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

  • $\begingroup$ Tested for $n\le 4000$? If somebody claimed "For a positive integer $y$, $61y^2+1$ is never a square of an integer, tested for $y<226153980$", that would still be wrong, because $61\cdot226153980^2+1=1766319049^2$ $\endgroup$ – Professor Vector Aug 19 '17 at 12:11
  • $\begingroup$ You should add the positivity condition to your question, it may get overlooked in the comments. $\endgroup$ – Professor Vector Aug 19 '17 at 13:22
  • 2
    $\begingroup$ The fact that there are exceptions rules out an identity as proof. However, just an observation, the simple identity $$(n - 1)^2 + n^2 + (n^2 - n)^2 = (n^2 - n + 1)^2$$ shows that $a+b+c+d = 2n^2$, hence every twice a square can be so expressed. $\endgroup$ – Tito Piezas III Aug 21 '17 at 8:26
  • $\begingroup$ @TitoPiezasIII Identities are not entirely ruled out. There might be one that produces negative values for one or more of $a,b,c,d$ in those exceptional cases. $\endgroup$ – Jaap Scherphuis Aug 21 '17 at 8:40
  • 2
    $\begingroup$ @TitoPiezasIII Furthermore, twice every non-squarefree number can be expressed (via scaling), $\endgroup$ – Carl Schildkraut Aug 21 '17 at 19:25

If you rewrite your sum of squares as


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


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 $u^2+v^2$.

  • $\begingroup$ The OP isn't looking for any solution of $a^2+b^2+c^2=d^2,$ only for those where $a+b+c+d$ is a given even number $n$. $\endgroup$ – Professor Vector Aug 19 '17 at 13:38
  • $\begingroup$ @ProfessorVector: I admit my answer could be better worded and arranged, but I did write "To fit your constraint above $\left( \frac{b+1}{2}\right)^2$ is even and $\left( \frac{b-1}{2}\right)^2$ is odd" to acknowledge this requirement. $\endgroup$ – James Arathoon Aug 19 '17 at 13:49
  • $\begingroup$ How would that give a representation of, say, $n=36$ as $a+b+c+d$ satisfying $a^2+b^2+c^2=d^2$? $\endgroup$ – Professor Vector Aug 19 '17 at 13:56
  • $\begingroup$ @ProfessorVector: I realise now that the constraint above is not correct. all that is required is that $a+b+c+d$ is even, $\left( \frac{b+1}{2}\right) $ and $\left( \frac{b-1}{2}\right)$ can be odd and even either way around as long as $a_1$ and $a_2$ are an odd-even pair. $\endgroup$ – James Arathoon Aug 20 '17 at 12:15
  • $\begingroup$ @ProfessorVector: in regard to your question there are two answers to $n=36$; $14^2-12^2=6^2+4^2$ and $15^2-5^2=14^2+2^2$, (a) and (b) respectively. (a) is simply a multiple of $7^2-6^2=13=3^2+2^2$, however (b) factorizes as $5^2(3^2-1^2)=2^2(7^2+1^2)$ with $5^2=\frac{1}{2}(7^2+1^2)$ and $3^2-1^2=2^3$. $\endgroup$ – James Arathoon Aug 20 '17 at 12:24


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



Lay on multipliers and pick the right.

Or other item.







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

  • $\begingroup$ How can you guarantee $c>0$? $\endgroup$ – Professor Vector Aug 19 '17 at 13:31
  • $\begingroup$ No, it remains to prove that a choice with $p>k$ $p^2+k^2-2s^2>0$ is possible. And don't forget to tell me if you succeed, because I'd revoke my downvote, then, naturally. $\endgroup$ – Professor Vector Aug 19 '17 at 21:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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