,b,c in mathbb{Z} \ b^2-4ac=D}(az^2+bz+c)^{-k}$$ converges absolutely.










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  • What does $Dequiv 0,1 pmod4$ mean?
    – zhw.
    Aug 2 at 20:40












  • $D equiv 0 pmod 4$ or $D equiv 1 pmod 4$ sorry if that was cloudy
    – Deavor
    Aug 2 at 20:50












  • You might wanted to put $b^2-4ac$ in two places, one under the sum, the other after your attempted proof.
    – i707107
    Aug 4 at 19:38










  • Oh yes thanks for noticing the typo
    – Deavor
    Aug 4 at 20:18










  • My answer had serious flaws. I will come back if I resolve all issues.
    – i707107
    Aug 9 at 5:04















up vote
9
down vote

favorite
3












I want to show that for $D>0, Dequiv 0,1 pmod4$, $z in mathbb{C}$ with $ operatorname{Im}(z)>0$ and $k>1$ the sum



$$sum_{a,b,c in mathbb{Z} \ b^2-4ac=D}(az^2+bz+c)^{-k}$$ converges absolutely.










share|cite|improve this question
























  • What does $Dequiv 0,1 pmod4$ mean?
    – zhw.
    Aug 2 at 20:40












  • $D equiv 0 pmod 4$ or $D equiv 1 pmod 4$ sorry if that was cloudy
    – Deavor
    Aug 2 at 20:50












  • You might wanted to put $b^2-4ac$ in two places, one under the sum, the other after your attempted proof.
    – i707107
    Aug 4 at 19:38










  • Oh yes thanks for noticing the typo
    – Deavor
    Aug 4 at 20:18










  • My answer had serious flaws. I will come back if I resolve all issues.
    – i707107
    Aug 9 at 5:04













up vote
9
down vote

favorite
3









up vote
9
down vote

favorite
3






3





I want to show that for $D>0, Dequiv 0,1 pmod4$, $z in mathbb{C}$ with $ operatorname{Im}(z)>0$ and $k>1$ the sum



$$sum_{a,b,c in mathbb{Z} \ b^2-4ac=D}(az^2+bz+c)^{-k}$$ converges absolutely.










share|cite|improve this question















I want to show that for $D>0, Dequiv 0,1 pmod4$, $z in mathbb{C}$ with $ operatorname{Im}(z)>0$ and $k>1$ the sum



$$sum_{a,b,c in mathbb{Z} \ b^2-4ac=D}(az^2+bz+c)^{-k}$$ converges absolutely.







sequences-and-series complex-analysis convergence






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edited Dec 3 at 17:02

























asked Aug 2 at 17:57









Deavor

479514




479514












  • What does $Dequiv 0,1 pmod4$ mean?
    – zhw.
    Aug 2 at 20:40












  • $D equiv 0 pmod 4$ or $D equiv 1 pmod 4$ sorry if that was cloudy
    – Deavor
    Aug 2 at 20:50












  • You might wanted to put $b^2-4ac$ in two places, one under the sum, the other after your attempted proof.
    – i707107
    Aug 4 at 19:38










  • Oh yes thanks for noticing the typo
    – Deavor
    Aug 4 at 20:18










  • My answer had serious flaws. I will come back if I resolve all issues.
    – i707107
    Aug 9 at 5:04


















  • What does $Dequiv 0,1 pmod4$ mean?
    – zhw.
    Aug 2 at 20:40












  • $D equiv 0 pmod 4$ or $D equiv 1 pmod 4$ sorry if that was cloudy
    – Deavor
    Aug 2 at 20:50












  • You might wanted to put $b^2-4ac$ in two places, one under the sum, the other after your attempted proof.
    – i707107
    Aug 4 at 19:38










  • Oh yes thanks for noticing the typo
    – Deavor
    Aug 4 at 20:18










  • My answer had serious flaws. I will come back if I resolve all issues.
    – i707107
    Aug 9 at 5:04
















What does $Dequiv 0,1 pmod4$ mean?
– zhw.
Aug 2 at 20:40






What does $Dequiv 0,1 pmod4$ mean?
– zhw.
Aug 2 at 20:40














$D equiv 0 pmod 4$ or $D equiv 1 pmod 4$ sorry if that was cloudy
– Deavor
Aug 2 at 20:50






$D equiv 0 pmod 4$ or $D equiv 1 pmod 4$ sorry if that was cloudy
– Deavor
Aug 2 at 20:50














You might wanted to put $b^2-4ac$ in two places, one under the sum, the other after your attempted proof.
– i707107
Aug 4 at 19:38




You might wanted to put $b^2-4ac$ in two places, one under the sum, the other after your attempted proof.
– i707107
Aug 4 at 19:38












Oh yes thanks for noticing the typo
– Deavor
Aug 4 at 20:18




Oh yes thanks for noticing the typo
– Deavor
Aug 4 at 20:18












My answer had serious flaws. I will come back if I resolve all issues.
– i707107
Aug 9 at 5:04




My answer had serious flaws. I will come back if I resolve all issues.
– i707107
Aug 9 at 5:04










2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted
+150










It is actually rather straightforward. If $a=0$, then $b$ can take at most 2 values, none of them $0$, and we are left with the absolutely convergent sum $sum_c |c+bi|^{-k}$. If $c=0$, we can run a similar argument. So it is enough to consider $a,b,cne 0$. In this case the roots $u,v$ of the corresponding quadratic polynomial $P(x)=ax^2+bx+c$ are real and lie at the distance $sqrt D$ from $-frac b{2a}$. Thus $|P(z)|=|a||z-u||z-v|$. This is never less than $|a||Im z|^2$ and for $|b|>(6sqrt D+4|Re z|)|a|$ is not less than $|a||frac b{4a}|^2approx frac{|b|^2}{|a|}$. Thus, $|P(z)|gtrsimmax(|a|,frac{|b|^2}{|a|})ge |b|$ and we only need to consider $sum_{a,b,c}|b|^{-k}$. However, for fixed $b$, the number of non-zero pairs $(a,c)$ with $0ne b^2-D=4ac$ is at most the number of divisors of $b^2-D$, which grows slower than any positive power of $|b|$.






share|cite|improve this answer





















  • It is not clear why we can assume $|b|>(6sqrt D+4|Re z|)|a|$. Would you please explain this?
    – i707107
    Oct 6 at 4:48










  • @i707107 If $|b|$ is less, we just use the lower bound of $|a|$, which is better than $|b|^2/|a|$ in that case.
    – fedja
    Oct 6 at 10:37












  • This works great! Now, I think I actually made the problem unnecessarily difficult. Do you have any suggestion on the bound of orbit points $gamma z$ in my last figure?
    – i707107
    Oct 6 at 15:21










  • One small typo is $u, v$ are real and lie at the distance $sqrt D /2a$ from $-b/2a$. But, it does not affect the validity of this argument.
    – i707107
    Oct 6 at 15:25






  • 1




    @Deavor Yes, we just choose $varepsilon<k-1$.
    – fedja
    Dec 3 at 21:08


















up vote
3
down vote













This is an incomplete answer using the Theory of Quadratic Forms. The constants $C$ appearing in this solution may not be the same. We begin with



Definition




Let $ax^2+bxy+cy^2$ be a quadratic form with $a, b, cin mathbb{Z}$ and $b^2-4ac=D$. We say that a quadratic form $a_1x_1^2+b_1x_1y_1+c_1y_1^2$ with $a_1,b_1,c_1inmathbb{Z}$ and $b_1^2-4a_1c_1=D$ is equivalent to $ax^2+bxy+cy^2$ if there is $begin{bmatrix}p & q \ r & s end{bmatrix}inmathrm{SL}_2(mathbb{Z})$ such that
$$
a_1x_1^2+b_1x_1y_1+c_1y_1^2=a(px_1+qy_1)^2+b(px_1+qy_1)(rx_1+sy_1)+c(rx_1+sy_1)^2.
$$




This can be written again with matrices
$$
begin{bmatrix}a_1 & frac{b_1}2\ frac{b_1}2 & c_1 end{bmatrix} = begin{bmatrix}p & r \ q & s end{bmatrix} begin{bmatrix}a & frac{b}2\ frac{b}2 & c end{bmatrix} begin{bmatrix}p & q \ r & s end{bmatrix}
$$

and
$$
begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}p & q \ r & s end{bmatrix}begin{bmatrix}x_1\y_1end{bmatrix}.
$$



The next one is the reducing step.



Lemma 1




Any integer quadratic form with discriminant $D$ is equivalent to $ax^2+bxy+cy^2$ with $|b|leq |a|leq |c|$.




The proof is a repeated application of the matrices in $mathrm{SL}_2(mathbb{Z})$.
$$
begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix}, begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}.
$$



Corollary 1
The number of equivalence classes of integer quadratic forms with discriminant $D$ is finite.



If $b^2-4ac=D$ and $|b|leq |a|leq |c|$, then $|D|geq 4|ac|-|b^2| geq 3|ac|$ gives only finitely many choices for $a$ and $c$, subsequently for $b$.



Then it suffices to prove the following for $A=begin{bmatrix}a& b/2\ b/2&cend{bmatrix}$ with $b^2-4ac=D$ and $|b|leq |a|leq |c|$, as the sum can be written as a finite sum of the following kind.




The sum is absolutely convergent for $k>1$:
$$sum_{gammain mathrm{SL}_2(mathbb{Z})} frac1{left(| gamma z - alpha | |gamma z - alpha'|right)^k|rz+s|^{2k}} (*)
$$

where $gamma z=frac{pz+q}{rz+s}$, $alpha=frac{-b-sqrt{D}}{2a}$, and $alpha'=frac{-b+sqrt{D}}{2a}$.




Note that after the reduction, we have $ac<0$ and $|alpha-alpha'|>1$.



Let $Delta_{gamma}$ be the area of a triangle with vertices at $alpha, alpha',gamma z$. Let $theta_{gamma}$ be the angle formed by the corner $alpha, gamma z, alpha'$. Then $Delta_{gamma}=frac12| gamma z - alpha | |gamma z - alpha'|sin theta_{gamma}=frac12 |alpha-alpha'|Im(gamma z)=frac12|alpha-alpha'| frac{Im(z)}{|rz+s|^2}$.



Thus, it suffices to prove the convergence of
$$
sum_{gammain mathrm{SL}_2(mathbb{Z})} (sin theta_{gamma})^k
$$



Consider $Gamma=mathrm{SL}_2(mathbb{Z})$ and $H=left{pmbegin{bmatrix} 1&2m_0m\0&1end{bmatrix}, minmathbb{Z}right}$. Fix a fundamental domain $F_{infty}$ for $H$ as $F_{infty}={z| u-m_0 le Re z < u+m_0}$ and $u=frac{alpha+alpha'}2$. Here we set the integer $m_0$ large enough that $u-m_0<min(alpha,alpha')<max(alpha,alpha')<u+m_0$.
For $gamma znotin F_{infty}$, $gamma z$ is sufficiently away from any of $alpha$, $alpha'$. Thus, we have
$$
sum_{(r,s)=1, begin{bmatrix} *&*\r&send{bmatrix} z notin F_{infty}}frac1{left(| gamma z - alpha | |gamma z - alpha'|right)^k|rz+s|^{2k}}ll sum_{(r,s)=1} frac1{|rz+s|^{2k}}
$$

due to $sum frac 1{n^k}$ is convergent. Moreover, the sum on the right side is
$$
ll sum_{(r,s)=1, rsneq 0} frac1{|rs|^k}
$$

which is also convergent. The implied constants depend on $k,alpha,alpha',m_0,z$.



Thus, we consider the following sum
$$
sum_{gammain Hbackslash Gamma} (sintheta_{gamma})^k=sum_{gammainGamma, gamma z in F_{infty}} (sin theta_{gamma})^k.
$$

We need a bound for the number of orbit points $gamma z$ in the following figure. enter image description here






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  • Thank you for your detailed answer. I hope you can answer two more questions for my comprehension. The first is, what is the motivation behind substituting $A$ by $begin{bmatrix}alpha \ betaend{bmatrix} * begin{bmatrix}alpha' &beta'end{bmatrix}$, since by multiplying them I got $4a*begin{bmatrix}a & frac{b-sqrt{D}}{2} \ frac{b+sqrt{D}}{2} & cend{bmatrix}$. And $D$ has not to be a square number. Did I make a mistake? The second question is why do we have to inspect the case $t=0$ seperately.
    – Deavor
    Aug 7 at 11:30












  • Since we can interpret quadratic forms as matrices:$$begin{bmatrix}x&yend{bmatrix}begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}x\yend{bmatrix}=ax^2+(b+c)xy+dy^2$$, we may have that decomposition. Please postpone accepting, because I found an error in Lemma 2 and need to fix it.
    – i707107
    Aug 7 at 12:46













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2 Answers
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2 Answers
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active

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active

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active

oldest

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up vote
2
down vote



accepted
+150










It is actually rather straightforward. If $a=0$, then $b$ can take at most 2 values, none of them $0$, and we are left with the absolutely convergent sum $sum_c |c+bi|^{-k}$. If $c=0$, we can run a similar argument. So it is enough to consider $a,b,cne 0$. In this case the roots $u,v$ of the corresponding quadratic polynomial $P(x)=ax^2+bx+c$ are real and lie at the distance $sqrt D$ from $-frac b{2a}$. Thus $|P(z)|=|a||z-u||z-v|$. This is never less than $|a||Im z|^2$ and for $|b|>(6sqrt D+4|Re z|)|a|$ is not less than $|a||frac b{4a}|^2approx frac{|b|^2}{|a|}$. Thus, $|P(z)|gtrsimmax(|a|,frac{|b|^2}{|a|})ge |b|$ and we only need to consider $sum_{a,b,c}|b|^{-k}$. However, for fixed $b$, the number of non-zero pairs $(a,c)$ with $0ne b^2-D=4ac$ is at most the number of divisors of $b^2-D$, which grows slower than any positive power of $|b|$.






share|cite|improve this answer





















  • It is not clear why we can assume $|b|>(6sqrt D+4|Re z|)|a|$. Would you please explain this?
    – i707107
    Oct 6 at 4:48










  • @i707107 If $|b|$ is less, we just use the lower bound of $|a|$, which is better than $|b|^2/|a|$ in that case.
    – fedja
    Oct 6 at 10:37












  • This works great! Now, I think I actually made the problem unnecessarily difficult. Do you have any suggestion on the bound of orbit points $gamma z$ in my last figure?
    – i707107
    Oct 6 at 15:21










  • One small typo is $u, v$ are real and lie at the distance $sqrt D /2a$ from $-b/2a$. But, it does not affect the validity of this argument.
    – i707107
    Oct 6 at 15:25






  • 1




    @Deavor Yes, we just choose $varepsilon<k-1$.
    – fedja
    Dec 3 at 21:08















up vote
2
down vote



accepted
+150










It is actually rather straightforward. If $a=0$, then $b$ can take at most 2 values, none of them $0$, and we are left with the absolutely convergent sum $sum_c |c+bi|^{-k}$. If $c=0$, we can run a similar argument. So it is enough to consider $a,b,cne 0$. In this case the roots $u,v$ of the corresponding quadratic polynomial $P(x)=ax^2+bx+c$ are real and lie at the distance $sqrt D$ from $-frac b{2a}$. Thus $|P(z)|=|a||z-u||z-v|$. This is never less than $|a||Im z|^2$ and for $|b|>(6sqrt D+4|Re z|)|a|$ is not less than $|a||frac b{4a}|^2approx frac{|b|^2}{|a|}$. Thus, $|P(z)|gtrsimmax(|a|,frac{|b|^2}{|a|})ge |b|$ and we only need to consider $sum_{a,b,c}|b|^{-k}$. However, for fixed $b$, the number of non-zero pairs $(a,c)$ with $0ne b^2-D=4ac$ is at most the number of divisors of $b^2-D$, which grows slower than any positive power of $|b|$.






share|cite|improve this answer





















  • It is not clear why we can assume $|b|>(6sqrt D+4|Re z|)|a|$. Would you please explain this?
    – i707107
    Oct 6 at 4:48










  • @i707107 If $|b|$ is less, we just use the lower bound of $|a|$, which is better than $|b|^2/|a|$ in that case.
    – fedja
    Oct 6 at 10:37












  • This works great! Now, I think I actually made the problem unnecessarily difficult. Do you have any suggestion on the bound of orbit points $gamma z$ in my last figure?
    – i707107
    Oct 6 at 15:21










  • One small typo is $u, v$ are real and lie at the distance $sqrt D /2a$ from $-b/2a$. But, it does not affect the validity of this argument.
    – i707107
    Oct 6 at 15:25






  • 1




    @Deavor Yes, we just choose $varepsilon<k-1$.
    – fedja
    Dec 3 at 21:08













up vote
2
down vote



accepted
+150







up vote
2
down vote



accepted
+150




+150




It is actually rather straightforward. If $a=0$, then $b$ can take at most 2 values, none of them $0$, and we are left with the absolutely convergent sum $sum_c |c+bi|^{-k}$. If $c=0$, we can run a similar argument. So it is enough to consider $a,b,cne 0$. In this case the roots $u,v$ of the corresponding quadratic polynomial $P(x)=ax^2+bx+c$ are real and lie at the distance $sqrt D$ from $-frac b{2a}$. Thus $|P(z)|=|a||z-u||z-v|$. This is never less than $|a||Im z|^2$ and for $|b|>(6sqrt D+4|Re z|)|a|$ is not less than $|a||frac b{4a}|^2approx frac{|b|^2}{|a|}$. Thus, $|P(z)|gtrsimmax(|a|,frac{|b|^2}{|a|})ge |b|$ and we only need to consider $sum_{a,b,c}|b|^{-k}$. However, for fixed $b$, the number of non-zero pairs $(a,c)$ with $0ne b^2-D=4ac$ is at most the number of divisors of $b^2-D$, which grows slower than any positive power of $|b|$.






share|cite|improve this answer












It is actually rather straightforward. If $a=0$, then $b$ can take at most 2 values, none of them $0$, and we are left with the absolutely convergent sum $sum_c |c+bi|^{-k}$. If $c=0$, we can run a similar argument. So it is enough to consider $a,b,cne 0$. In this case the roots $u,v$ of the corresponding quadratic polynomial $P(x)=ax^2+bx+c$ are real and lie at the distance $sqrt D$ from $-frac b{2a}$. Thus $|P(z)|=|a||z-u||z-v|$. This is never less than $|a||Im z|^2$ and for $|b|>(6sqrt D+4|Re z|)|a|$ is not less than $|a||frac b{4a}|^2approx frac{|b|^2}{|a|}$. Thus, $|P(z)|gtrsimmax(|a|,frac{|b|^2}{|a|})ge |b|$ and we only need to consider $sum_{a,b,c}|b|^{-k}$. However, for fixed $b$, the number of non-zero pairs $(a,c)$ with $0ne b^2-D=4ac$ is at most the number of divisors of $b^2-D$, which grows slower than any positive power of $|b|$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 6 at 3:42









fedja

9,10011327




9,10011327












  • It is not clear why we can assume $|b|>(6sqrt D+4|Re z|)|a|$. Would you please explain this?
    – i707107
    Oct 6 at 4:48










  • @i707107 If $|b|$ is less, we just use the lower bound of $|a|$, which is better than $|b|^2/|a|$ in that case.
    – fedja
    Oct 6 at 10:37












  • This works great! Now, I think I actually made the problem unnecessarily difficult. Do you have any suggestion on the bound of orbit points $gamma z$ in my last figure?
    – i707107
    Oct 6 at 15:21










  • One small typo is $u, v$ are real and lie at the distance $sqrt D /2a$ from $-b/2a$. But, it does not affect the validity of this argument.
    – i707107
    Oct 6 at 15:25






  • 1




    @Deavor Yes, we just choose $varepsilon<k-1$.
    – fedja
    Dec 3 at 21:08


















  • It is not clear why we can assume $|b|>(6sqrt D+4|Re z|)|a|$. Would you please explain this?
    – i707107
    Oct 6 at 4:48










  • @i707107 If $|b|$ is less, we just use the lower bound of $|a|$, which is better than $|b|^2/|a|$ in that case.
    – fedja
    Oct 6 at 10:37












  • This works great! Now, I think I actually made the problem unnecessarily difficult. Do you have any suggestion on the bound of orbit points $gamma z$ in my last figure?
    – i707107
    Oct 6 at 15:21










  • One small typo is $u, v$ are real and lie at the distance $sqrt D /2a$ from $-b/2a$. But, it does not affect the validity of this argument.
    – i707107
    Oct 6 at 15:25






  • 1




    @Deavor Yes, we just choose $varepsilon<k-1$.
    – fedja
    Dec 3 at 21:08
















It is not clear why we can assume $|b|>(6sqrt D+4|Re z|)|a|$. Would you please explain this?
– i707107
Oct 6 at 4:48




It is not clear why we can assume $|b|>(6sqrt D+4|Re z|)|a|$. Would you please explain this?
– i707107
Oct 6 at 4:48












@i707107 If $|b|$ is less, we just use the lower bound of $|a|$, which is better than $|b|^2/|a|$ in that case.
– fedja
Oct 6 at 10:37






@i707107 If $|b|$ is less, we just use the lower bound of $|a|$, which is better than $|b|^2/|a|$ in that case.
– fedja
Oct 6 at 10:37














This works great! Now, I think I actually made the problem unnecessarily difficult. Do you have any suggestion on the bound of orbit points $gamma z$ in my last figure?
– i707107
Oct 6 at 15:21




This works great! Now, I think I actually made the problem unnecessarily difficult. Do you have any suggestion on the bound of orbit points $gamma z$ in my last figure?
– i707107
Oct 6 at 15:21












One small typo is $u, v$ are real and lie at the distance $sqrt D /2a$ from $-b/2a$. But, it does not affect the validity of this argument.
– i707107
Oct 6 at 15:25




One small typo is $u, v$ are real and lie at the distance $sqrt D /2a$ from $-b/2a$. But, it does not affect the validity of this argument.
– i707107
Oct 6 at 15:25




1




1




@Deavor Yes, we just choose $varepsilon<k-1$.
– fedja
Dec 3 at 21:08




@Deavor Yes, we just choose $varepsilon<k-1$.
– fedja
Dec 3 at 21:08










up vote
3
down vote













This is an incomplete answer using the Theory of Quadratic Forms. The constants $C$ appearing in this solution may not be the same. We begin with



Definition




Let $ax^2+bxy+cy^2$ be a quadratic form with $a, b, cin mathbb{Z}$ and $b^2-4ac=D$. We say that a quadratic form $a_1x_1^2+b_1x_1y_1+c_1y_1^2$ with $a_1,b_1,c_1inmathbb{Z}$ and $b_1^2-4a_1c_1=D$ is equivalent to $ax^2+bxy+cy^2$ if there is $begin{bmatrix}p & q \ r & s end{bmatrix}inmathrm{SL}_2(mathbb{Z})$ such that
$$
a_1x_1^2+b_1x_1y_1+c_1y_1^2=a(px_1+qy_1)^2+b(px_1+qy_1)(rx_1+sy_1)+c(rx_1+sy_1)^2.
$$




This can be written again with matrices
$$
begin{bmatrix}a_1 & frac{b_1}2\ frac{b_1}2 & c_1 end{bmatrix} = begin{bmatrix}p & r \ q & s end{bmatrix} begin{bmatrix}a & frac{b}2\ frac{b}2 & c end{bmatrix} begin{bmatrix}p & q \ r & s end{bmatrix}
$$

and
$$
begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}p & q \ r & s end{bmatrix}begin{bmatrix}x_1\y_1end{bmatrix}.
$$



The next one is the reducing step.



Lemma 1




Any integer quadratic form with discriminant $D$ is equivalent to $ax^2+bxy+cy^2$ with $|b|leq |a|leq |c|$.




The proof is a repeated application of the matrices in $mathrm{SL}_2(mathbb{Z})$.
$$
begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix}, begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}.
$$



Corollary 1
The number of equivalence classes of integer quadratic forms with discriminant $D$ is finite.



If $b^2-4ac=D$ and $|b|leq |a|leq |c|$, then $|D|geq 4|ac|-|b^2| geq 3|ac|$ gives only finitely many choices for $a$ and $c$, subsequently for $b$.



Then it suffices to prove the following for $A=begin{bmatrix}a& b/2\ b/2&cend{bmatrix}$ with $b^2-4ac=D$ and $|b|leq |a|leq |c|$, as the sum can be written as a finite sum of the following kind.




The sum is absolutely convergent for $k>1$:
$$sum_{gammain mathrm{SL}_2(mathbb{Z})} frac1{left(| gamma z - alpha | |gamma z - alpha'|right)^k|rz+s|^{2k}} (*)
$$

where $gamma z=frac{pz+q}{rz+s}$, $alpha=frac{-b-sqrt{D}}{2a}$, and $alpha'=frac{-b+sqrt{D}}{2a}$.




Note that after the reduction, we have $ac<0$ and $|alpha-alpha'|>1$.



Let $Delta_{gamma}$ be the area of a triangle with vertices at $alpha, alpha',gamma z$. Let $theta_{gamma}$ be the angle formed by the corner $alpha, gamma z, alpha'$. Then $Delta_{gamma}=frac12| gamma z - alpha | |gamma z - alpha'|sin theta_{gamma}=frac12 |alpha-alpha'|Im(gamma z)=frac12|alpha-alpha'| frac{Im(z)}{|rz+s|^2}$.



Thus, it suffices to prove the convergence of
$$
sum_{gammain mathrm{SL}_2(mathbb{Z})} (sin theta_{gamma})^k
$$



Consider $Gamma=mathrm{SL}_2(mathbb{Z})$ and $H=left{pmbegin{bmatrix} 1&2m_0m\0&1end{bmatrix}, minmathbb{Z}right}$. Fix a fundamental domain $F_{infty}$ for $H$ as $F_{infty}={z| u-m_0 le Re z < u+m_0}$ and $u=frac{alpha+alpha'}2$. Here we set the integer $m_0$ large enough that $u-m_0<min(alpha,alpha')<max(alpha,alpha')<u+m_0$.
For $gamma znotin F_{infty}$, $gamma z$ is sufficiently away from any of $alpha$, $alpha'$. Thus, we have
$$
sum_{(r,s)=1, begin{bmatrix} *&*\r&send{bmatrix} z notin F_{infty}}frac1{left(| gamma z - alpha | |gamma z - alpha'|right)^k|rz+s|^{2k}}ll sum_{(r,s)=1} frac1{|rz+s|^{2k}}
$$

due to $sum frac 1{n^k}$ is convergent. Moreover, the sum on the right side is
$$
ll sum_{(r,s)=1, rsneq 0} frac1{|rs|^k}
$$

which is also convergent. The implied constants depend on $k,alpha,alpha',m_0,z$.



Thus, we consider the following sum
$$
sum_{gammain Hbackslash Gamma} (sintheta_{gamma})^k=sum_{gammainGamma, gamma z in F_{infty}} (sin theta_{gamma})^k.
$$

We need a bound for the number of orbit points $gamma z$ in the following figure. enter image description here






share|cite|improve this answer























  • Thank you for your detailed answer. I hope you can answer two more questions for my comprehension. The first is, what is the motivation behind substituting $A$ by $begin{bmatrix}alpha \ betaend{bmatrix} * begin{bmatrix}alpha' &beta'end{bmatrix}$, since by multiplying them I got $4a*begin{bmatrix}a & frac{b-sqrt{D}}{2} \ frac{b+sqrt{D}}{2} & cend{bmatrix}$. And $D$ has not to be a square number. Did I make a mistake? The second question is why do we have to inspect the case $t=0$ seperately.
    – Deavor
    Aug 7 at 11:30












  • Since we can interpret quadratic forms as matrices:$$begin{bmatrix}x&yend{bmatrix}begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}x\yend{bmatrix}=ax^2+(b+c)xy+dy^2$$, we may have that decomposition. Please postpone accepting, because I found an error in Lemma 2 and need to fix it.
    – i707107
    Aug 7 at 12:46

















up vote
3
down vote













This is an incomplete answer using the Theory of Quadratic Forms. The constants $C$ appearing in this solution may not be the same. We begin with



Definition




Let $ax^2+bxy+cy^2$ be a quadratic form with $a, b, cin mathbb{Z}$ and $b^2-4ac=D$. We say that a quadratic form $a_1x_1^2+b_1x_1y_1+c_1y_1^2$ with $a_1,b_1,c_1inmathbb{Z}$ and $b_1^2-4a_1c_1=D$ is equivalent to $ax^2+bxy+cy^2$ if there is $begin{bmatrix}p & q \ r & s end{bmatrix}inmathrm{SL}_2(mathbb{Z})$ such that
$$
a_1x_1^2+b_1x_1y_1+c_1y_1^2=a(px_1+qy_1)^2+b(px_1+qy_1)(rx_1+sy_1)+c(rx_1+sy_1)^2.
$$




This can be written again with matrices
$$
begin{bmatrix}a_1 & frac{b_1}2\ frac{b_1}2 & c_1 end{bmatrix} = begin{bmatrix}p & r \ q & s end{bmatrix} begin{bmatrix}a & frac{b}2\ frac{b}2 & c end{bmatrix} begin{bmatrix}p & q \ r & s end{bmatrix}
$$

and
$$
begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}p & q \ r & s end{bmatrix}begin{bmatrix}x_1\y_1end{bmatrix}.
$$



The next one is the reducing step.



Lemma 1




Any integer quadratic form with discriminant $D$ is equivalent to $ax^2+bxy+cy^2$ with $|b|leq |a|leq |c|$.




The proof is a repeated application of the matrices in $mathrm{SL}_2(mathbb{Z})$.
$$
begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix}, begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}.
$$



Corollary 1
The number of equivalence classes of integer quadratic forms with discriminant $D$ is finite.



If $b^2-4ac=D$ and $|b|leq |a|leq |c|$, then $|D|geq 4|ac|-|b^2| geq 3|ac|$ gives only finitely many choices for $a$ and $c$, subsequently for $b$.



Then it suffices to prove the following for $A=begin{bmatrix}a& b/2\ b/2&cend{bmatrix}$ with $b^2-4ac=D$ and $|b|leq |a|leq |c|$, as the sum can be written as a finite sum of the following kind.




The sum is absolutely convergent for $k>1$:
$$sum_{gammain mathrm{SL}_2(mathbb{Z})} frac1{left(| gamma z - alpha | |gamma z - alpha'|right)^k|rz+s|^{2k}} (*)
$$

where $gamma z=frac{pz+q}{rz+s}$, $alpha=frac{-b-sqrt{D}}{2a}$, and $alpha'=frac{-b+sqrt{D}}{2a}$.




Note that after the reduction, we have $ac<0$ and $|alpha-alpha'|>1$.



Let $Delta_{gamma}$ be the area of a triangle with vertices at $alpha, alpha',gamma z$. Let $theta_{gamma}$ be the angle formed by the corner $alpha, gamma z, alpha'$. Then $Delta_{gamma}=frac12| gamma z - alpha | |gamma z - alpha'|sin theta_{gamma}=frac12 |alpha-alpha'|Im(gamma z)=frac12|alpha-alpha'| frac{Im(z)}{|rz+s|^2}$.



Thus, it suffices to prove the convergence of
$$
sum_{gammain mathrm{SL}_2(mathbb{Z})} (sin theta_{gamma})^k
$$



Consider $Gamma=mathrm{SL}_2(mathbb{Z})$ and $H=left{pmbegin{bmatrix} 1&2m_0m\0&1end{bmatrix}, minmathbb{Z}right}$. Fix a fundamental domain $F_{infty}$ for $H$ as $F_{infty}={z| u-m_0 le Re z < u+m_0}$ and $u=frac{alpha+alpha'}2$. Here we set the integer $m_0$ large enough that $u-m_0<min(alpha,alpha')<max(alpha,alpha')<u+m_0$.
For $gamma znotin F_{infty}$, $gamma z$ is sufficiently away from any of $alpha$, $alpha'$. Thus, we have
$$
sum_{(r,s)=1, begin{bmatrix} *&*\r&send{bmatrix} z notin F_{infty}}frac1{left(| gamma z - alpha | |gamma z - alpha'|right)^k|rz+s|^{2k}}ll sum_{(r,s)=1} frac1{|rz+s|^{2k}}
$$

due to $sum frac 1{n^k}$ is convergent. Moreover, the sum on the right side is
$$
ll sum_{(r,s)=1, rsneq 0} frac1{|rs|^k}
$$

which is also convergent. The implied constants depend on $k,alpha,alpha',m_0,z$.



Thus, we consider the following sum
$$
sum_{gammain Hbackslash Gamma} (sintheta_{gamma})^k=sum_{gammainGamma, gamma z in F_{infty}} (sin theta_{gamma})^k.
$$

We need a bound for the number of orbit points $gamma z$ in the following figure. enter image description here






share|cite|improve this answer























  • Thank you for your detailed answer. I hope you can answer two more questions for my comprehension. The first is, what is the motivation behind substituting $A$ by $begin{bmatrix}alpha \ betaend{bmatrix} * begin{bmatrix}alpha' &beta'end{bmatrix}$, since by multiplying them I got $4a*begin{bmatrix}a & frac{b-sqrt{D}}{2} \ frac{b+sqrt{D}}{2} & cend{bmatrix}$. And $D$ has not to be a square number. Did I make a mistake? The second question is why do we have to inspect the case $t=0$ seperately.
    – Deavor
    Aug 7 at 11:30












  • Since we can interpret quadratic forms as matrices:$$begin{bmatrix}x&yend{bmatrix}begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}x\yend{bmatrix}=ax^2+(b+c)xy+dy^2$$, we may have that decomposition. Please postpone accepting, because I found an error in Lemma 2 and need to fix it.
    – i707107
    Aug 7 at 12:46















up vote
3
down vote










up vote
3
down vote









This is an incomplete answer using the Theory of Quadratic Forms. The constants $C$ appearing in this solution may not be the same. We begin with



Definition




Let $ax^2+bxy+cy^2$ be a quadratic form with $a, b, cin mathbb{Z}$ and $b^2-4ac=D$. We say that a quadratic form $a_1x_1^2+b_1x_1y_1+c_1y_1^2$ with $a_1,b_1,c_1inmathbb{Z}$ and $b_1^2-4a_1c_1=D$ is equivalent to $ax^2+bxy+cy^2$ if there is $begin{bmatrix}p & q \ r & s end{bmatrix}inmathrm{SL}_2(mathbb{Z})$ such that
$$
a_1x_1^2+b_1x_1y_1+c_1y_1^2=a(px_1+qy_1)^2+b(px_1+qy_1)(rx_1+sy_1)+c(rx_1+sy_1)^2.
$$




This can be written again with matrices
$$
begin{bmatrix}a_1 & frac{b_1}2\ frac{b_1}2 & c_1 end{bmatrix} = begin{bmatrix}p & r \ q & s end{bmatrix} begin{bmatrix}a & frac{b}2\ frac{b}2 & c end{bmatrix} begin{bmatrix}p & q \ r & s end{bmatrix}
$$

and
$$
begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}p & q \ r & s end{bmatrix}begin{bmatrix}x_1\y_1end{bmatrix}.
$$



The next one is the reducing step.



Lemma 1




Any integer quadratic form with discriminant $D$ is equivalent to $ax^2+bxy+cy^2$ with $|b|leq |a|leq |c|$.




The proof is a repeated application of the matrices in $mathrm{SL}_2(mathbb{Z})$.
$$
begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix}, begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}.
$$



Corollary 1
The number of equivalence classes of integer quadratic forms with discriminant $D$ is finite.



If $b^2-4ac=D$ and $|b|leq |a|leq |c|$, then $|D|geq 4|ac|-|b^2| geq 3|ac|$ gives only finitely many choices for $a$ and $c$, subsequently for $b$.



Then it suffices to prove the following for $A=begin{bmatrix}a& b/2\ b/2&cend{bmatrix}$ with $b^2-4ac=D$ and $|b|leq |a|leq |c|$, as the sum can be written as a finite sum of the following kind.




The sum is absolutely convergent for $k>1$:
$$sum_{gammain mathrm{SL}_2(mathbb{Z})} frac1{left(| gamma z - alpha | |gamma z - alpha'|right)^k|rz+s|^{2k}} (*)
$$

where $gamma z=frac{pz+q}{rz+s}$, $alpha=frac{-b-sqrt{D}}{2a}$, and $alpha'=frac{-b+sqrt{D}}{2a}$.




Note that after the reduction, we have $ac<0$ and $|alpha-alpha'|>1$.



Let $Delta_{gamma}$ be the area of a triangle with vertices at $alpha, alpha',gamma z$. Let $theta_{gamma}$ be the angle formed by the corner $alpha, gamma z, alpha'$. Then $Delta_{gamma}=frac12| gamma z - alpha | |gamma z - alpha'|sin theta_{gamma}=frac12 |alpha-alpha'|Im(gamma z)=frac12|alpha-alpha'| frac{Im(z)}{|rz+s|^2}$.



Thus, it suffices to prove the convergence of
$$
sum_{gammain mathrm{SL}_2(mathbb{Z})} (sin theta_{gamma})^k
$$



Consider $Gamma=mathrm{SL}_2(mathbb{Z})$ and $H=left{pmbegin{bmatrix} 1&2m_0m\0&1end{bmatrix}, minmathbb{Z}right}$. Fix a fundamental domain $F_{infty}$ for $H$ as $F_{infty}={z| u-m_0 le Re z < u+m_0}$ and $u=frac{alpha+alpha'}2$. Here we set the integer $m_0$ large enough that $u-m_0<min(alpha,alpha')<max(alpha,alpha')<u+m_0$.
For $gamma znotin F_{infty}$, $gamma z$ is sufficiently away from any of $alpha$, $alpha'$. Thus, we have
$$
sum_{(r,s)=1, begin{bmatrix} *&*\r&send{bmatrix} z notin F_{infty}}frac1{left(| gamma z - alpha | |gamma z - alpha'|right)^k|rz+s|^{2k}}ll sum_{(r,s)=1} frac1{|rz+s|^{2k}}
$$

due to $sum frac 1{n^k}$ is convergent. Moreover, the sum on the right side is
$$
ll sum_{(r,s)=1, rsneq 0} frac1{|rs|^k}
$$

which is also convergent. The implied constants depend on $k,alpha,alpha',m_0,z$.



Thus, we consider the following sum
$$
sum_{gammain Hbackslash Gamma} (sintheta_{gamma})^k=sum_{gammainGamma, gamma z in F_{infty}} (sin theta_{gamma})^k.
$$

We need a bound for the number of orbit points $gamma z$ in the following figure. enter image description here






share|cite|improve this answer














This is an incomplete answer using the Theory of Quadratic Forms. The constants $C$ appearing in this solution may not be the same. We begin with



Definition




Let $ax^2+bxy+cy^2$ be a quadratic form with $a, b, cin mathbb{Z}$ and $b^2-4ac=D$. We say that a quadratic form $a_1x_1^2+b_1x_1y_1+c_1y_1^2$ with $a_1,b_1,c_1inmathbb{Z}$ and $b_1^2-4a_1c_1=D$ is equivalent to $ax^2+bxy+cy^2$ if there is $begin{bmatrix}p & q \ r & s end{bmatrix}inmathrm{SL}_2(mathbb{Z})$ such that
$$
a_1x_1^2+b_1x_1y_1+c_1y_1^2=a(px_1+qy_1)^2+b(px_1+qy_1)(rx_1+sy_1)+c(rx_1+sy_1)^2.
$$




This can be written again with matrices
$$
begin{bmatrix}a_1 & frac{b_1}2\ frac{b_1}2 & c_1 end{bmatrix} = begin{bmatrix}p & r \ q & s end{bmatrix} begin{bmatrix}a & frac{b}2\ frac{b}2 & c end{bmatrix} begin{bmatrix}p & q \ r & s end{bmatrix}
$$

and
$$
begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}p & q \ r & s end{bmatrix}begin{bmatrix}x_1\y_1end{bmatrix}.
$$



The next one is the reducing step.



Lemma 1




Any integer quadratic form with discriminant $D$ is equivalent to $ax^2+bxy+cy^2$ with $|b|leq |a|leq |c|$.




The proof is a repeated application of the matrices in $mathrm{SL}_2(mathbb{Z})$.
$$
begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix}, begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}.
$$



Corollary 1
The number of equivalence classes of integer quadratic forms with discriminant $D$ is finite.



If $b^2-4ac=D$ and $|b|leq |a|leq |c|$, then $|D|geq 4|ac|-|b^2| geq 3|ac|$ gives only finitely many choices for $a$ and $c$, subsequently for $b$.



Then it suffices to prove the following for $A=begin{bmatrix}a& b/2\ b/2&cend{bmatrix}$ with $b^2-4ac=D$ and $|b|leq |a|leq |c|$, as the sum can be written as a finite sum of the following kind.




The sum is absolutely convergent for $k>1$:
$$sum_{gammain mathrm{SL}_2(mathbb{Z})} frac1{left(| gamma z - alpha | |gamma z - alpha'|right)^k|rz+s|^{2k}} (*)
$$

where $gamma z=frac{pz+q}{rz+s}$, $alpha=frac{-b-sqrt{D}}{2a}$, and $alpha'=frac{-b+sqrt{D}}{2a}$.




Note that after the reduction, we have $ac<0$ and $|alpha-alpha'|>1$.



Let $Delta_{gamma}$ be the area of a triangle with vertices at $alpha, alpha',gamma z$. Let $theta_{gamma}$ be the angle formed by the corner $alpha, gamma z, alpha'$. Then $Delta_{gamma}=frac12| gamma z - alpha | |gamma z - alpha'|sin theta_{gamma}=frac12 |alpha-alpha'|Im(gamma z)=frac12|alpha-alpha'| frac{Im(z)}{|rz+s|^2}$.



Thus, it suffices to prove the convergence of
$$
sum_{gammain mathrm{SL}_2(mathbb{Z})} (sin theta_{gamma})^k
$$



Consider $Gamma=mathrm{SL}_2(mathbb{Z})$ and $H=left{pmbegin{bmatrix} 1&2m_0m\0&1end{bmatrix}, minmathbb{Z}right}$. Fix a fundamental domain $F_{infty}$ for $H$ as $F_{infty}={z| u-m_0 le Re z < u+m_0}$ and $u=frac{alpha+alpha'}2$. Here we set the integer $m_0$ large enough that $u-m_0<min(alpha,alpha')<max(alpha,alpha')<u+m_0$.
For $gamma znotin F_{infty}$, $gamma z$ is sufficiently away from any of $alpha$, $alpha'$. Thus, we have
$$
sum_{(r,s)=1, begin{bmatrix} *&*\r&send{bmatrix} z notin F_{infty}}frac1{left(| gamma z - alpha | |gamma z - alpha'|right)^k|rz+s|^{2k}}ll sum_{(r,s)=1} frac1{|rz+s|^{2k}}
$$

due to $sum frac 1{n^k}$ is convergent. Moreover, the sum on the right side is
$$
ll sum_{(r,s)=1, rsneq 0} frac1{|rs|^k}
$$

which is also convergent. The implied constants depend on $k,alpha,alpha',m_0,z$.



Thus, we consider the following sum
$$
sum_{gammain Hbackslash Gamma} (sintheta_{gamma})^k=sum_{gammainGamma, gamma z in F_{infty}} (sin theta_{gamma})^k.
$$

We need a bound for the number of orbit points $gamma z$ in the following figure. enter image description here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Oct 4 at 2:07

























answered Aug 7 at 5:37









i707107

11.9k21447




11.9k21447












  • Thank you for your detailed answer. I hope you can answer two more questions for my comprehension. The first is, what is the motivation behind substituting $A$ by $begin{bmatrix}alpha \ betaend{bmatrix} * begin{bmatrix}alpha' &beta'end{bmatrix}$, since by multiplying them I got $4a*begin{bmatrix}a & frac{b-sqrt{D}}{2} \ frac{b+sqrt{D}}{2} & cend{bmatrix}$. And $D$ has not to be a square number. Did I make a mistake? The second question is why do we have to inspect the case $t=0$ seperately.
    – Deavor
    Aug 7 at 11:30












  • Since we can interpret quadratic forms as matrices:$$begin{bmatrix}x&yend{bmatrix}begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}x\yend{bmatrix}=ax^2+(b+c)xy+dy^2$$, we may have that decomposition. Please postpone accepting, because I found an error in Lemma 2 and need to fix it.
    – i707107
    Aug 7 at 12:46




















  • Thank you for your detailed answer. I hope you can answer two more questions for my comprehension. The first is, what is the motivation behind substituting $A$ by $begin{bmatrix}alpha \ betaend{bmatrix} * begin{bmatrix}alpha' &beta'end{bmatrix}$, since by multiplying them I got $4a*begin{bmatrix}a & frac{b-sqrt{D}}{2} \ frac{b+sqrt{D}}{2} & cend{bmatrix}$. And $D$ has not to be a square number. Did I make a mistake? The second question is why do we have to inspect the case $t=0$ seperately.
    – Deavor
    Aug 7 at 11:30












  • Since we can interpret quadratic forms as matrices:$$begin{bmatrix}x&yend{bmatrix}begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}x\yend{bmatrix}=ax^2+(b+c)xy+dy^2$$, we may have that decomposition. Please postpone accepting, because I found an error in Lemma 2 and need to fix it.
    – i707107
    Aug 7 at 12:46


















Thank you for your detailed answer. I hope you can answer two more questions for my comprehension. The first is, what is the motivation behind substituting $A$ by $begin{bmatrix}alpha \ betaend{bmatrix} * begin{bmatrix}alpha' &beta'end{bmatrix}$, since by multiplying them I got $4a*begin{bmatrix}a & frac{b-sqrt{D}}{2} \ frac{b+sqrt{D}}{2} & cend{bmatrix}$. And $D$ has not to be a square number. Did I make a mistake? The second question is why do we have to inspect the case $t=0$ seperately.
– Deavor
Aug 7 at 11:30






Thank you for your detailed answer. I hope you can answer two more questions for my comprehension. The first is, what is the motivation behind substituting $A$ by $begin{bmatrix}alpha \ betaend{bmatrix} * begin{bmatrix}alpha' &beta'end{bmatrix}$, since by multiplying them I got $4a*begin{bmatrix}a & frac{b-sqrt{D}}{2} \ frac{b+sqrt{D}}{2} & cend{bmatrix}$. And $D$ has not to be a square number. Did I make a mistake? The second question is why do we have to inspect the case $t=0$ seperately.
– Deavor
Aug 7 at 11:30














Since we can interpret quadratic forms as matrices:$$begin{bmatrix}x&yend{bmatrix}begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}x\yend{bmatrix}=ax^2+(b+c)xy+dy^2$$, we may have that decomposition. Please postpone accepting, because I found an error in Lemma 2 and need to fix it.
– i707107
Aug 7 at 12:46






Since we can interpret quadratic forms as matrices:$$begin{bmatrix}x&yend{bmatrix}begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}x\yend{bmatrix}=ax^2+(b+c)xy+dy^2$$, we may have that decomposition. Please postpone accepting, because I found an error in Lemma 2 and need to fix it.
– i707107
Aug 7 at 12:46




















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