How can I generate solutions to the Diophantine equation: $a^2 + b^2 + 2c^2 = d^2$

Question

Is there a way to reduce or quickly find integer solutions to the equation $a^2 + b^2 + 2c^2 = d^2$ (where $a, b, c$ and $d$ are distinct natural numbers) ?

Sorry I’m really bad at Diophantine equations. Thanks for any help.

Answer 1

$b^2 + 2 c^2 = d^2 - a^2 = (d-a)(d+a)$ is the product of two numbers which are either both even or both odd. Take any odd $b$ and any $c$, or even $b$ and even $c$, and you should get some solutions (not necessarily with all values distinct, but usually). For example, $b = 1$ and $c = 2$ makes $b^2 + 2 c^2 = 9 = 1 \cdot 9$ (and $d-a = 1$, $d+a = 9$ gives you $d = 5$, $a = 4$, or $3 \cdot 3$ (and then $d-a = d+a = 3$ means $d = 3$, $a=0$). Of $b = 2$ and $c=4$ makes $b^2 + 2 c^2 = 36 = 2 \cdot 18 = 6 \cdot 6$ so $d=10$, $a = 8$ or $d = 6$, $a=0$.

Answer 2

It is a homogenous equation, so it is sufficient to find rational solutions of the following equation :

\begin{equation} x^2+y^2+2z^2=1 \end{equation} where we have used $x\equiv\frac{a}{d},~y\equiv\frac{b}{d}$ and $z\equiv\frac{c}{d}$.

It is an squared equation, so we are trying to find a special solution to the above equation and then use the old trick in Diophantine equations. The solution is $(x,y,z)=(1,0,0)$. We pass the line with rational slope $(t_0,t_1,t_2)$ through $(x,y,z)$ which is as follows : \begin{equation} \frac{x-1}{t_0}=\frac{y}{t_1}=\frac{z}{t_2} \end{equation}

so we have $y=\frac{t_1}{t_0}(x-1)\equiv\alpha (x-1)$ and $z=\frac{t_2}{t_0}(x-1)\equiv\beta(x-1)$, we have the following formula :

\begin{equation} x^2(1+\alpha^2+2\beta^2)-2x(\alpha^2+2\beta^2)+\alpha^2+2\beta^2-1=0 \end{equation} where the solutions are $x=1$ and $x=\frac{\alpha^2+2\beta^2-1}{1+\alpha^2+2\beta^2}$, So you have the generic solutions as follows :

\begin{equation} x=\frac{\alpha^2+2\beta^2-1}{1+\alpha^2+2\beta^2},~~~~~y=\frac{-2\alpha}{1+\alpha^2+2\beta^2},~~~~~z=\frac{-2\beta}{1+\alpha^2+2\beta^2} \end{equation}

for example, by choosing $\alpha=2$ and $\beta=1$, you have $x=\frac{5}{7}$, $y=\frac{-4}{7}$ and $z=\frac{-2}{7}$. In the original natural numbers, you have $a=5,~b=4,~c=2,~d=7$

Answer 3

Notice that if $a_0, b_0, c_0$ and $d_0$ is a solution to the equation then also $$ a_0t, b_0t, c_0t, d_0t$$ is a solution for every integer $t$. So you have to find only thi one particular soution and you will get infinitely solutions. In particulary, we see that $a_0=1 =b_0, c_0=7$ and $d_0=10$ works.

Also $a_0=b_0 =c_0=1$ and $d_0=2$ works.