7
$\begingroup$

I've come upon the following question, and I'm a bit stumped. Any help would be greatly appreciated.

Find all integer solutions $(x,y,z)$ to the equation

$$ 2x^4 + 2x^2 y^2 + y^4 = z^2 $$

Here is the progress I have. I note that $(0,k,k^2)$ is a solution for any $k$ (including $k=0$ which gives the trivial solution). I conjecture that these are all the solutions, but I could be wrong about that.

Let's look at the equation mod 8. If $x$ and $y$ are both odd then $x^2, y^2$ are 1 (mod 8), so the left side is 5, and this is a contradiction. If both are even, then we can divide $x$ and $y$ by 2, and divide $z$ by 4, to obtain a new solution, and keep doing this until at least one of them is odd. Thus, we can assume that exactly one is odd. If $x$ is odd and $y$ is even, then the left side is 2 (mod 8), but $z^2$ must be either 0 or 4 (mod 8), contradiction. Thus we are reduced to the case that $x$ is even, $y$ is odd, and then $z$ is odd. But I can't see how to show that $x$ must be $0$ (if that is true).

Can anyone help with this?

Greg

$\endgroup$
6
  • 7
    $\begingroup$ Perhaps use that $x^2, x^2 + y^2, z$ is a pythagorean triple? $\endgroup$ Apr 3 at 8:51
  • 2
    $\begingroup$ Nice observation! Since $x$ is even and the trio is primitive, there must exist integer $m,n$ such that $x^2=2mn, x^2+y^2 = m^2 -n^2, z= m^2+n^2$. Looking at that second equation mod 4 shows that $m$ must be odd (since if it is even then $n$ is odd, but then LHS = 1 and RHS = -1), and since $gcd(m,n)=1$ we see from $x^2 = 2mn$ that $m$ is an odd square, say $r^2$, and $n=2s^2$. All we need or this to exist is that $y^2 = r^4 - 4s^4 - 4r^2 s^2$ must be a perfect square. So we've exchanged our first D. equation for another. But, I have to admit, I don't see how to solve this one, either :) $\endgroup$
    – user387394
    Apr 3 at 10:55
  • 1
    $\begingroup$ You could rewrite it as $y^2 + (2 s^2 + r^2)^2 = 2 r^4$ and try to use math.stackexchange.com/questions/1250912/… $\endgroup$ Apr 3 at 11:11
  • 1
    $\begingroup$ OK, cool, what this argument seems to me that this gives is that $$(s^2+(r^2+y)/2)^2 + (s^2+(r^2-y)/2)^2 = r^4$$, and we have another Pythagorean triple. I'm not sure where to go from here, though. It tells us a few things, like (by looking at it mod 16) that the odd term on the LHS is $\pm 1 (mod 8)$ and the even term is $0 (mod 4)$ (but I don't know which is odd and which is even), also that $r^2 = a^2 + b^2$ for some other integers $a,b$, so also $r = c^2 + d^2$ for some other integers. But still not seeing the final step. $\endgroup$
    – user387394
    Apr 3 at 14:00
  • 4
    $\begingroup$ Elementary methods for $ax^4 + b x^2 y^2 + c y^4 = d z^2 $ are shown in chapter 4 of Mordell, Diophantine Equations. Pages 16-29 $\endgroup$
    – Will Jagy
    Apr 3 at 20:07

3 Answers 3

3
$\begingroup$

The only integer solutions of $$2x^4 + 2x^2 y^2 + y^4 = z^2\tag1$$ are $(x,y,z)=(0,s,\pm s^2)$ where $s$ is any integer.

Proof :

Let us first consider the case $xyz=0$.

  • If $x=0$, then $z=\pm y^2$.

  • If $y=0$, then $2x^4=z^2$ implies $x=z=0$.

  • If $z=0$, then $x=y=0$.

So, $(x,y,z)=(0,s,\pm s^2)$ are solutions where $s$ is any integer.

In the following, let us prove that there is no solution $(x,y,z)$ satisfying $xyz\not=0$.

Suppose that there is a solution $(x,y,z)$ satisfying $xyz\not=0$.

We can see that if $(x,y,z)=(a,b,c)$ is a solution, then $(x,y,z)=(\pm a,\pm b,\pm c)$ are also solutions.

So, there has to be a solution $(x,y,z)$ such that $x,y,z$ are positive integers.

Let $(x,y,z)=(X,Y,Z)$ (where $X,Y,Z$ are positive integers) be a solution such that $Z$ is the smallest.

Here, we use your good observations.

  • Suppose that $X,Y$ are odd. Then, we have $5\equiv Z^2\pmod 8$ which is impossible.

  • Suppose that $X$ is odd and $Y$ is even. Then, we have $2\equiv Z^2\pmod 8$ which is impossible.

  • Suppose that $\gcd(X,Y)\gt 1$. Then, there is a prime number $p$ such that $p\mid X$ and $p\mid Y$. Then, we have to have $p^2\mid Z$, and we see that $(\frac Xp,\frac Yp,\frac Z{p^2})$ is also a solution (even when $p=2$). This contradicts that $Z$ is the smallest.

So, we can say that $X$ is even, $Y$ is odd, $Z$ is odd and $\gcd(X,Y)=1$.

As commented by Gribouillis, we can write $$(X^2)^2+(X^2+Y^2)^2=Z^2$$ So, we can write $$X^2=2mn,X^2+Y^2=m^2-n^2,Z=m^2+n^2$$ where $m,n$ are positive integers satisfying $m\gt n,\gcd(m,n)=1$ and $m\not\equiv n\pmod 2$.

We have $Y^2+2n^2=(m-n)^2$.

It follows from this answer that we can write $$Y=k|b^2-2a^2|,n=2abk,m-n=k(b^2+2a^2)$$ where $a,b,k$ are positive integers.

  • $k=1$ since $\gcd(m,n)=1$.

  • $b$ is odd since $Y$ is odd.

  • $\gcd(a,b)=1$ since $\gcd(m,n)=1$

We have $$X^2=2mn=4ab(2a^2+2ab+b^2)$$

So, there has to be a positive integer $c$ such that $$ab(2a^2+2ab+b^2)=c^2$$ We have the followings :

  • $\gcd(a,b)=1$

  • $\gcd(a,2a^2+2ab+b^2)=\gcd(a,b)=1$

  • $\gcd(b,2a^2+2ab+b^2)=\gcd(b,a)=1$

So, there have to be positive integers $u,v,w$ such that $$a=u^2,b=v^2, 2a^2+2ab+b^2=w^2$$ So, $$2u^4+2u^2v^2+v^4=w^2$$ which means that $(x,y,z)=(u,v,w)$ is also a solution.

Here, we have $$w^2=2a^2+2ab+b^2=m$$ So, $$\begin{align}Z^2-w^2&=(m^2+n^2)^2-m \\\\&=m^4-m+2m^2n^2+n^4 \\\\&=\underbrace{m(m-1)(m^2+m+1)}_{\text{non-negative}}+\underbrace{2m^2n^2+n^4}_{\text{positive}} \\\\&\gt 0\end{align}$$ This contradicts that $Z$ is the smallest.

So, we can say that there is no solution $(x,y,z)$ satisfying $xyz\not=0$.

Therefore, the only integer solutions are $(x,y,z)=(0,s,\pm s^2)$ where $s$ is any integer.$\ \blacksquare$

$\endgroup$
1
  • $\begingroup$ Thanks so much for your answer, I'm sorry that I only seem to be able to award the bounty to one answer, and I could understand the last one the easiest, otherwise I would have given to you as well. $\endgroup$
    – user387394
    Apr 16 at 13:48
2
$\begingroup$

The question is

Find all integer solutions $(x,y,z)$ to the equation $$ 2x^4 + 2x^2 y^2 + y^4 = z^2 $$

Divide both sides by $y^4$ to get $$ 2r^4 + 2r^2 + 1 = s^2 $$

where $\,r=x/y\,$ and $\,s=z/y^2.\,$ The problem now is to find rational solutions of this new equation. It is equivalent to an elliptic curve according to PARI/GP:

? E = ellfromeqn(2*r^4 + 2*r^2 + 1 - s^2)
[0, 2, 0, -8, -16]
? ellidentify(ellinit(E))
[["128c2", [0, -1, 0, -9, -7], []], [1, -1, 0, 0]]

The LMFDB entry for this curve is at URL https://www.lmfdb.org/EllipticCurve/Q/128c2/

It states that the Mordell-Weil group structure is $\,\mathbb{Z}/2\mathbb{Z}\,$ and thus there is only one rational point which is a 2-torsion element. It corresponds to the $(0,k,k^2)$ solution you already knew about. This is not a proof, but It seems to me that you did not ask for one, but only for help in finding all integer solutions.

$\endgroup$
1
  • $\begingroup$ As with the other answer, I'm sorry that I only seem to be able to award the bounty to one answer, and I could understand the last one the easiest, so I gave it there. But thank you for solving it :) $\endgroup$
    – user387394
    Apr 16 at 13:49
2
+100
$\begingroup$

$2x^4+2x^2y^2+y^4=z^2 \text{ for integers } x, y, z$

Quite long solution, and lots to edit…


Case 1. $x=0$:
$y^4=z^2, (x, y, z) = (0, t, \pm t^2)$ is a solution.


Case 2. $x\ne0, y=0:$
$2x^4=z^2$, No solution.


Case 3. $x\ne0, y\ne0:$
$z\ne0.$

\begin{align} &\text{let } (x, y, z) = d. \\ \Rightarrow \; & x=dx_1, y=dy_1, z=dz_1. \\ \Rightarrow \; & 2d^4{x_1}^4+2d^4{x_1}^2{y_1}^2+d^4{y_1}^4=d^2{z_1}^2. \\ \Rightarrow \; & d^4|d^2{z_1}^2, d^2|{z_1}^2, d|z_1. \\ \therefore \; & 2\left(\frac x d\right)^4 + 2\left(\frac x d\right)^2\left(\frac y d \right)^2+\left(\frac y d\right)^4=\left(\frac z {d^2}\right)^2. \\ \therefore \; & \left(\frac x d, \frac y d, \frac z {d^2}\right) \text{ is also a solution.} \\ \ \\ \Rightarrow \; & \text{let } (x, y, z) \text{ has a smallest value of } x.\\ \Rightarrow \; & (x, y, z) = 1. \\ & (x^2)^2+(x^2+y^2)^2=z^2 \\ & (x^2, x^2+y^2, z): \text{ Pythagorean Triple.} \\ \Rightarrow \; & \begin{cases} i. & x^2=m^2-n^2, x^2+y^2=2mn, z=m^2+n^2.\\ ii. & x^2=2mn, x^2+y^2=m^2-n^2, z=m^2+n^2.\\ \end{cases} \\ & \left((m, n) = 1, \text{not } 2\not|m \text{ and } 2\not|n\text{, by Pythagorean Triple.}\right)\\ \ \\ i. \; & x^2=m^2-n^2, x^2+y^2=2mn. \\ \Rightarrow \; & x^2\equiv 1 (\mod 4), 2\not|m, 2|n. \\ \therefore \; & x^2+y^2 = 2mn \equiv 0 (\mod 4), 2|x, 2|y, \text{Contradiction.} \\ \ \\ ii. \; & x^2=2mn, x^2+y^2=m^2-n^2. \\ \Rightarrow \; & 2|x, 2\not|y. \\ \therefore \; & m^2-n^2=x^2+y^2\equiv 1 (\mod 4). \\ \Rightarrow \; & 2\not|m, 2|n. \\ & (m, n) = 1 \Rightarrow m = s^2, n = 2t^2. \\ \therefore \; & 4s^2t^2+y^2=s^4-4t^4. \\ & s^4-4s^2t^2-4t^4=y^2. \\ \Rightarrow \; & (s^2-2t^2)^2-y^2=8t^4, 8t^4=(s^2-2t^2-y)(s^2-2t^2+y). \\ &s^2-2t^2\pm y \equiv 1-0\pm1 \equiv 0 (\mod 2). \\ \ \\ &\text{let } (s^2-2t^2+y, s^2-2t^2-y)=d. \\ \Rightarrow \; & 2|d, d|(2s^2-4t^2). \\ & d^2|(s^2-2t^2+y)(s^2-2t^2-y)=8t^4 \Rightarrow d|4t^2 \\ \therefore \; & d|4t^2 \text{ and } d|2s^2, (t, s)=1. \\ \therefore \; & d=2.(\because 2\not|s.) \\ \ \\ \text{let} \; & t=2^{\alpha}l, 2\not|l \in \Bbb{Z}. \\ \Rightarrow \; & (s^2-2t^2+y)(s^2-2t^2-y)=2^{4\alpha+3}l^4 \\ \text{let}\; & s^2-2t^2+y=2p, s^2-2t^2-y=2q. \\ \Rightarrow\; & pq=2^{4\alpha+1}l^4. \\ & (p, q)=1, \begin{cases} p=2^{4\alpha+1}u^4, q=v^4 \\ \text{or} \\ p=v^4, q=2^{4\alpha+1}u^4 \end{cases} (\gcd(u, v)=1, uv=l) \\ &p+q=s^2-2(2^{\alpha}u)^2v^2=2(2^{\alpha}u)^4+v^4. \\ & 2(2^{\alpha}u)^4+2(2^{\alpha}u)^2v^2+v^4=s^2 \\ \Rightarrow \; & 2^{\alpha}u \leqslant 2^{\alpha}uv = 2^{\alpha}l=t<2st=x \\ &\text{Contradiction of minimum of $x$.} \\ \ \\ & \therefore \text{ Contradiction, No Solution for Case 3.} \end{align}



$\therefore (0, t, \pm t^2)$ is the only solution.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you! That makes perfect sense. But, after seeing the solution, I don't feel bad that I missed it, it's pretty involved. $\endgroup$
    – user387394
    Apr 16 at 13:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .