What are the positive integer solutions to $x^2-x+1 = y^3$?












6












$begingroup$


The only solutions that I know of till now are $(x,y) = (1,1) space , (19,7)$. We can note that:
$$x^2-x+1 = y^3 implies (2x-1)^2 = 4y^3-3$$
Thus, if odd prime $p mid y$, then $(2x-1)^2 equiv -3 pmod{p}$ and thus, $-3$ is a quadratic residue. This implies that $p equiv 1 pmod{6}$ .



How can we further proceed into this problem?



Note: As mentioned in one of the links in the comments, if we instead replace $x$ by $x+1$, we get $x^2+x+1 = y^3$. Thus, the solutions for this is $(x,y) = (0,1) space, (18,7)$.






number-theory diophantine-equations elliptic-curves eisenstein-integers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think that the fact that $y^3equiv 0,1pmod{8}$ is useful here.
    $endgroup$
    – Don Thousand
    Dec 17 '18 at 16:01






  • 5




    $begingroup$
    This is also an elliptic curve equation (in long Weierstrass form).
    $endgroup$
    – Dietrich Burde
    Dec 17 '18 at 16:34






  • 1




    $begingroup$
    I've confirmed by computer search that there are no further solutions for $y leq 160, 000$.
    $endgroup$
    – Connor Harris
    Dec 17 '18 at 17:27






  • 1




    $begingroup$
    Magma "says" they are the unique solutions.
    $endgroup$
    – xarles
    Dec 17 '18 at 17:43






  • 2




    $begingroup$
    lmfdb.org/EllipticCurve/Q/243/a/1
    $endgroup$
    – Álvaro Lozano-Robledo
    Dec 19 '18 at 17:51
















6












$begingroup$


The only solutions that I know of till now are $(x,y) = (1,1) space , (19,7)$. We can note that:
$$x^2-x+1 = y^3 implies (2x-1)^2 = 4y^3-3$$
Thus, if odd prime $p mid y$, then $(2x-1)^2 equiv -3 pmod{p}$ and thus, $-3$ is a quadratic residue. This implies that $p equiv 1 pmod{6}$ .



How can we further proceed into this problem?



Note: As mentioned in one of the links in the comments, if we instead replace $x$ by $x+1$, we get $x^2+x+1 = y^3$. Thus, the solutions for this is $(x,y) = (0,1) space, (18,7)$.






number-theory diophantine-equations elliptic-curves eisenstein-integers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think that the fact that $y^3equiv 0,1pmod{8}$ is useful here.
    $endgroup$
    – Don Thousand
    Dec 17 '18 at 16:01






  • 5




    $begingroup$
    This is also an elliptic curve equation (in long Weierstrass form).
    $endgroup$
    – Dietrich Burde
    Dec 17 '18 at 16:34






  • 1




    $begingroup$
    I've confirmed by computer search that there are no further solutions for $y leq 160, 000$.
    $endgroup$
    – Connor Harris
    Dec 17 '18 at 17:27






  • 1




    $begingroup$
    Magma "says" they are the unique solutions.
    $endgroup$
    – xarles
    Dec 17 '18 at 17:43






  • 2




    $begingroup$
    lmfdb.org/EllipticCurve/Q/243/a/1
    $endgroup$
    – Álvaro Lozano-Robledo
    Dec 19 '18 at 17:51














6












6








6


4



$begingroup$


The only solutions that I know of till now are $(x,y) = (1,1) space , (19,7)$. We can note that:
$$x^2-x+1 = y^3 implies (2x-1)^2 = 4y^3-3$$
Thus, if odd prime $p mid y$, then $(2x-1)^2 equiv -3 pmod{p}$ and thus, $-3$ is a quadratic residue. This implies that $p equiv 1 pmod{6}$ .



How can we further proceed into this problem?



Note: As mentioned in one of the links in the comments, if we instead replace $x$ by $x+1$, we get $x^2+x+1 = y^3$. Thus, the solutions for this is $(x,y) = (0,1) space, (18,7)$.






number-theory diophantine-equations elliptic-curves eisenstein-integers







share|cite|improve this question











$endgroup$




The only solutions that I know of till now are $(x,y) = (1,1) space , (19,7)$. We can note that:
$$x^2-x+1 = y^3 implies (2x-1)^2 = 4y^3-3$$
Thus, if odd prime $p mid y$, then $(2x-1)^2 equiv -3 pmod{p}$ and thus, $-3$ is a quadratic residue. This implies that $p equiv 1 pmod{6}$ .



How can we further proceed into this problem?



Note: As mentioned in one of the links in the comments, if we instead replace $x$ by $x+1$, we get $x^2+x+1 = y^3$. Thus, the solutions for this is $(x,y) = (0,1) space, (18,7)$.






number-theory diophantine-equations elliptic-curves eisenstein-integers




number-theory diophantine-equations elliptic-curves eisenstein-integers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 5:19







Haran

















asked Dec 17 '18 at 15:41









HaranHaran

1,086324




1,086324








  • 1




    $begingroup$
    I think that the fact that $y^3equiv 0,1pmod{8}$ is useful here.
    $endgroup$
    – Don Thousand
    Dec 17 '18 at 16:01






  • 5




    $begingroup$
    This is also an elliptic curve equation (in long Weierstrass form).
    $endgroup$
    – Dietrich Burde
    Dec 17 '18 at 16:34






  • 1




    $begingroup$
    I've confirmed by computer search that there are no further solutions for $y leq 160, 000$.
    $endgroup$
    – Connor Harris
    Dec 17 '18 at 17:27






  • 1




    $begingroup$
    Magma "says" they are the unique solutions.
    $endgroup$
    – xarles
    Dec 17 '18 at 17:43






  • 2




    $begingroup$
    lmfdb.org/EllipticCurve/Q/243/a/1
    $endgroup$
    – Álvaro Lozano-Robledo
    Dec 19 '18 at 17:51














  • 1




    $begingroup$
    I think that the fact that $y^3equiv 0,1pmod{8}$ is useful here.
    $endgroup$
    – Don Thousand
    Dec 17 '18 at 16:01






  • 5




    $begingroup$
    This is also an elliptic curve equation (in long Weierstrass form).
    $endgroup$
    – Dietrich Burde
    Dec 17 '18 at 16:34






  • 1




    $begingroup$
    I've confirmed by computer search that there are no further solutions for $y leq 160, 000$.
    $endgroup$
    – Connor Harris
    Dec 17 '18 at 17:27






  • 1




    $begingroup$
    Magma "says" they are the unique solutions.
    $endgroup$
    – xarles
    Dec 17 '18 at 17:43






  • 2




    $begingroup$
    lmfdb.org/EllipticCurve/Q/243/a/1
    $endgroup$
    – Álvaro Lozano-Robledo
    Dec 19 '18 at 17:51








1




1




$begingroup$
I think that the fact that $y^3equiv 0,1pmod{8}$ is useful here.
$endgroup$
– Don Thousand
Dec 17 '18 at 16:01




$begingroup$
I think that the fact that $y^3equiv 0,1pmod{8}$ is useful here.
$endgroup$
– Don Thousand
Dec 17 '18 at 16:01




5




5




$begingroup$
This is also an elliptic curve equation (in long Weierstrass form).
$endgroup$
– Dietrich Burde
Dec 17 '18 at 16:34




$begingroup$
This is also an elliptic curve equation (in long Weierstrass form).
$endgroup$
– Dietrich Burde
Dec 17 '18 at 16:34




1




1




$begingroup$
I've confirmed by computer search that there are no further solutions for $y leq 160, 000$.
$endgroup$
– Connor Harris
Dec 17 '18 at 17:27




$begingroup$
I've confirmed by computer search that there are no further solutions for $y leq 160, 000$.
$endgroup$
– Connor Harris
Dec 17 '18 at 17:27




1




1




$begingroup$
Magma "says" they are the unique solutions.
$endgroup$
– xarles
Dec 17 '18 at 17:43




$begingroup$
Magma "says" they are the unique solutions.
$endgroup$
– xarles
Dec 17 '18 at 17:43




2




2




$begingroup$
lmfdb.org/EllipticCurve/Q/243/a/1
$endgroup$
– Álvaro Lozano-Robledo
Dec 19 '18 at 17:51




$begingroup$
lmfdb.org/EllipticCurve/Q/243/a/1
$endgroup$
– Álvaro Lozano-Robledo
Dec 19 '18 at 17:51










1 Answer
1






active

oldest

votes


















2












$begingroup$


Theorem. The diophantine equation
$$
X^2+3=4Y^3
$$
has only the solutions $(X,Y) = (pm 1,1), (pm 37,7)$.




Proof. From this reference.




Adapting to this theorem, we rewrite as
$$
(2x-1)^2+3 = 4y^3
$$
Then the solutions are $(2x-1,y) = (pm 1,1),(pm 37,7)$. Therefore
$$
(x,y) in {(0,1),(1,1),(-18,7),(19,7)}
$$




Algebraic Number Theory proof



We shall use some Algebraic Number Theory to show that




Proposition. Let $x,y$ be integers such that
$$x^2-x+1 = y^3$$
Then $x$ is a polynomial in $u,v$ such that
$$u^3+3u^2v-v^3=-1,quad x = frac{1}{2}(1 + u^3 - 3 u^2 v - 6 u v^2 - v^3)$$
or
$$u^3-3uv^2-v^3=-1,quad x = frac{1}{2}(1 - u^3 - 6 u^2 v - 3 u v^2 + v^3)$$




It is straightforward to see that solutions of equation 1 and 2 are bijective:
$$
(u,v) longleftrightarrow (-v,-u)
$$
So the main difficulty lies in solving the first cubic Thue equation
$$
u^3+3u^2v-v^3=-1
$$
In particular we know that there are only finitely many solutions and they are all below a certain bound: $|u|,|v| < B$. So a (smart) search suffices if we know $B$.



Unfortunately I was unable to find any elementary solutions, nor was I able to find a way to compute $B$ by pen and pencil.



Remark: This is a known Thue equation that is already solved in a few papers, say both in the reference paper earlier, or as in here. All the solutions look rather advanced/involved. More explicitly,




Theorem. The only integer solutions to
$$
x^3-3xy^2-y^3=1
$$
are
$$
(1,0), (0,-1), (-1,1), (1,-3), (-3,2), (2,1)
$$





We now derive the proposition.



Proof. Let $K$ be the number field $mathbb Q(sqrt{-3})$. Letting $w=(1+sqrt{-3})/2$, the ring of integers is $mathcal O_K = mathbb Z[w]$ and the units are $U = {pm 1,pm w,pm w^2}$. $K$ has class number $1$ and hence unique factorization.



Now we solve the problem in $mathcal O_K$:
$$
(x-w)(x+w-1) = y^3
$$
Since
$$
(x+w-1)-(x-w) = 2w-1 = sqrt{-3}
$$
and $sqrt{-3}$ is prime (since it has prime norm $3$), either $x-w$ and $x+w-1$ are relatively prime or they share a common prime factor $sqrt{-3}$.



For the latter case, since $sqrt{-3}$ has norm $3$, we require norm $N(x-w)$ to be divisible by $3$. This gives
$$
N(x-w) = N(frac{2x-1-sqrt{-3}}{2})=frac{(2x-1)^2+3}{4} equiv 0pmod 3
$$
so $(2x-1)equiv 0pmod 3$. From
$$
(2x-1)^2+3 = 4y^2,
$$
we get $yequiv 0pmod 3$. Now taking $pmod 9$ we get a contradiction:
$$
(2x-1)^2+3 = 4y^2 implies 3 equiv 0 pmod 9
$$
Therefore $x+w-1$ and $x-w$ are coprime.




Since $x-w$ and $x+w-1$ are relatively prime, we must have
$$
begin{align}
x-w &= mu (u+v w)^3\
x+w-1 &= mu^{-1} (s+tw)^3
end{align}
$$
for some unit $mu in U$ and integers $u,v,s,t$. By absorbing the negative sign into the cube, we may assume $muin{1,w,w^2}$.



If $mu=1$, equation 1 becomes
$$
0 = frac{1 + 3 u^2 v + 3 u v^2}{2} sqrt{-3} + frac{1 + 2 u^3 + 3 u^2 v - 3 u v^2 - 2 v^3 - 2 x}{2}
$$
This requires $1+3u^2v + 3uv^2=0$, which is not possible (check $pmod 3$).



For the other two cases, $mu = w$ and $mu=w^2$, checking the coefficient of $sqrt{-3}$ for $x-w=mu(u+vw)^3$ gives us the two respective equations:
$$
begin{align}
u^3 + 3 u^2 v - v^3 &= -1\
u^3 - 3 u v^2 - v^3 &= -1
end{align}
$$
This gives the first part of the proposition. Checking the coefficient of the constants will give the other part containing $x$.



$$tag*{$square$}$$




Computer solution of the Thue equations



We can solve the first using the online PARI/GP with command:



thue(thueinit(u^3 + 3*u^2 - 1,1),-1)



giving us the solutions
$$
(u,v) = (-3, 1), (-1, 0), (0, 1), (1, -1), (1, 2), (2, -3)
$$
Then putting in back $u,v$ into $x-w=w(u+v w)^3$ we get
$$
x = -18,0,0,0,-18,-18
$$
corresponding to $(x,y) = (-18,7),(0,1)$.



For the other case, the command



thue(thueinit(u^3 + 3*u^2 - 1,1),-1)



gives solutions
$$
(u,v) = (-2, -1), (-1, 0), (-1, 3), (0, 1), (1, -1), (3, -2)
$$
and solving $x-w=w^2(u+vw)^3$ gives
$$
x = 19,1,19,1,1,19
$$
corresponding to $(x,y) = (19,7), (1,1)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Beautiful! Thanks for the solution.
    $endgroup$
    – Haran
    Jan 31 at 16:43











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1 Answer
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oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$


Theorem. The diophantine equation
$$
X^2+3=4Y^3
$$
has only the solutions $(X,Y) = (pm 1,1), (pm 37,7)$.




Proof. From this reference.




Adapting to this theorem, we rewrite as
$$
(2x-1)^2+3 = 4y^3
$$
Then the solutions are $(2x-1,y) = (pm 1,1),(pm 37,7)$. Therefore
$$
(x,y) in {(0,1),(1,1),(-18,7),(19,7)}
$$




Algebraic Number Theory proof



We shall use some Algebraic Number Theory to show that




Proposition. Let $x,y$ be integers such that
$$x^2-x+1 = y^3$$
Then $x$ is a polynomial in $u,v$ such that
$$u^3+3u^2v-v^3=-1,quad x = frac{1}{2}(1 + u^3 - 3 u^2 v - 6 u v^2 - v^3)$$
or
$$u^3-3uv^2-v^3=-1,quad x = frac{1}{2}(1 - u^3 - 6 u^2 v - 3 u v^2 + v^3)$$




It is straightforward to see that solutions of equation 1 and 2 are bijective:
$$
(u,v) longleftrightarrow (-v,-u)
$$
So the main difficulty lies in solving the first cubic Thue equation
$$
u^3+3u^2v-v^3=-1
$$
In particular we know that there are only finitely many solutions and they are all below a certain bound: $|u|,|v| < B$. So a (smart) search suffices if we know $B$.



Unfortunately I was unable to find any elementary solutions, nor was I able to find a way to compute $B$ by pen and pencil.



Remark: This is a known Thue equation that is already solved in a few papers, say both in the reference paper earlier, or as in here. All the solutions look rather advanced/involved. More explicitly,




Theorem. The only integer solutions to
$$
x^3-3xy^2-y^3=1
$$
are
$$
(1,0), (0,-1), (-1,1), (1,-3), (-3,2), (2,1)
$$





We now derive the proposition.



Proof. Let $K$ be the number field $mathbb Q(sqrt{-3})$. Letting $w=(1+sqrt{-3})/2$, the ring of integers is $mathcal O_K = mathbb Z[w]$ and the units are $U = {pm 1,pm w,pm w^2}$. $K$ has class number $1$ and hence unique factorization.



Now we solve the problem in $mathcal O_K$:
$$
(x-w)(x+w-1) = y^3
$$
Since
$$
(x+w-1)-(x-w) = 2w-1 = sqrt{-3}
$$
and $sqrt{-3}$ is prime (since it has prime norm $3$), either $x-w$ and $x+w-1$ are relatively prime or they share a common prime factor $sqrt{-3}$.



For the latter case, since $sqrt{-3}$ has norm $3$, we require norm $N(x-w)$ to be divisible by $3$. This gives
$$
N(x-w) = N(frac{2x-1-sqrt{-3}}{2})=frac{(2x-1)^2+3}{4} equiv 0pmod 3
$$
so $(2x-1)equiv 0pmod 3$. From
$$
(2x-1)^2+3 = 4y^2,
$$
we get $yequiv 0pmod 3$. Now taking $pmod 9$ we get a contradiction:
$$
(2x-1)^2+3 = 4y^2 implies 3 equiv 0 pmod 9
$$
Therefore $x+w-1$ and $x-w$ are coprime.




Since $x-w$ and $x+w-1$ are relatively prime, we must have
$$
begin{align}
x-w &= mu (u+v w)^3\
x+w-1 &= mu^{-1} (s+tw)^3
end{align}
$$
for some unit $mu in U$ and integers $u,v,s,t$. By absorbing the negative sign into the cube, we may assume $muin{1,w,w^2}$.



If $mu=1$, equation 1 becomes
$$
0 = frac{1 + 3 u^2 v + 3 u v^2}{2} sqrt{-3} + frac{1 + 2 u^3 + 3 u^2 v - 3 u v^2 - 2 v^3 - 2 x}{2}
$$
This requires $1+3u^2v + 3uv^2=0$, which is not possible (check $pmod 3$).



For the other two cases, $mu = w$ and $mu=w^2$, checking the coefficient of $sqrt{-3}$ for $x-w=mu(u+vw)^3$ gives us the two respective equations:
$$
begin{align}
u^3 + 3 u^2 v - v^3 &= -1\
u^3 - 3 u v^2 - v^3 &= -1
end{align}
$$
This gives the first part of the proposition. Checking the coefficient of the constants will give the other part containing $x$.



$$tag*{$square$}$$




Computer solution of the Thue equations



We can solve the first using the online PARI/GP with command:



thue(thueinit(u^3 + 3*u^2 - 1,1),-1)



giving us the solutions
$$
(u,v) = (-3, 1), (-1, 0), (0, 1), (1, -1), (1, 2), (2, -3)
$$
Then putting in back $u,v$ into $x-w=w(u+v w)^3$ we get
$$
x = -18,0,0,0,-18,-18
$$
corresponding to $(x,y) = (-18,7),(0,1)$.



For the other case, the command



thue(thueinit(u^3 + 3*u^2 - 1,1),-1)



gives solutions
$$
(u,v) = (-2, -1), (-1, 0), (-1, 3), (0, 1), (1, -1), (3, -2)
$$
and solving $x-w=w^2(u+vw)^3$ gives
$$
x = 19,1,19,1,1,19
$$
corresponding to $(x,y) = (19,7), (1,1)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Beautiful! Thanks for the solution.
    $endgroup$
    – Haran
    Jan 31 at 16:43
















2












$begingroup$


Theorem. The diophantine equation
$$
X^2+3=4Y^3
$$
has only the solutions $(X,Y) = (pm 1,1), (pm 37,7)$.




Proof. From this reference.




Adapting to this theorem, we rewrite as
$$
(2x-1)^2+3 = 4y^3
$$
Then the solutions are $(2x-1,y) = (pm 1,1),(pm 37,7)$. Therefore
$$
(x,y) in {(0,1),(1,1),(-18,7),(19,7)}
$$




Algebraic Number Theory proof



We shall use some Algebraic Number Theory to show that




Proposition. Let $x,y$ be integers such that
$$x^2-x+1 = y^3$$
Then $x$ is a polynomial in $u,v$ such that
$$u^3+3u^2v-v^3=-1,quad x = frac{1}{2}(1 + u^3 - 3 u^2 v - 6 u v^2 - v^3)$$
or
$$u^3-3uv^2-v^3=-1,quad x = frac{1}{2}(1 - u^3 - 6 u^2 v - 3 u v^2 + v^3)$$




It is straightforward to see that solutions of equation 1 and 2 are bijective:
$$
(u,v) longleftrightarrow (-v,-u)
$$
So the main difficulty lies in solving the first cubic Thue equation
$$
u^3+3u^2v-v^3=-1
$$
In particular we know that there are only finitely many solutions and they are all below a certain bound: $|u|,|v| < B$. So a (smart) search suffices if we know $B$.



Unfortunately I was unable to find any elementary solutions, nor was I able to find a way to compute $B$ by pen and pencil.



Remark: This is a known Thue equation that is already solved in a few papers, say both in the reference paper earlier, or as in here. All the solutions look rather advanced/involved. More explicitly,




Theorem. The only integer solutions to
$$
x^3-3xy^2-y^3=1
$$
are
$$
(1,0), (0,-1), (-1,1), (1,-3), (-3,2), (2,1)
$$





We now derive the proposition.



Proof. Let $K$ be the number field $mathbb Q(sqrt{-3})$. Letting $w=(1+sqrt{-3})/2$, the ring of integers is $mathcal O_K = mathbb Z[w]$ and the units are $U = {pm 1,pm w,pm w^2}$. $K$ has class number $1$ and hence unique factorization.



Now we solve the problem in $mathcal O_K$:
$$
(x-w)(x+w-1) = y^3
$$
Since
$$
(x+w-1)-(x-w) = 2w-1 = sqrt{-3}
$$
and $sqrt{-3}$ is prime (since it has prime norm $3$), either $x-w$ and $x+w-1$ are relatively prime or they share a common prime factor $sqrt{-3}$.



For the latter case, since $sqrt{-3}$ has norm $3$, we require norm $N(x-w)$ to be divisible by $3$. This gives
$$
N(x-w) = N(frac{2x-1-sqrt{-3}}{2})=frac{(2x-1)^2+3}{4} equiv 0pmod 3
$$
so $(2x-1)equiv 0pmod 3$. From
$$
(2x-1)^2+3 = 4y^2,
$$
we get $yequiv 0pmod 3$. Now taking $pmod 9$ we get a contradiction:
$$
(2x-1)^2+3 = 4y^2 implies 3 equiv 0 pmod 9
$$
Therefore $x+w-1$ and $x-w$ are coprime.




Since $x-w$ and $x+w-1$ are relatively prime, we must have
$$
begin{align}
x-w &= mu (u+v w)^3\
x+w-1 &= mu^{-1} (s+tw)^3
end{align}
$$
for some unit $mu in U$ and integers $u,v,s,t$. By absorbing the negative sign into the cube, we may assume $muin{1,w,w^2}$.



If $mu=1$, equation 1 becomes
$$
0 = frac{1 + 3 u^2 v + 3 u v^2}{2} sqrt{-3} + frac{1 + 2 u^3 + 3 u^2 v - 3 u v^2 - 2 v^3 - 2 x}{2}
$$
This requires $1+3u^2v + 3uv^2=0$, which is not possible (check $pmod 3$).



For the other two cases, $mu = w$ and $mu=w^2$, checking the coefficient of $sqrt{-3}$ for $x-w=mu(u+vw)^3$ gives us the two respective equations:
$$
begin{align}
u^3 + 3 u^2 v - v^3 &= -1\
u^3 - 3 u v^2 - v^3 &= -1
end{align}
$$
This gives the first part of the proposition. Checking the coefficient of the constants will give the other part containing $x$.



$$tag*{$square$}$$




Computer solution of the Thue equations



We can solve the first using the online PARI/GP with command:



thue(thueinit(u^3 + 3*u^2 - 1,1),-1)



giving us the solutions
$$
(u,v) = (-3, 1), (-1, 0), (0, 1), (1, -1), (1, 2), (2, -3)
$$
Then putting in back $u,v$ into $x-w=w(u+v w)^3$ we get
$$
x = -18,0,0,0,-18,-18
$$
corresponding to $(x,y) = (-18,7),(0,1)$.



For the other case, the command



thue(thueinit(u^3 + 3*u^2 - 1,1),-1)



gives solutions
$$
(u,v) = (-2, -1), (-1, 0), (-1, 3), (0, 1), (1, -1), (3, -2)
$$
and solving $x-w=w^2(u+vw)^3$ gives
$$
x = 19,1,19,1,1,19
$$
corresponding to $(x,y) = (19,7), (1,1)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Beautiful! Thanks for the solution.
    $endgroup$
    – Haran
    Jan 31 at 16:43














2












2








2





$begingroup$


Theorem. The diophantine equation
$$
X^2+3=4Y^3
$$
has only the solutions $(X,Y) = (pm 1,1), (pm 37,7)$.




Proof. From this reference.




Adapting to this theorem, we rewrite as
$$
(2x-1)^2+3 = 4y^3
$$
Then the solutions are $(2x-1,y) = (pm 1,1),(pm 37,7)$. Therefore
$$
(x,y) in {(0,1),(1,1),(-18,7),(19,7)}
$$




Algebraic Number Theory proof



We shall use some Algebraic Number Theory to show that




Proposition. Let $x,y$ be integers such that
$$x^2-x+1 = y^3$$
Then $x$ is a polynomial in $u,v$ such that
$$u^3+3u^2v-v^3=-1,quad x = frac{1}{2}(1 + u^3 - 3 u^2 v - 6 u v^2 - v^3)$$
or
$$u^3-3uv^2-v^3=-1,quad x = frac{1}{2}(1 - u^3 - 6 u^2 v - 3 u v^2 + v^3)$$




It is straightforward to see that solutions of equation 1 and 2 are bijective:
$$
(u,v) longleftrightarrow (-v,-u)
$$
So the main difficulty lies in solving the first cubic Thue equation
$$
u^3+3u^2v-v^3=-1
$$
In particular we know that there are only finitely many solutions and they are all below a certain bound: $|u|,|v| < B$. So a (smart) search suffices if we know $B$.



Unfortunately I was unable to find any elementary solutions, nor was I able to find a way to compute $B$ by pen and pencil.



Remark: This is a known Thue equation that is already solved in a few papers, say both in the reference paper earlier, or as in here. All the solutions look rather advanced/involved. More explicitly,




Theorem. The only integer solutions to
$$
x^3-3xy^2-y^3=1
$$
are
$$
(1,0), (0,-1), (-1,1), (1,-3), (-3,2), (2,1)
$$





We now derive the proposition.



Proof. Let $K$ be the number field $mathbb Q(sqrt{-3})$. Letting $w=(1+sqrt{-3})/2$, the ring of integers is $mathcal O_K = mathbb Z[w]$ and the units are $U = {pm 1,pm w,pm w^2}$. $K$ has class number $1$ and hence unique factorization.



Now we solve the problem in $mathcal O_K$:
$$
(x-w)(x+w-1) = y^3
$$
Since
$$
(x+w-1)-(x-w) = 2w-1 = sqrt{-3}
$$
and $sqrt{-3}$ is prime (since it has prime norm $3$), either $x-w$ and $x+w-1$ are relatively prime or they share a common prime factor $sqrt{-3}$.



For the latter case, since $sqrt{-3}$ has norm $3$, we require norm $N(x-w)$ to be divisible by $3$. This gives
$$
N(x-w) = N(frac{2x-1-sqrt{-3}}{2})=frac{(2x-1)^2+3}{4} equiv 0pmod 3
$$
so $(2x-1)equiv 0pmod 3$. From
$$
(2x-1)^2+3 = 4y^2,
$$
we get $yequiv 0pmod 3$. Now taking $pmod 9$ we get a contradiction:
$$
(2x-1)^2+3 = 4y^2 implies 3 equiv 0 pmod 9
$$
Therefore $x+w-1$ and $x-w$ are coprime.




Since $x-w$ and $x+w-1$ are relatively prime, we must have
$$
begin{align}
x-w &= mu (u+v w)^3\
x+w-1 &= mu^{-1} (s+tw)^3
end{align}
$$
for some unit $mu in U$ and integers $u,v,s,t$. By absorbing the negative sign into the cube, we may assume $muin{1,w,w^2}$.



If $mu=1$, equation 1 becomes
$$
0 = frac{1 + 3 u^2 v + 3 u v^2}{2} sqrt{-3} + frac{1 + 2 u^3 + 3 u^2 v - 3 u v^2 - 2 v^3 - 2 x}{2}
$$
This requires $1+3u^2v + 3uv^2=0$, which is not possible (check $pmod 3$).



For the other two cases, $mu = w$ and $mu=w^2$, checking the coefficient of $sqrt{-3}$ for $x-w=mu(u+vw)^3$ gives us the two respective equations:
$$
begin{align}
u^3 + 3 u^2 v - v^3 &= -1\
u^3 - 3 u v^2 - v^3 &= -1
end{align}
$$
This gives the first part of the proposition. Checking the coefficient of the constants will give the other part containing $x$.



$$tag*{$square$}$$




Computer solution of the Thue equations



We can solve the first using the online PARI/GP with command:



thue(thueinit(u^3 + 3*u^2 - 1,1),-1)



giving us the solutions
$$
(u,v) = (-3, 1), (-1, 0), (0, 1), (1, -1), (1, 2), (2, -3)
$$
Then putting in back $u,v$ into $x-w=w(u+v w)^3$ we get
$$
x = -18,0,0,0,-18,-18
$$
corresponding to $(x,y) = (-18,7),(0,1)$.



For the other case, the command



thue(thueinit(u^3 + 3*u^2 - 1,1),-1)



gives solutions
$$
(u,v) = (-2, -1), (-1, 0), (-1, 3), (0, 1), (1, -1), (3, -2)
$$
and solving $x-w=w^2(u+vw)^3$ gives
$$
x = 19,1,19,1,1,19
$$
corresponding to $(x,y) = (19,7), (1,1)$.






share|cite|improve this answer









$endgroup$




Theorem. The diophantine equation
$$
X^2+3=4Y^3
$$
has only the solutions $(X,Y) = (pm 1,1), (pm 37,7)$.




Proof. From this reference.




Adapting to this theorem, we rewrite as
$$
(2x-1)^2+3 = 4y^3
$$
Then the solutions are $(2x-1,y) = (pm 1,1),(pm 37,7)$. Therefore
$$
(x,y) in {(0,1),(1,1),(-18,7),(19,7)}
$$




Algebraic Number Theory proof



We shall use some Algebraic Number Theory to show that




Proposition. Let $x,y$ be integers such that
$$x^2-x+1 = y^3$$
Then $x$ is a polynomial in $u,v$ such that
$$u^3+3u^2v-v^3=-1,quad x = frac{1}{2}(1 + u^3 - 3 u^2 v - 6 u v^2 - v^3)$$
or
$$u^3-3uv^2-v^3=-1,quad x = frac{1}{2}(1 - u^3 - 6 u^2 v - 3 u v^2 + v^3)$$




It is straightforward to see that solutions of equation 1 and 2 are bijective:
$$
(u,v) longleftrightarrow (-v,-u)
$$
So the main difficulty lies in solving the first cubic Thue equation
$$
u^3+3u^2v-v^3=-1
$$
In particular we know that there are only finitely many solutions and they are all below a certain bound: $|u|,|v| < B$. So a (smart) search suffices if we know $B$.



Unfortunately I was unable to find any elementary solutions, nor was I able to find a way to compute $B$ by pen and pencil.



Remark: This is a known Thue equation that is already solved in a few papers, say both in the reference paper earlier, or as in here. All the solutions look rather advanced/involved. More explicitly,




Theorem. The only integer solutions to
$$
x^3-3xy^2-y^3=1
$$
are
$$
(1,0), (0,-1), (-1,1), (1,-3), (-3,2), (2,1)
$$





We now derive the proposition.



Proof. Let $K$ be the number field $mathbb Q(sqrt{-3})$. Letting $w=(1+sqrt{-3})/2$, the ring of integers is $mathcal O_K = mathbb Z[w]$ and the units are $U = {pm 1,pm w,pm w^2}$. $K$ has class number $1$ and hence unique factorization.



Now we solve the problem in $mathcal O_K$:
$$
(x-w)(x+w-1) = y^3
$$
Since
$$
(x+w-1)-(x-w) = 2w-1 = sqrt{-3}
$$
and $sqrt{-3}$ is prime (since it has prime norm $3$), either $x-w$ and $x+w-1$ are relatively prime or they share a common prime factor $sqrt{-3}$.



For the latter case, since $sqrt{-3}$ has norm $3$, we require norm $N(x-w)$ to be divisible by $3$. This gives
$$
N(x-w) = N(frac{2x-1-sqrt{-3}}{2})=frac{(2x-1)^2+3}{4} equiv 0pmod 3
$$
so $(2x-1)equiv 0pmod 3$. From
$$
(2x-1)^2+3 = 4y^2,
$$
we get $yequiv 0pmod 3$. Now taking $pmod 9$ we get a contradiction:
$$
(2x-1)^2+3 = 4y^2 implies 3 equiv 0 pmod 9
$$
Therefore $x+w-1$ and $x-w$ are coprime.




Since $x-w$ and $x+w-1$ are relatively prime, we must have
$$
begin{align}
x-w &= mu (u+v w)^3\
x+w-1 &= mu^{-1} (s+tw)^3
end{align}
$$
for some unit $mu in U$ and integers $u,v,s,t$. By absorbing the negative sign into the cube, we may assume $muin{1,w,w^2}$.



If $mu=1$, equation 1 becomes
$$
0 = frac{1 + 3 u^2 v + 3 u v^2}{2} sqrt{-3} + frac{1 + 2 u^3 + 3 u^2 v - 3 u v^2 - 2 v^3 - 2 x}{2}
$$
This requires $1+3u^2v + 3uv^2=0$, which is not possible (check $pmod 3$).



For the other two cases, $mu = w$ and $mu=w^2$, checking the coefficient of $sqrt{-3}$ for $x-w=mu(u+vw)^3$ gives us the two respective equations:
$$
begin{align}
u^3 + 3 u^2 v - v^3 &= -1\
u^3 - 3 u v^2 - v^3 &= -1
end{align}
$$
This gives the first part of the proposition. Checking the coefficient of the constants will give the other part containing $x$.



$$tag*{$square$}$$




Computer solution of the Thue equations



We can solve the first using the online PARI/GP with command:



thue(thueinit(u^3 + 3*u^2 - 1,1),-1)



giving us the solutions
$$
(u,v) = (-3, 1), (-1, 0), (0, 1), (1, -1), (1, 2), (2, -3)
$$
Then putting in back $u,v$ into $x-w=w(u+v w)^3$ we get
$$
x = -18,0,0,0,-18,-18
$$
corresponding to $(x,y) = (-18,7),(0,1)$.



For the other case, the command



thue(thueinit(u^3 + 3*u^2 - 1,1),-1)



gives solutions
$$
(u,v) = (-2, -1), (-1, 0), (-1, 3), (0, 1), (1, -1), (3, -2)
$$
and solving $x-w=w^2(u+vw)^3$ gives
$$
x = 19,1,19,1,1,19
$$
corresponding to $(x,y) = (19,7), (1,1)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 31 at 11:17









Yong Hao NgYong Hao Ng

3,5691222




3,5691222












  • $begingroup$
    Beautiful! Thanks for the solution.
    $endgroup$
    – Haran
    Jan 31 at 16:43


















  • $begingroup$
    Beautiful! Thanks for the solution.
    $endgroup$
    – Haran
    Jan 31 at 16:43
















$begingroup$
Beautiful! Thanks for the solution.
$endgroup$
– Haran
Jan 31 at 16:43




$begingroup$
Beautiful! Thanks for the solution.
$endgroup$
– Haran
Jan 31 at 16:43


















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