Solve in positive integers $x^2+3y^2=z^2$ and $x^2+y^2=5z^2$

Question

Solve the following equations in postive integers:

1. $x^2+3y^2=z^2$

2. $x^2+y^2=5z^2$

I have solved them using this as a reference, but I am interested in other solutions, more "elegant" ones. The equations are to be solved separate!!!

Answer 1

People are not careful with these. With $x^2 + 3 y^2 = z^2;$ with odd $z,$ we indeed get $$ (r^2 - 3 s^2)^2 + 3(2rs)^2 = (r^2 + 3 s^2)^2. $$ This does not give primitive solutions with even $z,$ which come from $$ \left( \frac{r^2 - 3 s^2}{2} \right)^2 + 3 (rs)^2 = \left( \frac{r^2 + 3 s^2}{2} \right)^2 $$ with both $r,s$ odd

For $x^2 + y^2 = 5 z^2$ primitive means $z$ odd, we take $x$ to be the even one. $$ (2u^2 - 2 u v - 2 v^2)^2 + (u^2 + 4 uv - v^2)^2 = 5 (u^2 + v^2)^2 $$

Answer 2

In general, given one solution to,

$$a_1y_1^2+a_2y_2^2+\dots+ a_ny_n^2 = 0$$

then an infinite more can be found without scaling. For example,

$n=3$

$$ax_1^2+bx_2^2+cx_3^2 = (ay_1^2+by_2^2+cy_3^2)(az_1^2+bz_2^2+cz_3^2)^2$$

where,

$$x_1, x_2, x_3 = uy_1-vz_1,\; uy_2-vz_2,\; uy_3-vz_3\\ \text{and}\\ u,v = az_1^2+bz_2^2+cz_3^2,\; 2(ay_1z_1+by_2z_2+cy_3z_3)$$

If you have an initial solution $ay_1^2+by_2^2+cy_3^2 = 0$, then the identity gives you an infinite more with three free variables $z_1, z_2, z_3$.

$n=4$

$$ax_1^2+bx_2^2+cx_3^2+dx_4^2 = (ay_1^2+by_2^2+cy_3^2+dy_4^2)(az_1^2+bz_2^2+cz_3^2+dz_4^2)^2$$

where,

$$x_1, x_2, x_3, x_4 = uy_1-vz_1,\; uy_2-vz_2,\; uy_3-vz_3,\; uy_4-vz_4\\ \text{and}\\ u,v = az_1^2+bz_2^2+cz_3^2+dz_4^2,\; 2(ay_1z_1+by_2z_2+cy_3z_3+dy_4z_4)$$

Likewise, you now have four free variables $z_1, z_2, z_3, z_4$.

And so on for any $n$. I trust the pattern is easy to see?

Example:

For $x^2+y^2 = 5z^2$, we have $a,b,c = 1,1,-5$, and initial solution $y_1, y_2, y_3 = 1,2,1$. Using the formula, we get,

$$x = z_1^2 + z_2^2 - 5 z_3^2 - 2 z_1 (z_1 + 2 z_2 - 5 z_3)\\ y = 2 ( z_1^2 + z_2^2 - 5 z_3^2) - 2 z_2(z_1 + 2 z_2 - 5 z_3)\\ z = z_1^2 + z_2^2 - 5 z_3^2 - 2 z_3(z_1 + 2 z_2 - 5 z_3)$$

for three free variables $z_i$. Let, $z_2 = u + 2 v + 2 z_1,\; z_3 = v + z_1$, then $z_1$ cancels out and we recover W. Jagy's version in the other answer.

Answer 3

There appear to be infinite solutions to these equations but only two patterns jump out for both of them.

$\text{For equation 1. }$ are $x=z, y=0$ because no integer $y$ yields $3y^2$ is a perfect square. \begin{equation} y=\sqrt{\frac{z^2-x^2}{3}}\quad \text{for}\quad x,z\in\mathbb{N}\land z\ge x \end{equation} some solutions are $$(x,0,x)\qquad (x,x,2x)$$

other solutions do not show a simple pattern

\begin{equation} (1,4,7)\quad (1,15,26)\quad (2,8,14)\quad (2,30,52)\quad (3,12,21)\quad (4,16,28)\quad (6,24,42)\quad (7,28,49)\quad (7,4,1)\quad (8,32,56)\quad (11,21,38)\quad (11,24,43)\quad (11,4,13)\quad (11,5,14)\quad (13,20,37)\quad (13,3,14)\quad (13,8,19)\quad (22,10,28)\quad (22,8,26)\quad (23,12,31)\quad (23,7,26)\quad (26,16,38)\quad (26,6,28)\quad (33,12,39)\quad (33,15,42)\quad (37,5,38)\quad (39,24,57)\quad (39,9,42)\quad (44,16,52)\quad (44,20,56)\quad (46,14,52)\quad (52,12,56)\quad \end{equation}

$\\ \text{For equation 2.}$ $$z = \sqrt{\frac{x^2 + y^2}{5}}\quad\text{for}\quad x\in\mathbb{N}$$ some solutions are \begin{equation} (x,2x,x)\qquad \bigg(x,\frac{11x}{2},\frac{5x}{2}\bigg) \end{equation}

other solutions do not show a simple pattern \begin{equation} (1,38,17)\quad (2,29,13)\quad (2,76,34)\quad (4,58,26)\quad (6,87,39)\quad (11,82,37)\quad (19,62,29)\quad (22,19,13)\quad (22,31,17)\quad (22,61,29)\quad (22,89,41)\quad (31,22,17)\quad (38,41,25)\quad (41,38,25)\quad (44,38,26)\quad (44,62,34)\quad (55,10,25)\quad (57,66,39)\quad (58,4,26)\quad (58,59,37)\quad (58,71,41)\quad \end{equation}