1
$\begingroup$

Solve the following equations in postivie 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!!!

$\endgroup$
  • $\begingroup$ I don't see how there can be any non-trivial solution. If you subtract the two equations, you get 2y^2 = -4z^2. There can be no solution without making y or z imaginary, so only x=y=z=0 remains. $\endgroup$ – Aganju Jan 21 '16 at 1:04
  • 3
    $\begingroup$ @Aganju I expect that this is two separate questions bundled together, not a simultaneous system. $\endgroup$ – T. Bongers Jan 21 '16 at 1:05
  • $\begingroup$ Sorry for the confusion, I edited it $\endgroup$ – HeatTheIce Jan 21 '16 at 1:05
  • $\begingroup$ The complete primitive solution to $x^2+dy^2= z^2$ is given by $(p^2-dq^2)^2+d(2pq)^2 = (p^2+dq^2)^2$. However, the one for $x^2+y^2 = dz^2$ needs an initial solution. $\endgroup$ – Tito Piezas III Jan 21 '16 at 1:17
  • 1
    $\begingroup$ @WillJagy: Ah, I stand corrected. The complete rational parameterization is $(p^2-dq^2)^2r^2+d(2pqr)^2=(p^2+dq^2)^2r^2$. I should not have dismissed the scaling factor $r$. $\endgroup$ – Tito Piezas III Jan 21 '16 at 4:16
3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ I like the solution of yours. It is indeed one way to generate solutions to the equation I gave, but I should also have a proof that those are maybe the only solutions to it, bcs I am interested in all solutions $\endgroup$ – HeatTheIce Jan 22 '16 at 13:56
  • $\begingroup$ @HeatTheIce: I'm not completely sure, but I assume the three degrees of freedom $z_1, z_2, z_3$ are enough to cover all solutions $x,y,z$ if one takes into account sign changes and the permutations $x,y$ and $y,x$. I edited to give an example. $\endgroup$ – Tito Piezas III Jan 22 '16 at 14:30
2
$\begingroup$

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. $$ (2u^2 - 2 u v - 2 v^2)^2 + (u^2 + 4 uv - v^2)^2 = 5 (u^2 + v^2)^2 $$

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.