Is there any solutions to the diophantine equation $x^2 + y^2 = 2z^3, x \neq y$?

Question

So far, I have been unable to find any solutions to the diophantine $ x^2 + y^2 = 2z^3, x \neq y $, and I believe there to be none for natural x,y,z, but I cannot find any ways to prove this.

Answer 1

So in the interest of promoting a thinking process for this sort of problem, here's how I went about finding a solution:

"Well, that factor of two on the right is a little awkward. Can I get rid of it? Well, if I take $x=4x'$ and $y=4y'$, then $x^2+y^2=16(x'^2+y'^2)=2\cdot8(x'^2+y'^2)$, so if I take $z=2z'$ then we have $x'^2+y'^2=z'^3$. This won't be all the solutions but it might be some!"

"Lots of numbers are sums of squares. What're the obstructions to a number being a sum of squares? Well, it can't be $3\bmod 4$. And of course no cubes are $\equiv 2\bmod 4$. So it's either $0$ — and I want to be done with factors of $2$ — or it's $1\bmod 4$. A cube is $1\bmod 4$ iff the number itself is. So what's the first 'medium' number that's $1\bmod 4$? Well, let's try $5$..."

"$5^3=125$ and that has a pretty obvious dissection as $10^2+5^2$. So if I take $x=4\cdot 10=40$ and $y=4\cdot 5=20$, then $x^2+y^2=40^2+20^2=2000=2(10)^3$."

Answer 2

There are a ton of solutions.

Suppose $x^2 + y^2 = 2z^3$ where $(x,y) = 1$. (I am introducing the condition $(x,y) = 1$ in order to relate this equation to the same equation without the coefficient $2$ on the right side.) Then $x$ and $y$ are odd. Factor the left side in the Gaussian integers: $$ (x+yi)(x-yi) = 2z^3 = (1+i)(1-i)z^3. $$ Since $x$ and $y$ are odd, $x+yi$ is divisible by $1+i$ in the Gaussian integers: $x+yi = (1+i)(a+bi)$ for some integers $a$ and $b$. Then $a$ and $b$ are relatively prime (any common factor of them is a common factor of $x$ and $y$), and since $x = a+b$ and $y = a-b$, $a \not\equiv b \bmod 2$ (otherwise $x$ and $y$ would be even). Conjugating the equation $x+yi = (1+i)(a+bi)$ gives us $x-yi = (1-i)(a-bi)$, so $$ (1+i)(a+bi)(1-i)(a-bi) = (1+i)(1-i)z^3 \Longrightarrow a^2 + b^2 = z^3. $$ Conversely, if $a^2 + b^2 = z^3$ where $a$ and $b$ are relatively prime with $a \not\equiv b \bmod 2$, define $x$ and $y$ by $x+yi = (1+i)(a+bi)$, so $x = a - b$ and $y = a + b$. Then $x$ and $y$ are odd and $$ x^2 + y^2 = (x+yi)(x-yi) = (1+i)(a+bi)(1-i)(a-bi) = 2(a^2 + b^2) = 2z^3. $$ In summary, the integral solutions of $x^2 + y^2 = 2z^3$ where $(x,y) = 1$ are in bijection with the integral solutions of $a^2 + b^2 = z^3$ where $(a,b) = 1$ and $a \not\equiv b \bmod 2$, by the relation $x + yi = (1+i)(a+bi)$.

A parametric description of all integral solutions of $a^2 + b^2 = z^3$ where $(a,b) = 1$ is in Theorem 8.4 here. The parametric description is $a = m^3 - 3mn^2$ and $b = 3m^2n - n^3$ where $(m,n) = 1$ and $m \not\equiv n \bmod 2$. Then $m+n$ is odd, so $$ a \equiv m^3 + mn^2 \equiv m(m^2 + n^2) \equiv m(m+n)^2 \equiv m \bmod 2 $$ and $$ b \equiv 3m^2n - n^3 \equiv n(m^2 + n^2) \equiv n(m+n)^2 \equiv n \bmod 2, $$ so $a \not\equiv b \bmod 2$. That is, the condition $a \not\equiv b \bmod 2$ is a consequence of $(a,b) = 1$.

The solutions in the comments, $(x,y) = (13,9)$ and $(55,37)$, occur when $(a,b)$ is $(11,-2)$ and $(46,-9)$, and these come from the parameter values $(m,n) = (-1,2)$ and $(-2,-3)$.

Answer 3

There are actually lots of solutions. To get a whole two-parameter family of solutions all you have to do is pick a value of $z$ that's a sum of two squares, call this $m^2+n^2$. We then have, using the identity $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$:

$z^2=(m^2+n^2)(m^2+n^2)=(m^2-n^2)^2+(2mn)^2$

$z^3=[(m^2-n^2)^2+(2mn)^2](m^2+n^2)=(m^3-3mn^2)^2+(3m^2n-n^3)^2$

$2z^3=[(m^3-3mn^2)^2+(3m^2n-n^3)^2](1^2+1^2)=(m^3-3mn^2-3m^2n+n^3)^2+(m^3-3mn^2+3m^2n-n^3)^2$

So

$x=|m^3-3mn^2-3m^2n+n^3|$

$y=|m^3-3mn^2+3m^2n-n^3|$

$z=m^2+n^2$

will give a solution. For instance, $m=2, n=1$ gives $x=9,y=13,z=5$ (these being relatively prime; compare with what is below), corresponding to the sum $81+169=250$.

We can also obtain a second,simpler family of two-parameter solutions. Start again with $z=m^2+n^2$ and this time multiply both sides directly by $z^2=(m^2+n^2)^2$:

$z^3=(m^3+mn^2)^2+(m^2n+n^3)^2$

and then

$2z^2=[(m^3+mn^2)^2+(m^2n+n^3)^2](1^2+1^2)=(n^2+mn^2-m^2n-n^3)^2+(n^2+mn^2+m^2n+n^3)$

giving the solution set

$x=|n^2+mn^2-m^2n-n^3|=|m-n|(m^2+n^2)$

$y=|n^2+mn^2+m^2n+n^3|=|m+n|(m^2+n^2)$

$z=m^2+n^2$

With this family of solutions $m=2,n=1$ gives $x=5, y=15, z=5$ corresponding to $25+225=250$. Note that this family does not give solutions where all terms are relatively prime (except for $x=y=z=1$), whereas the first family can do so.

Answer 4

We begin with Euclid's formula, referring to $\space (A,B,C)\space$ instead of $\space (x,y,z)\space$ shown here as $$ A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2=\big(2(z)^3\big)^2$$ We then solve the $\space C$-function for $\space k\space$ and, given a candidate $\space C$-value, we test to see if any $\space m$-values in a defined range yield an integer.

Values that work? $\space z\in\{5,10,13,17,20,25,26,29,37,40,41,45,50,\cdots\}$ Here is the procedure.

$$C=m^2+k^2\implies k=\sqrt{C-m^2}\\ \qquad\text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor$$ The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$. $$C=\big(2(5)^3\big)^2=62500\implies\\ \bigg\lfloor\frac{ 1+\sqrt{125000-1}}{2}\bigg\rfloor=177 \le m \le \lfloor\sqrt{62500-1}\rfloor=249\quad \\ \text{and we find} \quad m\in\{200,234,240\}\ \implies k\in\{150,80,70\}\\$$ $$ f(200,150)=\bigg(17500,60000,62500=\big(2(5)^3\big)^2\bigg)\\ f(234,88)=\bigg(47012,41184,62500=\big(2(5)^3\big)^2\bigg)\\ f(240,70)=\bigg(52700,33600,62500=\big(2(5)^3\big)^2\bigg)\\ $$