I tried working out a solution to satisfy this equation and I got that this has no solution, however:

$$1^2 + 2^2 + 2^2 = 3^2$$

so it does have a solution.

I started off with the equation:

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

Therefore $a^2 + b^2 = d^2 - c^2$ and therefore:

$$(a + b)^2 - 2ab = (d + c)^2 - 2ab - 2c^2$$

which simplifies to $(a + b)^2 = (d + c)^2 - 2c^2$. This means that $(a + b)^2 + 2c^2 = (d + c)^2$ and therefore:

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

Therefore: $a^2 + b^2 + 2ab + c^2 = d^2 + 2dc$ which means:

$$a^2 + b^2 + c^2 = d^2 + 2dc - 2ab$$

However, we just established that $a^2 + b^2 + c^2 = d^2$ so this means that $2dc = 2ab$. So now we have three options of what $a$, $b$, $c$, or $d$ could equal.

  1. $a = b = c = d = 0$

  2. $a = d$ and $b = c$

  3. $a = c$ and $b = d$

But when we test it, we see that our only option is $a = b = c = d = 0$ however we had a solution for this equation where $(a, b, c, d) > 0$ to satisfy this equation.

Could somebody please tell me where I am wrong, because I obviously did something wrong here but I can't identify where I made a mistake; there is probably something very obvious that I missed. I would appreciate if you showed me the steps to finding a solution to this equation.

EDIT: Yeah I think I should learn the basics first before I allow my curious mind to wander off onto other things.

  • $\begingroup$ I would have put (squared-numbers) or (perfect-squares) in my tags list but I need at least 1000 reputation to put it there. $\endgroup$ – George N. Missailidis Jul 24 '17 at 2:46
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    $\begingroup$ Hint: substitute $a=1,b=c=2,d=3$ into each of your equations and see which is the first one that is wrong. $\endgroup$ – Robert Israel Jul 24 '17 at 2:49
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    $\begingroup$ are we talking about integers only here? pls clarify that in your question $\endgroup$ – Agile_Eagle Jul 24 '17 at 2:50
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    $\begingroup$ the third (centered) equation is wrong $\endgroup$ – fonfonx Jul 24 '17 at 2:58
  • $\begingroup$ ... and there are additional mistakes in the subsequent "simplifications". $\endgroup$ – NickD Jul 24 '17 at 3:21

$$ \left( a^2 + b^2 - c^2 - d^2 \right)^2 + (-2ad+2bc)^2 +(2ac+2bd)^2 = \left( a^2 + b^2 + c^2 + d^2 \right)^2 $$

gives all possible solutions with gcd one, where we need $\gcd(a,b,c,d) = 1$ and $a+b+c+d$ odd.

You are welcome to take absolute values if any of the terms are negative.

This was surely known to Euler, and is the subject of a later article by V. A. Lebesgue about 1868. The first correct proof that this gives ALL coprime solutions was given by L. E. Dickson about 1920.

Alright, reading volume II of Dickson's History, I understand now. Euler (1748) gave the formula for taking the product of two sums of four squares and writing that as the sum of four squares. Later, this was written as multiplication of quaternions; quaternions were introduced by Hamilton in 1843, before Lebesgue. This is on Dickson page 277. Back on page 265, V. A. Lebesgue (1868) is discussed, and Dickson makes a point of showing how specific choices in Euler's formula leads to Lebesgue's version. Another 50 years and Dickson proved all solutions work this way. So, Wikipedia calls this thing Lebesgue's formula, which may or may not be historically valid.

A nice modern treatment, with clear proof, is in Spira (1962). Spira also gives a proof of a bunch of inequalities that give the solutions as positive numbers, once and only once for each quadruple (Steiger 1956). Alright, you can read Spira online for free.


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