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

Solution 1:

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 $u^2+v^2$.