How to find all solutions to
$$ a^2+b^2+c^2+d^2=e^2+2$$
where all variables $a$ to $e$ are positive integers and $e^2 \equiv 1 \mod 8$
I tried using parameterization similar to pythagoras equation, but no success so far. Any help will be appreciated.
Thanks!
On
For such equations, you can use the standard approach. One approach is to use equations Pell. For the beginning will talk about a more simple way.
I. Eq.1
$$a^2+b^2+c^2+d^2=(2q+1)^2+2$$
Solutions have the form:
$$a=x + k$$
$$b=x + p$$
$$c=x + s$$
$$d=x - z+1$$
$$q=x$$
where,
$$x = -1 -z+ (z^2 - k p - k s - p s),\quad z = k+p+s$$
for any $k,p,s$.
II. Eq.2
$$a^2+b^2+c^2+d^2=(4q+1)^2+2$$
Solutions have the form:
$$a=2(y + k)$$
$$b=2(y + p)+1$$
$$c=2(y + s)+1$$
$$d=2(y - z)-1$$
$$q=y$$
where,
$$y = -k + 2z + 2(z^2 - k p - k s - p s),\quad z = k+p+s$$
for any $k,p,s$.
III. Eq.3
$$a^2+b^2+c^2+d^2=(8q+1)^2+2$$
Solutions have the form:
$$a=16k^2+4p^2+4s^2+8pk+8ks+4ps+2s+2p-1$$
$$b=16k^2+4p^2+4s^2+8pk+8ks+4ps+4s+4p+8k$$
$$c=16k^2+4p^2+4s^2+8pk+8ks+4ps+4s+6p+4k+1$$
$$d=16k^2+4p^2+4s^2+8pk+8ks+4ps+6s+4p+4k+1$$
$$q=4k^2+p^2+s^2+2pk+2sk+ps+s+p+k$$
$k,p,s$ - integers asked us.
On
Now will show you a different approach. For,
$$a^2+b^2+c^2+d^2=(2q+1)^2+2$$
One can use the Pell equation,
$$x^2-ry^2=1$$
where,
$$r =k^2+(t^2+p^2+s^2)+(t+p+s)^2$$
for any $k,t,p,s$. Let,
$$u = x-2(t+p+s)y,\quad w = x-(t+p+s)y$$
Then solutions are,
$$a=2\, k\, u\, y$$
$$b=q - 2\, t\, u\, y + 1$$
$$c=q - 2\, p\, u\, y + 1$$
$$d=q - 2\, s\, u\, y + 1$$
$$q=2\,u\,w$$
And another solution:
$$a=2k(2(t+p+s)y-x)y$$
$$b=2y((p+s)x+(2(t^2+2(p+s)t+ps)-k^2)y)+1$$
$$c=2y((t+s)x+(2(p^2+2(t+s)p+ts)-k^2)y)+1$$
$$d=2y((p+t)x+(2(s^2+2(p+t)s+pt)-k^2)y)+1$$
$$q=2y((t+p+s)x+(2(tp+ts+ps)-k^2)y)$$
Be aware that if the ratio of the Pell equation $r$ - fold the square, can be reduced. In the formula, too, should be reduced.
Meanwhile, there certainly is an answer of sorts using stereographic projection. It has the virtue of seeming to be a parametrization in four new letters along with $e.$ Weakness is that solution quadruples may not always be primitive. Nope, stereographic needs homogeneous, does not apply here.
I do not see enough of a pattern to suggest a recipe for all solutions. $(0,1,1,e)$ always works, furthermore one number is divisible by 4 and the other three are odd, that is about it.