Most even numbers is a sum $a+b+c+d$ where $a^2+b^2+c^2=d^2$

I have no idea how hard this conjecture is to prove:

Any even number $n\ge 36$ can be written as $n=a+b+c+d$ where $a^2+b^2+c^2=d^2$ and $a,b,c,d\in\mathbb Z^+$.

Small exceptions are $n=2, 4, 6, 10, 12, 14, 20, 26, 34.\,$ Tested for $n\le 10,000$.

A counter-example would be as interesting as a proof.


$n$ as above must be even
$d-c,c+d|a^2+b^2$

Solutions Collecting From Web of "Most even numbers is a sum $a+b+c+d$ where $a^2+b^2+c^2=d^2$"

If you rewrite your sum of squares as

$$a_4^2-a_3^2=a_1^2+a_2^2$$

Then all you need to do is find a difference of two squares that equals a sum of two squares.

If the sum of two squares is an odd number $b$ with the form $2k+1$ that is

$$a_1^2+a_2^2=b=2k+1$$

Then since any odd number can be trivially written as the difference of two squares we have

$$\left( \frac{b+1}{2}\right)^2-\left( \frac{b-1}{2}\right)^2=b$$

which immediately gives

$$\left( \frac{b+1}{2}\right)^2-\left( \frac{b-1}{2}\right)^2=a_1^2+a_2^2$$

To fit your constraint above $\left( \frac{b+1}{2}\right)^2$ is even and $\left( \frac{b-1}{2}\right)^2$ is odd.

Most sums of two squares have the form $4k+1$ as you will see by adding these in turn
$$(4k_1+1)^2+(4k_2)^2$$
$$(4k_1+1)^2+(4k_2+1)^2$$
$$(4k_1+1)^2+(4k_2+2)^2$$
$$(4k_1+1)^2+(4k_2+3)^2$$
$$(4k_1+2)^2+(4k_2)^2$$
$$(4k_1+2)^2+(4k_2+1)^2$$
$$(4k_1+2)^2+(4k_2+2)^2$$
$$(4k_1+2)^2+(4k_2+3)^2$$
$$(4k_1+3)^2+(4k_2)^2$$
$$(4k_1+3)^2+(4k_2+3)^2$$

Find the results $(\mod 4)$ and see if you can find other sums of the form $a_4^2-a_3^2=a_1^2+a_2^2$

Note Added to help explain why this approach does not work:

I thought the above might help lead to a proof/disproof of the conjecture. Hopefully the comments below will help clarify why this approach does not work.

if $a_1=\left( \frac{u-v}{2}\right)$, $a_2=\left( \frac{u+v}{2}\right)$, $a_3=\left( \frac{w-x}{2}\right)$ and $a_4=\left( \frac{w+x}{2}\right)$ then

$$n=a_1+a_2+a_3+a_4$$
$$n=\left( \frac{u-v}{2}\right)+\left( \frac{u+v}{2}\right)+\left( \frac{w-x}{2}\right)+\left( \frac{w+x}{2}\right)$$
$$n=u+w \tag 1$$
and

$$a_4^2=a_1^2+a_2^2+a_3^2$$
$$\left( \frac{w+x}{2}\right)^2=\left( \frac{u+v}{2}\right)^2+\left( \frac{u-v}{2}\right)^2+\left( \frac{w-x}{2}\right)^2$$

$$wx= \left( \frac{u+v}{2}\right)^2+\left( \frac{u-v}{2}\right)^2$$
or
$$2wx=u^2+v^2$$

Substituting for $w$ using (1) gives
$$ n=u+\frac{u^2+v^2}{2x}$$

$2x$ being a factor of $u^2+v^2$.

This result I think shines a little light on why the problem of finding $n$ is not possible using this approach; that is the difficulty in finding a general solution to the factorization of $u^2+v^2$.

$$a^2+b^2+c^2=d^2$$

$$a=2ps$$
$$b=2ks$$
$$c=s^2-p^2-k^2$$
$$d=s^2+p^2+k^2$$

$$2n=2ps+2ks+s^2-p^2-k^2+s^2+p^2+k^2$$

$$qt=n=s(p+k+s)$$

Lay on multipliers and pick the right.

Or other item.

$$a=2s(p-k)$$

$$b=2s(p+k)$$

$$c=p^2+k^2-2s^2$$

$$d=p^2+k^2+2s^2$$

$$2n=2p^2+2k^2+4ps$$

$$p(p+2s)=n-k^2$$

It remains to try all possible $k$ . That expression was greater than $0$.