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prove there exist postive integers $a,b$ such $p^2|a^2+ab+b^2$
$begingroup$
Problem 1: Let prime $pequiv 1pmod 3$.show that:there exist postive integers $ale b<p$ such
$$p^2|a^2+ab+b^2$$
I have only prove there $a,b$ such $$p|a^2+ab+b^2$$
Problem 1 from this:
Problem 2.3 (Noam Elkies). Prove that there are infinitely many triples $(a,b,p)$ of integers, with $p$ prime and $0<aleq b<p$, for which $p^5$ divides $(a+b)^p-a^p-b^p$.
The key claim is that if $pequiv1pmod3$, then $$p(x^2+xy+y^2)^2;mathrm{divides};(x+y)^p-x^p-y^p$$ as polynomials in $x$ and $y$. $color{red}{underline{color{black}{text{Since it's known that}}}}$ one can select $a$ and $b$ such that $color{red}{underline{color{black}{p^2mid a^2+ab+b^2}}}$, the conclusion follows. (The theory of quadratic forms tells us we can do it with $p^2=a^2+ab+b^2$; Thue's lemma lets us do it by solving $x^2+x+1equiv0pmod{p^2}$.)
To prove this, it is the same to show that $$(x^2+x+1)^2;mathrm{divides};F(x)overset{mathrm{def}}=(x+1)^p-x^p-1.$$ since the binomial coefficients $binom pk$ are clearly divisible by $p$. Let $zeta$ be a third root of unity. Then $F(zeta)=(1+zeta)^p-zeta^p-1=-zeta^2-zeta-1=0$. Moreover, $F'(x)=p(x+1)^{p-1}-px^{p-1}$, so $F'(zeta)=p-p=0$. Hence $zeta$ is a double root of $F$ as needed.
(Incidentally, $p=2017$ works!)
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number-theory
edited Jan 15 at 8:58
manooooh
5681517
asked Dec 12 '18 at 13:18
geromtygeromty
977423
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add a comment |
$begingroup$
Problem 1: Let prime $pequiv 1pmod 3$.show that:there exist postive integers $ale b<p$ such
$$p^2|a^2+ab+b^2$$
I have only prove there $a,b$ such $$p|a^2+ab+b^2$$
Problem 1 from this:
Problem 2.3 (Noam Elkies). Prove that there are infinitely many triples $(a,b,p)$ of integers, with $p$ prime and $0<aleq b<p$, for which $p^5$ divides $(a+b)^p-a^p-b^p$.
The key claim is that if $pequiv1pmod3$, then $$p(x^2+xy+y^2)^2;mathrm{divides};(x+y)^p-x^p-y^p$$ as polynomials in $x$ and $y$. $color{red}{underline{color{black}{text{Since it's known that}}}}$ one can select $a$ and $b$ such that $color{red}{underline{color{black}{p^2mid a^2+ab+b^2}}}$, the conclusion follows. (The theory of quadratic forms tells us we can do it with $p^2=a^2+ab+b^2$; Thue's lemma lets us do it by solving $x^2+x+1equiv0pmod{p^2}$.)
To prove this, it is the same to show that $$(x^2+x+1)^2;mathrm{divides};F(x)overset{mathrm{def}}=(x+1)^p-x^p-1.$$ since the binomial coefficients $binom pk$ are clearly divisible by $p$. Let $zeta$ be a third root of unity. Then $F(zeta)=(1+zeta)^p-zeta^p-1=-zeta^2-zeta-1=0$. Moreover, $F'(x)=p(x+1)^{p-1}-px^{p-1}$, so $F'(zeta)=p-p=0$. Hence $zeta$ is a double root of $F$ as needed.
(Incidentally, $p=2017$ works!)
Image that replaced the text
number-theory
edited Jan 15 at 8:58
manooooh
5681517
asked Dec 12 '18 at 13:18
geromtygeromty
977423
$endgroup$
add a comment |
$begingroup$
Problem 1: Let prime $pequiv 1pmod 3$.show that:there exist postive integers $ale b<p$ such
$$p^2|a^2+ab+b^2$$
I have only prove there $a,b$ such $$p|a^2+ab+b^2$$
Problem 1 from this:
Problem 2.3 (Noam Elkies). Prove that there are infinitely many triples $(a,b,p)$ of integers, with $p$ prime and $0<aleq b<p$, for which $p^5$ divides $(a+b)^p-a^p-b^p$.
The key claim is that if $pequiv1pmod3$, then $$p(x^2+xy+y^2)^2;mathrm{divides};(x+y)^p-x^p-y^p$$ as polynomials in $x$ and $y$. $color{red}{underline{color{black}{text{Since it's known that}}}}$ one can select $a$ and $b$ such that $color{red}{underline{color{black}{p^2mid a^2+ab+b^2}}}$, the conclusion follows. (The theory of quadratic forms tells us we can do it with $p^2=a^2+ab+b^2$; Thue's lemma lets us do it by solving $x^2+x+1equiv0pmod{p^2}$.)
To prove this, it is the same to show that $$(x^2+x+1)^2;mathrm{divides};F(x)overset{mathrm{def}}=(x+1)^p-x^p-1.$$ since the binomial coefficients $binom pk$ are clearly divisible by $p$. Let $zeta$ be a third root of unity. Then $F(zeta)=(1+zeta)^p-zeta^p-1=-zeta^2-zeta-1=0$. Moreover, $F'(x)=p(x+1)^{p-1}-px^{p-1}$, so $F'(zeta)=p-p=0$. Hence $zeta$ is a double root of $F$ as needed.
(Incidentally, $p=2017$ works!)
Image that replaced the text
number-theory
edited Jan 15 at 8:58
manooooh
5681517
asked Dec 12 '18 at 13:18
geromtygeromty
977423
$endgroup$
Problem 1: Let prime $pequiv 1pmod 3$.show that:there exist postive integers $ale b<p$ such
$$p^2|a^2+ab+b^2$$
I have only prove there $a,b$ such $$p|a^2+ab+b^2$$
Problem 1 from this:
Problem 2.3 (Noam Elkies). Prove that there are infinitely many triples $(a,b,p)$ of integers, with $p$ prime and $0<aleq b<p$, for which $p^5$ divides $(a+b)^p-a^p-b^p$.
The key claim is that if $pequiv1pmod3$, then $$p(x^2+xy+y^2)^2;mathrm{divides};(x+y)^p-x^p-y^p$$ as polynomials in $x$ and $y$. $color{red}{underline{color{black}{text{Since it's known that}}}}$ one can select $a$ and $b$ such that $color{red}{underline{color{black}{p^2mid a^2+ab+b^2}}}$, the conclusion follows. (The theory of quadratic forms tells us we can do it with $p^2=a^2+ab+b^2$; Thue's lemma lets us do it by solving $x^2+x+1equiv0pmod{p^2}$.)
To prove this, it is the same to show that $$(x^2+x+1)^2;mathrm{divides};F(x)overset{mathrm{def}}=(x+1)^p-x^p-1.$$ since the binomial coefficients $binom pk$ are clearly divisible by $p$. Let $zeta$ be a third root of unity. Then $F(zeta)=(1+zeta)^p-zeta^p-1=-zeta^2-zeta-1=0$. Moreover, $F'(x)=p(x+1)^{p-1}-px^{p-1}$, so $F'(zeta)=p-p=0$. Hence $zeta$ is a double root of $F$ as needed.
(Incidentally, $p=2017$ works!)
Image that replaced the text
number-theory
number-theory
edited Jan 15 at 8:58
manooooh
5681517
asked Dec 12 '18 at 13:18
geromtygeromty
977423
edited Jan 15 at 8:58
manooooh
5681517
asked Dec 12 '18 at 13:18
geromtygeromty
977423
edited Jan 15 at 8:58
manooooh
5681517
edited Jan 15 at 8:58
manooooh
5681517
edited Jan 15 at 8:58
manooooh
5681517
5681517
asked Dec 12 '18 at 13:18
geromtygeromty
977423
asked Dec 12 '18 at 13:18
geromtygeromty
977423
asked Dec 12 '18 at 13:18
geromtygeromty
977423
977423
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Alternate solution via Thue's Lemma
Proposition. Let $pequiv 1pmod 3$ be a prime. Then there is a solution to
$$
x^2+x+1equiv 0 pmod{p^2}
$$
Proof. By Quadratic Reciprocity, $-3$ is a quadratic residue $pmod p$, hence
$$
u^2 equiv -3 pmod p
$$
for some integer $u$. Setting $uequiv 2c+1pmod p$, this becomes
$$
c^2+c+1 equiv 0 pmod p
$$
So $c$ is a solution to $f(x):=x^2+x+1equiv 0pmod p$. Hensel's Lemma tells us that if
$$
f'(c)notequiv 0 pmod p
$$
Then there exists a unique $d=c+kp$ such that
$$
f(d) equiv 0 pmod{p^2}
$$
Since
$$
f'(c)^2 = (2c+1)^2 equiv -3 pmod p
$$
If $f'(c)equiv 0 pmod p$ then $0equiv -3pmod p$ which implies $p=3$. This contradicts $pequiv 1pmod 3$, therefore $f'(c)notequiv 0 pmod p$.
Hence there is some $d$ such that
$$
d^2+d+1equiv 0 pmod{p^2}
$$
$$tag*{$square$}$$
We now solve our main problem.
Lemma. Let $pequiv 1pmod 3$ be a prime. Then there are integers $0<a,b<p$ such that
$$
a^2+ab+b^2equiv 0 pmod{p^2}
$$
Proof. We have shown that there is some integer $d$ such that
$$
d^2+d+1 equiv 0pmod{p^2}
$$
Now by Thue's Lemma, since $gcd(d,p^2)=1$, we can find $0<|a|,|b|<p$ such that
$$
db equiv a pmod{p^2}
$$
By rearranging the sign, we may assume that $b>0$ and $-p<a<p$. Now $a,b$ satisfies
$$
a^2+ab+b^2 equiv d^2b^2+db^2+b^2 equiv b^2(d^2+d+1)equiv 0 pmod{p^2}
$$
Therefore if $a> 0$ we have found our solution $0<a,b<p$. Otherwise, $a<0$ and we consider two cases ($-aneq b$, otherwise $dequiv -1pmod{p^2}$ contradicts $d^2+d+1equiv 0pmod{p^2}$):
Case 1: $-a > b$
We use the equality
$$
a^2+ab+b^2 = (-a-b)^2 + (-a-b)b + b^2
$$
and obtain a solution $(a',b') = (-a-b,b)$. Since $0<-a<p$ and $0<b<p$, we have $-a-b< p$. By assumption $0 <-a-b$, therefore $0 < a',b' < p$ and we are done.
Case 2: $-a < b$
We use another equality:
$$
a^2 + ab + b^2 = (-a)^2 + (-a)(a+b) + (a+b)^2
$$
and set the new solution as $(a',b') = (-a,a+b)$. Clearly $0 < a'=-a < p$. Since $-p < a < 0$ and $0 < b < p$, we get $a+b < p $. Finally, by assumption we have $0 < a+b$, therefore $0 < a+b=b' < p$. Hence we have a solution $0<a',b' < p$.
$$tag*{$square$}$$
This completes the proof.
Extra. Let $0<a,b<p$ and
$$a^2+ab+b^2equiv 0 pmod{p^2}$$
for some odd $p$. Then
$$a^2+ab+b^2 = p^2$$
Proof. Since $0<a,b< p$, we have
$$
a^2 + ab + b^2 < 3p^2
$$
Morever, since $a^2+ab+b^2equiv 0 pmod{p^2}$, we must have $a^2+ab+b^2 = p^2$ or $2p^2$. If $a^2+ab+b^2=2p^2$ then
$$
a^2+ab+b^2 equiv 0 pmod 2
$$
There is no solution if $a$ or $b$ is odd, therefore both $a,b$ are even. But this means
$$
a^2+ab+b^2equiv 0 pmod 4
$$
Since $a^2+ab+b^2$ is also divisible by odd $p^2$, it is divisible by $4p^2$. This contradicts that $a^2+ab+b^2 < 3p^2$, therefore we must have
$$
a^2+ab+b^2 = p^2
$$
$$tag*{$square$}$$
answered Jan 15 at 6:49
Yong Hao NgYong Hao Ng
3,5691222
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add a comment |
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This relies on the following statement: if some residue class $a$ is a square mod $p$, then it is a square mod $p^2$.
Indeed, if $b^2=a+tp [p^2]$, then $(b+kp)^2=a+(t+2k)p [p^2]$.
So you know that there is some $x$ such that $p|x^2+x+1$, thus $-3=(2x-1)^2$ is a square mod $p$, thus mod $p^2$, and if $y^2=-3[p^2]$, then $a=(1+y)/2$ (mod $p^2$) and $1$ work.
answered Dec 12 '18 at 13:32
MindlackMindlack
3,73518
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Your LaTeX is broken
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– Lucas Henrique
Dec 12 '18 at 13:33
1
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Thanks, corrected!
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– Mindlack
Dec 12 '18 at 13:34
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I can't undertand your unswer,first your step ,what's mean
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– math110
Dec 16 '18 at 15:22
add a comment |
$begingroup$
1. Solution to $p^2= X^2+3Y^2$
It is well known that $pequiv 1pmod 3$ if and only if
$$
p=x^2+3y^2
$$
for some integers $x,y$. So immediately we have a presentation of $p^2$ as
$$
begin{align}
p^2 &= (x^2+3y^2)^2\
&=(x+sqrt{-3}y)^2(x-sqrt{-3}y)^2\
&=((x^2-3y^2)+2xysqrt{-3})((x^2-3y^2)-2xysqrt{-3})\
&=(x^2-3y^2)^2+3(2xy)^2
end{align}
$$
2. Solution to $p^2=a^2+ab+b^2,quad 0leq a,b< p$
The expression $u^2+3v^2$ can be rewritten in the following three ways:
$$
begin{align}
u^2+3v^2 &= (pm u-v)^2 + (pm u-v)(2v) + (2v)^2\
u^2+3v^2 &= (u+v)^2 + (u+v)(-u+v) + (-u+v)^2
end{align}
$$
From (1), we have
$$
p^2 = (x^2-3y)^2 + 3(2xy)^2
$$
Therefore applying any of the three transformations will already give us
$$
p^2 = a^2 + ab+b^2
$$
We shall show that depending on whether $x<y$, $y<x<3y$ or $3y<x$, exactly one of the previous transformations will give us a solution such that $0leq a,b<p$. (Clearly $xneq y$ and $xneq 3y$, or $p$ is not prime.)
Case 2a: $x<y$
We use the transformation
$$
u^2+3v^2 = (-u-v)^2 + (-u-v)(2v) + (2v)^2
$$
Hence
$$
begin{align}
p^2 &= (x^2-3y^2)^2 + 3(2xy)^2\
&= (3y^2-x^2-2xy)^2 + (3y^2-x^2-2xy)(4xy) + (4xy)^2\
&= a^2+ab+b^2
end{align}
$$
Hence
$$
begin{align}
a &= 3y^2-x^2-2xy = (3y+x)(y-x)\
b &= 4xy
end{align}
$$
Since $0<x<y$, we get $0< a,b$. Next we check that
$$
begin{align}
p-a &= x^2+3y^2 - (3y^2-x^2-2xy) = 2x^2+2xy > 0\
p-b &= x^2+3y^2 - 4xy = (3y-x)(y-x) > 0
end{align}
$$
Therefore $0< a,b < p$.
Case 2b: $3y<x$
We use the transformation
$$
u^2+3v^2 = (u-v)^2 + (u-v)(2v) + (2v)^2
$$
Hence
$$
begin{align}
p^2 &= (x^2-3y^2)^2 + 3(2xy)^2\
&= (x^2-3y^2-2xy)^2 + (x^2-3y^2-2xy)(4xy) + (4xy)^2\
&= a^2+ab+b^2
end{align}
$$
Hence
$$
begin{align}
a &= x^2-3y^2-2xy = (x-3y)(x+y)\
b &= 4xy
end{align}
$$
Since $0<3y < x$, we get $0< a,b$. Next we check that
$$
begin{align}
p-a &= x^2+3y^2 - (x^2-3y^2-2xy) = 6y^2+2xy > 0\
p-b &= x^2+3y^2 - 4xy = (x-3y)(x-y) > 0
end{align}
$$
Therefore $0< a,b < p$.
Case 2c: $y<x<3y$
We use the transformation
$$
u^2+3v^2 = (u+v)^2 + (u+v)(-u+v) + (-u+v)^2
$$
Hence
$$
begin{align}
p^2 &= (x^2-3y^2)^2 + 3(2xy)^2\
&= (x^2-3y^2+2xy)^2 + (x^2-3y^2+2xy)(3y^2-x^2+2xy) + (3y^2-x^2+2xy)^2\
&= a^2+ab+b^2
end{align}
$$
Hence
$$
begin{align}
a &= x^2-3y^2+2xy = (x+3y)(x-y)\
b &= 3y^2-x^2+2xy = (3y-x)(y+x)
end{align}
$$
Since $y<x < 3y$, we get $0< a,b$. Next we check that
$$
begin{align}
p-a &= x^2+3y^2 - (x^2-3y^2+2xy) = 6y^2-2xy = 2y(3y-x) > 0\
p-b &= x^2+3y^2 - (3y^2-x^2+2xy) = 2x^2-2xy = 2x(x-y) > 0
end{align}
$$
Therefore $0< a,b < p$.
This completes the proof.
answered Jan 14 at 14:02
Yong Hao NgYong Hao Ng
3,5691222
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add a comment |
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3 Answers
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3 Answers
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$begingroup$
Alternate solution via Thue's Lemma
Proposition. Let $pequiv 1pmod 3$ be a prime. Then there is a solution to
$$
x^2+x+1equiv 0 pmod{p^2}
$$
Proof. By Quadratic Reciprocity, $-3$ is a quadratic residue $pmod p$, hence
$$
u^2 equiv -3 pmod p
$$
for some integer $u$. Setting $uequiv 2c+1pmod p$, this becomes
$$
c^2+c+1 equiv 0 pmod p
$$
So $c$ is a solution to $f(x):=x^2+x+1equiv 0pmod p$. Hensel's Lemma tells us that if
$$
f'(c)notequiv 0 pmod p
$$
Then there exists a unique $d=c+kp$ such that
$$
f(d) equiv 0 pmod{p^2}
$$
Since
$$
f'(c)^2 = (2c+1)^2 equiv -3 pmod p
$$
If $f'(c)equiv 0 pmod p$ then $0equiv -3pmod p$ which implies $p=3$. This contradicts $pequiv 1pmod 3$, therefore $f'(c)notequiv 0 pmod p$.
Hence there is some $d$ such that
$$
d^2+d+1equiv 0 pmod{p^2}
$$
$$tag*{$square$}$$
We now solve our main problem.
Lemma. Let $pequiv 1pmod 3$ be a prime. Then there are integers $0<a,b<p$ such that
$$
a^2+ab+b^2equiv 0 pmod{p^2}
$$
Proof. We have shown that there is some integer $d$ such that
$$
d^2+d+1 equiv 0pmod{p^2}
$$
Now by Thue's Lemma, since $gcd(d,p^2)=1$, we can find $0<|a|,|b|<p$ such that
$$
db equiv a pmod{p^2}
$$
By rearranging the sign, we may assume that $b>0$ and $-p<a<p$. Now $a,b$ satisfies
$$
a^2+ab+b^2 equiv d^2b^2+db^2+b^2 equiv b^2(d^2+d+1)equiv 0 pmod{p^2}
$$
Therefore if $a> 0$ we have found our solution $0<a,b<p$. Otherwise, $a<0$ and we consider two cases ($-aneq b$, otherwise $dequiv -1pmod{p^2}$ contradicts $d^2+d+1equiv 0pmod{p^2}$):
Case 1: $-a > b$
We use the equality
$$
a^2+ab+b^2 = (-a-b)^2 + (-a-b)b + b^2
$$
and obtain a solution $(a',b') = (-a-b,b)$. Since $0<-a<p$ and $0<b<p$, we have $-a-b< p$. By assumption $0 <-a-b$, therefore $0 < a',b' < p$ and we are done.
Case 2: $-a < b$
We use another equality:
$$
a^2 + ab + b^2 = (-a)^2 + (-a)(a+b) + (a+b)^2
$$
and set the new solution as $(a',b') = (-a,a+b)$. Clearly $0 < a'=-a < p$. Since $-p < a < 0$ and $0 < b < p$, we get $a+b < p $. Finally, by assumption we have $0 < a+b$, therefore $0 < a+b=b' < p$. Hence we have a solution $0<a',b' < p$.
$$tag*{$square$}$$
This completes the proof.
Extra. Let $0<a,b<p$ and
$$a^2+ab+b^2equiv 0 pmod{p^2}$$
for some odd $p$. Then
$$a^2+ab+b^2 = p^2$$
Proof. Since $0<a,b< p$, we have
$$
a^2 + ab + b^2 < 3p^2
$$
Morever, since $a^2+ab+b^2equiv 0 pmod{p^2}$, we must have $a^2+ab+b^2 = p^2$ or $2p^2$. If $a^2+ab+b^2=2p^2$ then
$$
a^2+ab+b^2 equiv 0 pmod 2
$$
There is no solution if $a$ or $b$ is odd, therefore both $a,b$ are even. But this means
$$
a^2+ab+b^2equiv 0 pmod 4
$$
Since $a^2+ab+b^2$ is also divisible by odd $p^2$, it is divisible by $4p^2$. This contradicts that $a^2+ab+b^2 < 3p^2$, therefore we must have
$$
a^2+ab+b^2 = p^2
$$
$$tag*{$square$}$$
answered Jan 15 at 6:49
Yong Hao NgYong Hao Ng
3,5691222
$endgroup$
add a comment |
$begingroup$
Alternate solution via Thue's Lemma
Proposition. Let $pequiv 1pmod 3$ be a prime. Then there is a solution to
$$
x^2+x+1equiv 0 pmod{p^2}
$$
Proof. By Quadratic Reciprocity, $-3$ is a quadratic residue $pmod p$, hence
$$
u^2 equiv -3 pmod p
$$
for some integer $u$. Setting $uequiv 2c+1pmod p$, this becomes
$$
c^2+c+1 equiv 0 pmod p
$$
So $c$ is a solution to $f(x):=x^2+x+1equiv 0pmod p$. Hensel's Lemma tells us that if
$$
f'(c)notequiv 0 pmod p
$$
Then there exists a unique $d=c+kp$ such that
$$
f(d) equiv 0 pmod{p^2}
$$
Since
$$
f'(c)^2 = (2c+1)^2 equiv -3 pmod p
$$
If $f'(c)equiv 0 pmod p$ then $0equiv -3pmod p$ which implies $p=3$. This contradicts $pequiv 1pmod 3$, therefore $f'(c)notequiv 0 pmod p$.
Hence there is some $d$ such that
$$
d^2+d+1equiv 0 pmod{p^2}
$$
$$tag*{$square$}$$
We now solve our main problem.
Lemma. Let $pequiv 1pmod 3$ be a prime. Then there are integers $0<a,b<p$ such that
$$
a^2+ab+b^2equiv 0 pmod{p^2}
$$
Proof. We have shown that there is some integer $d$ such that
$$
d^2+d+1 equiv 0pmod{p^2}
$$
Now by Thue's Lemma, since $gcd(d,p^2)=1$, we can find $0<|a|,|b|<p$ such that
$$
db equiv a pmod{p^2}
$$
By rearranging the sign, we may assume that $b>0$ and $-p<a<p$. Now $a,b$ satisfies
$$
a^2+ab+b^2 equiv d^2b^2+db^2+b^2 equiv b^2(d^2+d+1)equiv 0 pmod{p^2}
$$
Therefore if $a> 0$ we have found our solution $0<a,b<p$. Otherwise, $a<0$ and we consider two cases ($-aneq b$, otherwise $dequiv -1pmod{p^2}$ contradicts $d^2+d+1equiv 0pmod{p^2}$):
Case 1: $-a > b$
We use the equality
$$
a^2+ab+b^2 = (-a-b)^2 + (-a-b)b + b^2
$$
and obtain a solution $(a',b') = (-a-b,b)$. Since $0<-a<p$ and $0<b<p$, we have $-a-b< p$. By assumption $0 <-a-b$, therefore $0 < a',b' < p$ and we are done.
Case 2: $-a < b$
We use another equality:
$$
a^2 + ab + b^2 = (-a)^2 + (-a)(a+b) + (a+b)^2
$$
and set the new solution as $(a',b') = (-a,a+b)$. Clearly $0 < a'=-a < p$. Since $-p < a < 0$ and $0 < b < p$, we get $a+b < p $. Finally, by assumption we have $0 < a+b$, therefore $0 < a+b=b' < p$. Hence we have a solution $0<a',b' < p$.
$$tag*{$square$}$$
This completes the proof.
Extra. Let $0<a,b<p$ and
$$a^2+ab+b^2equiv 0 pmod{p^2}$$
for some odd $p$. Then
$$a^2+ab+b^2 = p^2$$
Proof. Since $0<a,b< p$, we have
$$
a^2 + ab + b^2 < 3p^2
$$
Morever, since $a^2+ab+b^2equiv 0 pmod{p^2}$, we must have $a^2+ab+b^2 = p^2$ or $2p^2$. If $a^2+ab+b^2=2p^2$ then
$$
a^2+ab+b^2 equiv 0 pmod 2
$$
There is no solution if $a$ or $b$ is odd, therefore both $a,b$ are even. But this means
$$
a^2+ab+b^2equiv 0 pmod 4
$$
Since $a^2+ab+b^2$ is also divisible by odd $p^2$, it is divisible by $4p^2$. This contradicts that $a^2+ab+b^2 < 3p^2$, therefore we must have
$$
a^2+ab+b^2 = p^2
$$
$$tag*{$square$}$$
answered Jan 15 at 6:49
Yong Hao NgYong Hao Ng
3,5691222
$endgroup$
add a comment |
$begingroup$
Alternate solution via Thue's Lemma
Proposition. Let $pequiv 1pmod 3$ be a prime. Then there is a solution to
$$
x^2+x+1equiv 0 pmod{p^2}
$$
Proof. By Quadratic Reciprocity, $-3$ is a quadratic residue $pmod p$, hence
$$
u^2 equiv -3 pmod p
$$
for some integer $u$. Setting $uequiv 2c+1pmod p$, this becomes
$$
c^2+c+1 equiv 0 pmod p
$$
So $c$ is a solution to $f(x):=x^2+x+1equiv 0pmod p$. Hensel's Lemma tells us that if
$$
f'(c)notequiv 0 pmod p
$$
Then there exists a unique $d=c+kp$ such that
$$
f(d) equiv 0 pmod{p^2}
$$
Since
$$
f'(c)^2 = (2c+1)^2 equiv -3 pmod p
$$
If $f'(c)equiv 0 pmod p$ then $0equiv -3pmod p$ which implies $p=3$. This contradicts $pequiv 1pmod 3$, therefore $f'(c)notequiv 0 pmod p$.
Hence there is some $d$ such that
$$
d^2+d+1equiv 0 pmod{p^2}
$$
$$tag*{$square$}$$
We now solve our main problem.
Lemma. Let $pequiv 1pmod 3$ be a prime. Then there are integers $0<a,b<p$ such that
$$
a^2+ab+b^2equiv 0 pmod{p^2}
$$
Proof. We have shown that there is some integer $d$ such that
$$
d^2+d+1 equiv 0pmod{p^2}
$$
Now by Thue's Lemma, since $gcd(d,p^2)=1$, we can find $0<|a|,|b|<p$ such that
$$
db equiv a pmod{p^2}
$$
By rearranging the sign, we may assume that $b>0$ and $-p<a<p$. Now $a,b$ satisfies
$$
a^2+ab+b^2 equiv d^2b^2+db^2+b^2 equiv b^2(d^2+d+1)equiv 0 pmod{p^2}
$$
Therefore if $a> 0$ we have found our solution $0<a,b<p$. Otherwise, $a<0$ and we consider two cases ($-aneq b$, otherwise $dequiv -1pmod{p^2}$ contradicts $d^2+d+1equiv 0pmod{p^2}$):
Case 1: $-a > b$
We use the equality
$$
a^2+ab+b^2 = (-a-b)^2 + (-a-b)b + b^2
$$
and obtain a solution $(a',b') = (-a-b,b)$. Since $0<-a<p$ and $0<b<p$, we have $-a-b< p$. By assumption $0 <-a-b$, therefore $0 < a',b' < p$ and we are done.
Case 2: $-a < b$
We use another equality:
$$
a^2 + ab + b^2 = (-a)^2 + (-a)(a+b) + (a+b)^2
$$
and set the new solution as $(a',b') = (-a,a+b)$. Clearly $0 < a'=-a < p$. Since $-p < a < 0$ and $0 < b < p$, we get $a+b < p $. Finally, by assumption we have $0 < a+b$, therefore $0 < a+b=b' < p$. Hence we have a solution $0<a',b' < p$.
$$tag*{$square$}$$
This completes the proof.
Extra. Let $0<a,b<p$ and
$$a^2+ab+b^2equiv 0 pmod{p^2}$$
for some odd $p$. Then
$$a^2+ab+b^2 = p^2$$
Proof. Since $0<a,b< p$, we have
$$
a^2 + ab + b^2 < 3p^2
$$
Morever, since $a^2+ab+b^2equiv 0 pmod{p^2}$, we must have $a^2+ab+b^2 = p^2$ or $2p^2$. If $a^2+ab+b^2=2p^2$ then
$$
a^2+ab+b^2 equiv 0 pmod 2
$$
There is no solution if $a$ or $b$ is odd, therefore both $a,b$ are even. But this means
$$
a^2+ab+b^2equiv 0 pmod 4
$$
Since $a^2+ab+b^2$ is also divisible by odd $p^2$, it is divisible by $4p^2$. This contradicts that $a^2+ab+b^2 < 3p^2$, therefore we must have
$$
a^2+ab+b^2 = p^2
$$
$$tag*{$square$}$$
answered Jan 15 at 6:49
Yong Hao NgYong Hao Ng
3,5691222
$endgroup$
Alternate solution via Thue's Lemma
Proposition. Let $pequiv 1pmod 3$ be a prime. Then there is a solution to
$$
x^2+x+1equiv 0 pmod{p^2}
$$
Proof. By Quadratic Reciprocity, $-3$ is a quadratic residue $pmod p$, hence
$$
u^2 equiv -3 pmod p
$$
for some integer $u$. Setting $uequiv 2c+1pmod p$, this becomes
$$
c^2+c+1 equiv 0 pmod p
$$
So $c$ is a solution to $f(x):=x^2+x+1equiv 0pmod p$. Hensel's Lemma tells us that if
$$
f'(c)notequiv 0 pmod p
$$
Then there exists a unique $d=c+kp$ such that
$$
f(d) equiv 0 pmod{p^2}
$$
Since
$$
f'(c)^2 = (2c+1)^2 equiv -3 pmod p
$$
If $f'(c)equiv 0 pmod p$ then $0equiv -3pmod p$ which implies $p=3$. This contradicts $pequiv 1pmod 3$, therefore $f'(c)notequiv 0 pmod p$.
Hence there is some $d$ such that
$$
d^2+d+1equiv 0 pmod{p^2}
$$
$$tag*{$square$}$$
We now solve our main problem.
Lemma. Let $pequiv 1pmod 3$ be a prime. Then there are integers $0<a,b<p$ such that
$$
a^2+ab+b^2equiv 0 pmod{p^2}
$$
Proof. We have shown that there is some integer $d$ such that
$$
d^2+d+1 equiv 0pmod{p^2}
$$
Now by Thue's Lemma, since $gcd(d,p^2)=1$, we can find $0<|a|,|b|<p$ such that
$$
db equiv a pmod{p^2}
$$
By rearranging the sign, we may assume that $b>0$ and $-p<a<p$. Now $a,b$ satisfies
$$
a^2+ab+b^2 equiv d^2b^2+db^2+b^2 equiv b^2(d^2+d+1)equiv 0 pmod{p^2}
$$
Therefore if $a> 0$ we have found our solution $0<a,b<p$. Otherwise, $a<0$ and we consider two cases ($-aneq b$, otherwise $dequiv -1pmod{p^2}$ contradicts $d^2+d+1equiv 0pmod{p^2}$):
Case 1: $-a > b$
We use the equality
$$
a^2+ab+b^2 = (-a-b)^2 + (-a-b)b + b^2
$$
and obtain a solution $(a',b') = (-a-b,b)$. Since $0<-a<p$ and $0<b<p$, we have $-a-b< p$. By assumption $0 <-a-b$, therefore $0 < a',b' < p$ and we are done.
Case 2: $-a < b$
We use another equality:
$$
a^2 + ab + b^2 = (-a)^2 + (-a)(a+b) + (a+b)^2
$$
and set the new solution as $(a',b') = (-a,a+b)$. Clearly $0 < a'=-a < p$. Since $-p < a < 0$ and $0 < b < p$, we get $a+b < p $. Finally, by assumption we have $0 < a+b$, therefore $0 < a+b=b' < p$. Hence we have a solution $0<a',b' < p$.
$$tag*{$square$}$$
This completes the proof.
Extra. Let $0<a,b<p$ and
$$a^2+ab+b^2equiv 0 pmod{p^2}$$
for some odd $p$. Then
$$a^2+ab+b^2 = p^2$$
Proof. Since $0<a,b< p$, we have
$$
a^2 + ab + b^2 < 3p^2
$$
Morever, since $a^2+ab+b^2equiv 0 pmod{p^2}$, we must have $a^2+ab+b^2 = p^2$ or $2p^2$. If $a^2+ab+b^2=2p^2$ then
$$
a^2+ab+b^2 equiv 0 pmod 2
$$
There is no solution if $a$ or $b$ is odd, therefore both $a,b$ are even. But this means
$$
a^2+ab+b^2equiv 0 pmod 4
$$
Since $a^2+ab+b^2$ is also divisible by odd $p^2$, it is divisible by $4p^2$. This contradicts that $a^2+ab+b^2 < 3p^2$, therefore we must have
$$
a^2+ab+b^2 = p^2
$$
$$tag*{$square$}$$
answered Jan 15 at 6:49
Yong Hao NgYong Hao Ng
3,5691222
answered Jan 15 at 6:49
Yong Hao NgYong Hao Ng
3,5691222
answered Jan 15 at 6:49
Yong Hao NgYong Hao Ng
3,5691222
answered Jan 15 at 6:49
Yong Hao NgYong Hao Ng
3,5691222
3,5691222
add a comment |
add a comment |
$begingroup$
This relies on the following statement: if some residue class $a$ is a square mod $p$, then it is a square mod $p^2$.
Indeed, if $b^2=a+tp [p^2]$, then $(b+kp)^2=a+(t+2k)p [p^2]$.
So you know that there is some $x$ such that $p|x^2+x+1$, thus $-3=(2x-1)^2$ is a square mod $p$, thus mod $p^2$, and if $y^2=-3[p^2]$, then $a=(1+y)/2$ (mod $p^2$) and $1$ work.
answered Dec 12 '18 at 13:32
MindlackMindlack
3,73518
$endgroup$
$begingroup$
Your LaTeX is broken
$endgroup$
– Lucas Henrique
Dec 12 '18 at 13:33
1
$begingroup$
Thanks, corrected!
$endgroup$
– Mindlack
Dec 12 '18 at 13:34
$begingroup$
I can't undertand your unswer,first your step ,what's mean
$endgroup$
– math110
Dec 16 '18 at 15:22
add a comment |
$begingroup$
This relies on the following statement: if some residue class $a$ is a square mod $p$, then it is a square mod $p^2$.
Indeed, if $b^2=a+tp [p^2]$, then $(b+kp)^2=a+(t+2k)p [p^2]$.
So you know that there is some $x$ such that $p|x^2+x+1$, thus $-3=(2x-1)^2$ is a square mod $p$, thus mod $p^2$, and if $y^2=-3[p^2]$, then $a=(1+y)/2$ (mod $p^2$) and $1$ work.
answered Dec 12 '18 at 13:32
MindlackMindlack
3,73518
$endgroup$
$begingroup$
Your LaTeX is broken
$endgroup$
– Lucas Henrique
Dec 12 '18 at 13:33
1
$begingroup$
Thanks, corrected!
$endgroup$
– Mindlack
Dec 12 '18 at 13:34
$begingroup$
I can't undertand your unswer,first your step ,what's mean
$endgroup$
– math110
Dec 16 '18 at 15:22
add a comment |
$begingroup$
This relies on the following statement: if some residue class $a$ is a square mod $p$, then it is a square mod $p^2$.
Indeed, if $b^2=a+tp [p^2]$, then $(b+kp)^2=a+(t+2k)p [p^2]$.
So you know that there is some $x$ such that $p|x^2+x+1$, thus $-3=(2x-1)^2$ is a square mod $p$, thus mod $p^2$, and if $y^2=-3[p^2]$, then $a=(1+y)/2$ (mod $p^2$) and $1$ work.
answered Dec 12 '18 at 13:32
MindlackMindlack
3,73518
$endgroup$
This relies on the following statement: if some residue class $a$ is a square mod $p$, then it is a square mod $p^2$.
Indeed, if $b^2=a+tp [p^2]$, then $(b+kp)^2=a+(t+2k)p [p^2]$.
So you know that there is some $x$ such that $p|x^2+x+1$, thus $-3=(2x-1)^2$ is a square mod $p$, thus mod $p^2$, and if $y^2=-3[p^2]$, then $a=(1+y)/2$ (mod $p^2$) and $1$ work.
answered Dec 12 '18 at 13:32
MindlackMindlack
3,73518
edited Dec 12 '18 at 13:34
answered Dec 12 '18 at 13:32
MindlackMindlack
3,73518
answered Dec 12 '18 at 13:32
MindlackMindlack
3,73518
answered Dec 12 '18 at 13:32
MindlackMindlack
3,73518
3,73518
$begingroup$
Your LaTeX is broken
$endgroup$
– Lucas Henrique
Dec 12 '18 at 13:33
1
$begingroup$
Thanks, corrected!
$endgroup$
– Mindlack
Dec 12 '18 at 13:34
$begingroup$
I can't undertand your unswer,first your step ,what's mean
$endgroup$
– math110
Dec 16 '18 at 15:22
add a comment |
$begingroup$
Your LaTeX is broken
$endgroup$
– Lucas Henrique
Dec 12 '18 at 13:33
1
$begingroup$
Thanks, corrected!
$endgroup$
– Mindlack
Dec 12 '18 at 13:34
$begingroup$
I can't undertand your unswer,first your step ,what's mean
$endgroup$
– math110
Dec 16 '18 at 15:22
$begingroup$
Your LaTeX is broken
$endgroup$
– Lucas Henrique
Dec 12 '18 at 13:33
$begingroup$
Your LaTeX is broken
$endgroup$
– Lucas Henrique
Dec 12 '18 at 13:33
1
1
$begingroup$
Thanks, corrected!
$endgroup$
– Mindlack
Dec 12 '18 at 13:34
$begingroup$
Thanks, corrected!
$endgroup$
– Mindlack
Dec 12 '18 at 13:34
$begingroup$
I can't undertand your unswer,first your step ,what's mean
$endgroup$
– math110
Dec 16 '18 at 15:22
$begingroup$
I can't undertand your unswer,first your step ,what's mean
$endgroup$
– math110
Dec 16 '18 at 15:22
add a comment |
$begingroup$
1. Solution to $p^2= X^2+3Y^2$
It is well known that $pequiv 1pmod 3$ if and only if
$$
p=x^2+3y^2
$$
for some integers $x,y$. So immediately we have a presentation of $p^2$ as
$$
begin{align}
p^2 &= (x^2+3y^2)^2\
&=(x+sqrt{-3}y)^2(x-sqrt{-3}y)^2\
&=((x^2-3y^2)+2xysqrt{-3})((x^2-3y^2)-2xysqrt{-3})\
&=(x^2-3y^2)^2+3(2xy)^2
end{align}
$$
2. Solution to $p^2=a^2+ab+b^2,quad 0leq a,b< p$
The expression $u^2+3v^2$ can be rewritten in the following three ways:
$$
begin{align}
u^2+3v^2 &= (pm u-v)^2 + (pm u-v)(2v) + (2v)^2\
u^2+3v^2 &= (u+v)^2 + (u+v)(-u+v) + (-u+v)^2
end{align}
$$
From (1), we have
$$
p^2 = (x^2-3y)^2 + 3(2xy)^2
$$
Therefore applying any of the three transformations will already give us
$$
p^2 = a^2 + ab+b^2
$$
We shall show that depending on whether $x<y$, $y<x<3y$ or $3y<x$, exactly one of the previous transformations will give us a solution such that $0leq a,b<p$. (Clearly $xneq y$ and $xneq 3y$, or $p$ is not prime.)
Case 2a: $x<y$
We use the transformation
$$
u^2+3v^2 = (-u-v)^2 + (-u-v)(2v) + (2v)^2
$$
Hence
$$
begin{align}
p^2 &= (x^2-3y^2)^2 + 3(2xy)^2\
&= (3y^2-x^2-2xy)^2 + (3y^2-x^2-2xy)(4xy) + (4xy)^2\
&= a^2+ab+b^2
end{align}
$$
Hence
$$
begin{align}
a &= 3y^2-x^2-2xy = (3y+x)(y-x)\
b &= 4xy
end{align}
$$
Since $0<x<y$, we get $0< a,b$. Next we check that
$$
begin{align}
p-a &= x^2+3y^2 - (3y^2-x^2-2xy) = 2x^2+2xy > 0\
p-b &= x^2+3y^2 - 4xy = (3y-x)(y-x) > 0
end{align}
$$
Therefore $0< a,b < p$.
Case 2b: $3y<x$
We use the transformation
$$
u^2+3v^2 = (u-v)^2 + (u-v)(2v) + (2v)^2
$$
Hence
$$
begin{align}
p^2 &= (x^2-3y^2)^2 + 3(2xy)^2\
&= (x^2-3y^2-2xy)^2 + (x^2-3y^2-2xy)(4xy) + (4xy)^2\
&= a^2+ab+b^2
end{align}
$$
Hence
$$
begin{align}
a &= x^2-3y^2-2xy = (x-3y)(x+y)\
b &= 4xy
end{align}
$$
Since $0<3y < x$, we get $0< a,b$. Next we check that
$$
begin{align}
p-a &= x^2+3y^2 - (x^2-3y^2-2xy) = 6y^2+2xy > 0\
p-b &= x^2+3y^2 - 4xy = (x-3y)(x-y) > 0
end{align}
$$
Therefore $0< a,b < p$.
Case 2c: $y<x<3y$
We use the transformation
$$
u^2+3v^2 = (u+v)^2 + (u+v)(-u+v) + (-u+v)^2
$$
Hence
$$
begin{align}
p^2 &= (x^2-3y^2)^2 + 3(2xy)^2\
&= (x^2-3y^2+2xy)^2 + (x^2-3y^2+2xy)(3y^2-x^2+2xy) + (3y^2-x^2+2xy)^2\
&= a^2+ab+b^2
end{align}
$$
Hence
$$
begin{align}
a &= x^2-3y^2+2xy = (x+3y)(x-y)\
b &= 3y^2-x^2+2xy = (3y-x)(y+x)
end{align}
$$
Since $y<x < 3y$, we get $0< a,b$. Next we check that
$$
begin{align}
p-a &= x^2+3y^2 - (x^2-3y^2+2xy) = 6y^2-2xy = 2y(3y-x) > 0\
p-b &= x^2+3y^2 - (3y^2-x^2+2xy) = 2x^2-2xy = 2x(x-y) > 0
end{align}
$$
Therefore $0< a,b < p$.
This completes the proof.
answered Jan 14 at 14:02
Yong Hao NgYong Hao Ng
3,5691222
$endgroup$
add a comment |
$begingroup$
1. Solution to $p^2= X^2+3Y^2$
It is well known that $pequiv 1pmod 3$ if and only if
$$
p=x^2+3y^2
$$
for some integers $x,y$. So immediately we have a presentation of $p^2$ as
$$
begin{align}
p^2 &= (x^2+3y^2)^2\
&=(x+sqrt{-3}y)^2(x-sqrt{-3}y)^2\
&=((x^2-3y^2)+2xysqrt{-3})((x^2-3y^2)-2xysqrt{-3})\
&=(x^2-3y^2)^2+3(2xy)^2
end{align}
$$
2. Solution to $p^2=a^2+ab+b^2,quad 0leq a,b< p$
The expression $u^2+3v^2$ can be rewritten in the following three ways:
$$
begin{align}
u^2+3v^2 &= (pm u-v)^2 + (pm u-v)(2v) + (2v)^2\
u^2+3v^2 &= (u+v)^2 + (u+v)(-u+v) + (-u+v)^2
end{align}
$$
From (1), we have
$$
p^2 = (x^2-3y)^2 + 3(2xy)^2
$$
Therefore applying any of the three transformations will already give us
$$
p^2 = a^2 + ab+b^2
$$
We shall show that depending on whether $x<y$, $y<x<3y$ or $3y<x$, exactly one of the previous transformations will give us a solution such that $0leq a,b<p$. (Clearly $xneq y$ and $xneq 3y$, or $p$ is not prime.)
Case 2a: $x<y$
We use the transformation
$$
u^2+3v^2 = (-u-v)^2 + (-u-v)(2v) + (2v)^2
$$
Hence
$$
begin{align}
p^2 &= (x^2-3y^2)^2 + 3(2xy)^2\
&= (3y^2-x^2-2xy)^2 + (3y^2-x^2-2xy)(4xy) + (4xy)^2\
&= a^2+ab+b^2
end{align}
$$
Hence
$$
begin{align}
a &= 3y^2-x^2-2xy = (3y+x)(y-x)\
b &= 4xy
end{align}
$$
Since $0<x<y$, we get $0< a,b$. Next we check that
$$
begin{align}
p-a &= x^2+3y^2 - (3y^2-x^2-2xy) = 2x^2+2xy > 0\
p-b &= x^2+3y^2 - 4xy = (3y-x)(y-x) > 0
end{align}
$$
Therefore $0< a,b < p$.
Case 2b: $3y<x$
We use the transformation
$$
u^2+3v^2 = (u-v)^2 + (u-v)(2v) + (2v)^2
$$
Hence
$$
begin{align}
p^2 &= (x^2-3y^2)^2 + 3(2xy)^2\
&= (x^2-3y^2-2xy)^2 + (x^2-3y^2-2xy)(4xy) + (4xy)^2\
&= a^2+ab+b^2
end{align}
$$
Hence
$$
begin{align}
a &= x^2-3y^2-2xy = (x-3y)(x+y)\
b &= 4xy
end{align}
$$
Since $0<3y < x$, we get $0< a,b$. Next we check that
$$
begin{align}
p-a &= x^2+3y^2 - (x^2-3y^2-2xy) = 6y^2+2xy > 0\
p-b &= x^2+3y^2 - 4xy = (x-3y)(x-y) > 0
end{align}
$$
Therefore $0< a,b < p$.
Case 2c: $y<x<3y$
We use the transformation
$$
u^2+3v^2 = (u+v)^2 + (u+v)(-u+v) + (-u+v)^2
$$
Hence
$$
begin{align}
p^2 &= (x^2-3y^2)^2 + 3(2xy)^2\
&= (x^2-3y^2+2xy)^2 + (x^2-3y^2+2xy)(3y^2-x^2+2xy) + (3y^2-x^2+2xy)^2\
&= a^2+ab+b^2
end{align}
$$
Hence
$$
begin{align}
a &= x^2-3y^2+2xy = (x+3y)(x-y)\
b &= 3y^2-x^2+2xy = (3y-x)(y+x)
end{align}
$$
Since $y<x < 3y$, we get $0< a,b$. Next we check that
$$
begin{align}
p-a &= x^2+3y^2 - (x^2-3y^2+2xy) = 6y^2-2xy = 2y(3y-x) > 0\
p-b &= x^2+3y^2 - (3y^2-x^2+2xy) = 2x^2-2xy = 2x(x-y) > 0
end{align}
$$
Therefore $0< a,b < p$.
This completes the proof.
answered Jan 14 at 14:02
Yong Hao NgYong Hao Ng
3,5691222
$endgroup$
add a comment |
$begingroup$
1. Solution to $p^2= X^2+3Y^2$
It is well known that $pequiv 1pmod 3$ if and only if
$$
p=x^2+3y^2
$$
for some integers $x,y$. So immediately we have a presentation of $p^2$ as
$$
begin{align}
p^2 &= (x^2+3y^2)^2\
&=(x+sqrt{-3}y)^2(x-sqrt{-3}y)^2\
&=((x^2-3y^2)+2xysqrt{-3})((x^2-3y^2)-2xysqrt{-3})\
&=(x^2-3y^2)^2+3(2xy)^2
end{align}
$$
2. Solution to $p^2=a^2+ab+b^2,quad 0leq a,b< p$
The expression $u^2+3v^2$ can be rewritten in the following three ways:
$$
begin{align}
u^2+3v^2 &= (pm u-v)^2 + (pm u-v)(2v) + (2v)^2\
u^2+3v^2 &= (u+v)^2 + (u+v)(-u+v) + (-u+v)^2
end{align}
$$
From (1), we have
$$
p^2 = (x^2-3y)^2 + 3(2xy)^2
$$
Therefore applying any of the three transformations will already give us
$$
p^2 = a^2 + ab+b^2
$$
We shall show that depending on whether $x<y$, $y<x<3y$ or $3y<x$, exactly one of the previous transformations will give us a solution such that $0leq a,b<p$. (Clearly $xneq y$ and $xneq 3y$, or $p$ is not prime.)
Case 2a: $x<y$
We use the transformation
$$
u^2+3v^2 = (-u-v)^2 + (-u-v)(2v) + (2v)^2
$$
Hence
$$
begin{align}
p^2 &= (x^2-3y^2)^2 + 3(2xy)^2\
&= (3y^2-x^2-2xy)^2 + (3y^2-x^2-2xy)(4xy) + (4xy)^2\
&= a^2+ab+b^2
end{align}
$$
Hence
$$
begin{align}
a &= 3y^2-x^2-2xy = (3y+x)(y-x)\
b &= 4xy
end{align}
$$
Since $0<x<y$, we get $0< a,b$. Next we check that
$$
begin{align}
p-a &= x^2+3y^2 - (3y^2-x^2-2xy) = 2x^2+2xy > 0\
p-b &= x^2+3y^2 - 4xy = (3y-x)(y-x) > 0
end{align}
$$
Therefore $0< a,b < p$.
Case 2b: $3y<x$
We use the transformation
$$
u^2+3v^2 = (u-v)^2 + (u-v)(2v) + (2v)^2
$$
Hence
$$
begin{align}
p^2 &= (x^2-3y^2)^2 + 3(2xy)^2\
&= (x^2-3y^2-2xy)^2 + (x^2-3y^2-2xy)(4xy) + (4xy)^2\
&= a^2+ab+b^2
end{align}
$$
Hence
$$
begin{align}
a &= x^2-3y^2-2xy = (x-3y)(x+y)\
b &= 4xy
end{align}
$$
Since $0<3y < x$, we get $0< a,b$. Next we check that
$$
begin{align}
p-a &= x^2+3y^2 - (x^2-3y^2-2xy) = 6y^2+2xy > 0\
p-b &= x^2+3y^2 - 4xy = (x-3y)(x-y) > 0
end{align}
$$
Therefore $0< a,b < p$.
Case 2c: $y<x<3y$
We use the transformation
$$
u^2+3v^2 = (u+v)^2 + (u+v)(-u+v) + (-u+v)^2
$$
Hence
$$
begin{align}
p^2 &= (x^2-3y^2)^2 + 3(2xy)^2\
&= (x^2-3y^2+2xy)^2 + (x^2-3y^2+2xy)(3y^2-x^2+2xy) + (3y^2-x^2+2xy)^2\
&= a^2+ab+b^2
end{align}
$$
Hence
$$
begin{align}
a &= x^2-3y^2+2xy = (x+3y)(x-y)\
b &= 3y^2-x^2+2xy = (3y-x)(y+x)
end{align}
$$
Since $y<x < 3y$, we get $0< a,b$. Next we check that
$$
begin{align}
p-a &= x^2+3y^2 - (x^2-3y^2+2xy) = 6y^2-2xy = 2y(3y-x) > 0\
p-b &= x^2+3y^2 - (3y^2-x^2+2xy) = 2x^2-2xy = 2x(x-y) > 0
end{align}
$$
Therefore $0< a,b < p$.
This completes the proof.
answered Jan 14 at 14:02
Yong Hao NgYong Hao Ng
3,5691222
$endgroup$
1. Solution to $p^2= X^2+3Y^2$
It is well known that $pequiv 1pmod 3$ if and only if
$$
p=x^2+3y^2
$$
for some integers $x,y$. So immediately we have a presentation of $p^2$ as
$$
begin{align}
p^2 &= (x^2+3y^2)^2\
&=(x+sqrt{-3}y)^2(x-sqrt{-3}y)^2\
&=((x^2-3y^2)+2xysqrt{-3})((x^2-3y^2)-2xysqrt{-3})\
&=(x^2-3y^2)^2+3(2xy)^2
end{align}
$$
2. Solution to $p^2=a^2+ab+b^2,quad 0leq a,b< p$
The expression $u^2+3v^2$ can be rewritten in the following three ways:
$$
begin{align}
u^2+3v^2 &= (pm u-v)^2 + (pm u-v)(2v) + (2v)^2\
u^2+3v^2 &= (u+v)^2 + (u+v)(-u+v) + (-u+v)^2
end{align}
$$
From (1), we have
$$
p^2 = (x^2-3y)^2 + 3(2xy)^2
$$
Therefore applying any of the three transformations will already give us
$$
p^2 = a^2 + ab+b^2
$$
We shall show that depending on whether $x<y$, $y<x<3y$ or $3y<x$, exactly one of the previous transformations will give us a solution such that $0leq a,b<p$. (Clearly $xneq y$ and $xneq 3y$, or $p$ is not prime.)
Case 2a: $x<y$
We use the transformation
$$
u^2+3v^2 = (-u-v)^2 + (-u-v)(2v) + (2v)^2
$$
Hence
$$
begin{align}
p^2 &= (x^2-3y^2)^2 + 3(2xy)^2\
&= (3y^2-x^2-2xy)^2 + (3y^2-x^2-2xy)(4xy) + (4xy)^2\
&= a^2+ab+b^2
end{align}
$$
Hence
$$
begin{align}
a &= 3y^2-x^2-2xy = (3y+x)(y-x)\
b &= 4xy
end{align}
$$
Since $0<x<y$, we get $0< a,b$. Next we check that
$$
begin{align}
p-a &= x^2+3y^2 - (3y^2-x^2-2xy) = 2x^2+2xy > 0\
p-b &= x^2+3y^2 - 4xy = (3y-x)(y-x) > 0
end{align}
$$
Therefore $0< a,b < p$.
Case 2b: $3y<x$
We use the transformation
$$
u^2+3v^2 = (u-v)^2 + (u-v)(2v) + (2v)^2
$$
Hence
$$
begin{align}
p^2 &= (x^2-3y^2)^2 + 3(2xy)^2\
&= (x^2-3y^2-2xy)^2 + (x^2-3y^2-2xy)(4xy) + (4xy)^2\
&= a^2+ab+b^2
end{align}
$$
Hence
$$
begin{align}
a &= x^2-3y^2-2xy = (x-3y)(x+y)\
b &= 4xy
end{align}
$$
Since $0<3y < x$, we get $0< a,b$. Next we check that
$$
begin{align}
p-a &= x^2+3y^2 - (x^2-3y^2-2xy) = 6y^2+2xy > 0\
p-b &= x^2+3y^2 - 4xy = (x-3y)(x-y) > 0
end{align}
$$
Therefore $0< a,b < p$.
Case 2c: $y<x<3y$
We use the transformation
$$
u^2+3v^2 = (u+v)^2 + (u+v)(-u+v) + (-u+v)^2
$$
Hence
$$
begin{align}
p^2 &= (x^2-3y^2)^2 + 3(2xy)^2\
&= (x^2-3y^2+2xy)^2 + (x^2-3y^2+2xy)(3y^2-x^2+2xy) + (3y^2-x^2+2xy)^2\
&= a^2+ab+b^2
end{align}
$$
Hence
$$
begin{align}
a &= x^2-3y^2+2xy = (x+3y)(x-y)\
b &= 3y^2-x^2+2xy = (3y-x)(y+x)
end{align}
$$
Since $y<x < 3y$, we get $0< a,b$. Next we check that
$$
begin{align}
p-a &= x^2+3y^2 - (x^2-3y^2+2xy) = 6y^2-2xy = 2y(3y-x) > 0\
p-b &= x^2+3y^2 - (3y^2-x^2+2xy) = 2x^2-2xy = 2x(x-y) > 0
end{align}
$$
Therefore $0< a,b < p$.
This completes the proof.
answered Jan 14 at 14:02
Yong Hao NgYong Hao Ng
3,5691222
answered Jan 14 at 14:02
Yong Hao NgYong Hao Ng
3,5691222
answered Jan 14 at 14:02
Yong Hao NgYong Hao Ng
3,5691222
answered Jan 14 at 14:02
Yong Hao NgYong Hao Ng
3,5691222
3,5691222
add a comment |
add a comment |
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