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

1,144424




1,144424








  • 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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044095%2fwhat-are-the-positive-integer-solutions-to-x2-x1-y3%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044095%2fwhat-are-the-positive-integer-solutions-to-x2-x1-y3%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Verónica Boquete

Image
Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more