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Solve the Diophantine equation $24x^4-5y^4=z^2$

A quartic diophantine equationConsecutive quadratic residuesPossible solutions of a diophantine equation: $p^2+pq+275p+10q=2008$Solutions to a quadratic diophantine modular equationEquation $x^3+2x+1=2^n$ in positive integers$2^n + 3^n = x^p$ has no solutions over the natural numbersIf $anotequiv 0mod{p}$ then there are $p-1$ solutions (ordered pairs) to $x^2-y^2equiv amod{p}$Quadratic equation solutions modulo prime pSolving the Diophantine Equation $ax^2 + bx + c = dy^2 + ey + f$?Is $u$ a quadratic residue mod $t + nu$?

1

1

$begingroup$

I want to solve $24x^4-5y^4=z^2$ in integers not all zero, and to fix ideas, I want to find them such that x,y are coprime.

I've tried plugging in small values of $x$ and $y$ and they don't return a square value for $24x^4-5y^4$. So I guess that no (nontrivial) solutions exist.

How would I prove this? I can't seem to derive a contradiction in $mathbb{R}$ nor in $mathbb{Q}_p$. So I guess that this equation violates Hasse's local-global principle. That is, I have to derive a contradiction by some other method than arguing in $mathbb{R}$ or mod $p$ for some prime $p$.

The only example I know of a method for solving an equation that violates Hasse's principle is for the example $2a^2+17b^4=c^4$. I shall now describe the method, and my question is, does this method generalize to my equation $24x^4-5y^4=z^2$? (And if not, how else would I go about solving it?)

Here is the method I know for showing that no integers solutions exist to $2a^2+17b^4=c^4$, where $a,b,c$ are not all zero, and $(b,c)=1$. For any prime $p$ dividing $a$, since $17b^4 equiv c^4$ modulo $p$, we get that $17$ is a quadratic residue mod $p$. By quadratic reciprocity, we get that $p$ is a quadratic residue mod $17$, where $p$ was any prime factor of $a$.
This implies that $a$ is a quadratic residue mod $17$. Write $a equiv x^2$ mod $17$. Then $2x^4 equiv c^4$ mod $17$, and then $2$ is a fourth power mod $17$. This is a contradiction.

number-theory prime-numbers diophantine-equations elliptic-curves

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edited Mar 19 at 5:41

SSF

asked Mar 19 at 5:33

SSFSSF

475111

$endgroup$

add a comment | 

1

1

$begingroup$

I want to solve $24x^4-5y^4=z^2$ in integers not all zero, and to fix ideas, I want to find them such that x,y are coprime.

I've tried plugging in small values of $x$ and $y$ and they don't return a square value for $24x^4-5y^4$. So I guess that no (nontrivial) solutions exist.

How would I prove this? I can't seem to derive a contradiction in $mathbb{R}$ nor in $mathbb{Q}_p$. So I guess that this equation violates Hasse's local-global principle. That is, I have to derive a contradiction by some other method than arguing in $mathbb{R}$ or mod $p$ for some prime $p$.

The only example I know of a method for solving an equation that violates Hasse's principle is for the example $2a^2+17b^4=c^4$. I shall now describe the method, and my question is, does this method generalize to my equation $24x^4-5y^4=z^2$? (And if not, how else would I go about solving it?)

Here is the method I know for showing that no integers solutions exist to $2a^2+17b^4=c^4$, where $a,b,c$ are not all zero, and $(b,c)=1$. For any prime $p$ dividing $a$, since $17b^4 equiv c^4$ modulo $p$, we get that $17$ is a quadratic residue mod $p$. By quadratic reciprocity, we get that $p$ is a quadratic residue mod $17$, where $p$ was any prime factor of $a$.
This implies that $a$ is a quadratic residue mod $17$. Write $a equiv x^2$ mod $17$. Then $2x^4 equiv c^4$ mod $17$, and then $2$ is a fourth power mod $17$. This is a contradiction.

number-theory prime-numbers diophantine-equations elliptic-curves

share|cite|improve this question

edited Mar 19 at 5:41

SSF

asked Mar 19 at 5:33

SSFSSF

475111

$endgroup$

add a comment | 

$begingroup$

I want to solve $24x^4-5y^4=z^2$ in integers not all zero, and to fix ideas, I want to find them such that x,y are coprime.

I've tried plugging in small values of $x$ and $y$ and they don't return a square value for $24x^4-5y^4$. So I guess that no (nontrivial) solutions exist.

How would I prove this? I can't seem to derive a contradiction in $mathbb{R}$ nor in $mathbb{Q}_p$. So I guess that this equation violates Hasse's local-global principle. That is, I have to derive a contradiction by some other method than arguing in $mathbb{R}$ or mod $p$ for some prime $p$.

The only example I know of a method for solving an equation that violates Hasse's principle is for the example $2a^2+17b^4=c^4$. I shall now describe the method, and my question is, does this method generalize to my equation $24x^4-5y^4=z^2$? (And if not, how else would I go about solving it?)

Here is the method I know for showing that no integers solutions exist to $2a^2+17b^4=c^4$, where $a,b,c$ are not all zero, and $(b,c)=1$. For any prime $p$ dividing $a$, since $17b^4 equiv c^4$ modulo $p$, we get that $17$ is a quadratic residue mod $p$. By quadratic reciprocity, we get that $p$ is a quadratic residue mod $17$, where $p$ was any prime factor of $a$.
This implies that $a$ is a quadratic residue mod $17$. Write $a equiv x^2$ mod $17$. Then $2x^4 equiv c^4$ mod $17$, and then $2$ is a fourth power mod $17$. This is a contradiction.

number-theory prime-numbers diophantine-equations elliptic-curves

share|cite|improve this question

edited Mar 19 at 5:41

SSF

asked Mar 19 at 5:33

SSFSSF

475111

$endgroup$

share|cite|improve this question

edited Mar 19 at 5:41

SSF

asked Mar 19 at 5:33

SSFSSF

475111

share|cite|improve this question

edited Mar 19 at 5:41

SSF

asked Mar 19 at 5:33

SSFSSF

475111

asked Mar 19 at 5:33

SSFSSF

475111

asked Mar 19 at 5:33

SSFSSF

475111

1 Answer
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3

$begingroup$

Here's a rudimentary proof. Suppose $z^2=24x^4-5y^4$. Then looking modulo $4$ gives
begin{align}
z^2 &= 24x^4-5y^4\
&equiv -y^4mod 4
end{align}

Since the only squares modulo $4$ are $0$ and $1$, this implies $z^2equiv y^4equiv 0mod 4$, so $2|z,y$. Then setting $overline{y}=frac{y}{2}$ and $overline{z}=frac{z}{2}$, our original equation becomes
begin{align}
4overline{z}^2 &= 24x^4-16cdot5overline{y}^4\
overline{z}^2 &= 6x^4-4cdot5overline{y}^4\
end{align}
This then immediately implies that $2|overline{z}$, so we can again set $w=frac{overline{z}}{2}$ and find
begin{align}
4w^2 &= 6x^4-4cdot5overline{y}^4\
2w^2 &= 3x^4-2cdot5overline{y}^4\
end{align}
Now this gives $2|x$, so we finally set $overline{x}=frac{x}{2}$ and get
begin{align}
2w^2 &= 16cdot3overline{x}^4-2cdot5overline{y}^4\
w^2 &= 24overline{x}^4-5overline{y}^4.
end{align}
Thus we get a new solution with every term strictly smaller, so by infinite descent no solutions can exist.

share|cite|improve this answer

answered Mar 19 at 6:19

Ethan MacBroughEthan MacBrough

1,241617

$endgroup$

add a comment | 

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3

$begingroup$

Here's a rudimentary proof. Suppose $z^2=24x^4-5y^4$. Then looking modulo $4$ gives
begin{align}
z^2 &= 24x^4-5y^4\
&equiv -y^4mod 4
end{align}

Since the only squares modulo $4$ are $0$ and $1$, this implies $z^2equiv y^4equiv 0mod 4$, so $2|z,y$. Then setting $overline{y}=frac{y}{2}$ and $overline{z}=frac{z}{2}$, our original equation becomes
begin{align}
4overline{z}^2 &= 24x^4-16cdot5overline{y}^4\
overline{z}^2 &= 6x^4-4cdot5overline{y}^4\
end{align}
This then immediately implies that $2|overline{z}$, so we can again set $w=frac{overline{z}}{2}$ and find
begin{align}
4w^2 &= 6x^4-4cdot5overline{y}^4\
2w^2 &= 3x^4-2cdot5overline{y}^4\
end{align}
Now this gives $2|x$, so we finally set $overline{x}=frac{x}{2}$ and get
begin{align}
2w^2 &= 16cdot3overline{x}^4-2cdot5overline{y}^4\
w^2 &= 24overline{x}^4-5overline{y}^4.
end{align}
Thus we get a new solution with every term strictly smaller, so by infinite descent no solutions can exist.

share|cite|improve this answer

answered Mar 19 at 6:19

Ethan MacBroughEthan MacBrough

1,241617

$endgroup$

add a comment | 

3

$begingroup$

Here's a rudimentary proof. Suppose $z^2=24x^4-5y^4$. Then looking modulo $4$ gives
begin{align}
z^2 &= 24x^4-5y^4\
&equiv -y^4mod 4
end{align}

Since the only squares modulo $4$ are $0$ and $1$, this implies $z^2equiv y^4equiv 0mod 4$, so $2|z,y$. Then setting $overline{y}=frac{y}{2}$ and $overline{z}=frac{z}{2}$, our original equation becomes
begin{align}
4overline{z}^2 &= 24x^4-16cdot5overline{y}^4\
overline{z}^2 &= 6x^4-4cdot5overline{y}^4\
end{align}
This then immediately implies that $2|overline{z}$, so we can again set $w=frac{overline{z}}{2}$ and find
begin{align}
4w^2 &= 6x^4-4cdot5overline{y}^4\
2w^2 &= 3x^4-2cdot5overline{y}^4\
end{align}
Now this gives $2|x$, so we finally set $overline{x}=frac{x}{2}$ and get
begin{align}
2w^2 &= 16cdot3overline{x}^4-2cdot5overline{y}^4\
w^2 &= 24overline{x}^4-5overline{y}^4.
end{align}
Thus we get a new solution with every term strictly smaller, so by infinite descent no solutions can exist.

share|cite|improve this answer

answered Mar 19 at 6:19

Ethan MacBroughEthan MacBrough

1,241617

$endgroup$

add a comment | 

$begingroup$

Here's a rudimentary proof. Suppose $z^2=24x^4-5y^4$. Then looking modulo $4$ gives
begin{align}
z^2 &= 24x^4-5y^4\
&equiv -y^4mod 4
end{align}

Since the only squares modulo $4$ are $0$ and $1$, this implies $z^2equiv y^4equiv 0mod 4$, so $2|z,y$. Then setting $overline{y}=frac{y}{2}$ and $overline{z}=frac{z}{2}$, our original equation becomes
begin{align}
4overline{z}^2 &= 24x^4-16cdot5overline{y}^4\
overline{z}^2 &= 6x^4-4cdot5overline{y}^4\
end{align}
This then immediately implies that $2|overline{z}$, so we can again set $w=frac{overline{z}}{2}$ and find
begin{align}
4w^2 &= 6x^4-4cdot5overline{y}^4\
2w^2 &= 3x^4-2cdot5overline{y}^4\
end{align}
Now this gives $2|x$, so we finally set $overline{x}=frac{x}{2}$ and get
begin{align}
2w^2 &= 16cdot3overline{x}^4-2cdot5overline{y}^4\
w^2 &= 24overline{x}^4-5overline{y}^4.
end{align}
Thus we get a new solution with every term strictly smaller, so by infinite descent no solutions can exist.

share|cite|improve this answer

answered Mar 19 at 6:19

Ethan MacBroughEthan MacBrough

1,241617

$endgroup$

share|cite|improve this answer

answered Mar 19 at 6:19

Ethan MacBroughEthan MacBrough

1,241617

answered Mar 19 at 6:19

Ethan MacBroughEthan MacBrough

1,241617

answered Mar 19 at 6:19

Ethan MacBroughEthan MacBrough

1,241617