If $n$ then $n^2$

Let $a \leq b \leq c$ . Then if $d = b^2 + c^2 - a^2$ , $e = 2ab$ , and $f = 2ac$ , then

$n^2$

$= (a^2 + b^2 + c^2)^2$

$= a^4 + b^4 + c^4 + 2a^2b^2 + 2a^2c^2 + 2b^2c^2$

$= (a^4 + b^4 + c^4 - 2a^2b^2 - 2a^2c^2 + 2b^2c^2) + 4a^2b^2 + 4a^2c^2$

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

$= d^2 + e^2 + f^2$

which makes the statement always true .

A similar solution to @David Vreken

Let $n=a^2+b^2+c^2$ .

Expand the square and rearrange:

$(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$

$=(a^4+b^4+c^4+2a^2b^2-2a^2c^2-2b^2c^2)+(2ac)^2+(2bc)^2$

$=(a^2+b^2-c^2)^2+(2ac)^2+(2bc)^2=d^2+e^2+f^2 =(a^2+b^2+c^2)^2=n^2.$

We can assume that $a\ge b \ge c > 0 \text { so } a^2 +b^2-c^2 > 0$ .

Thus, we have expressed $n^2$ as the sum of squares of three positive integers.

Remark For the sum of two squares the analogous statement is not true: $(1^2+1^2)^2=4$ which can't be expressed as the sum of two squares of positive integers, although an analogous identity is true:

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

But a similar approach to the one above can be used to prove the sum of four and more squares.

Any natural number that is not of the form $4^x(8y+7)$ is the sum of three squares. The square of the sum of any three squares is never of that form.

Legendre's Three Square Theorem

If all three squares are even, then the square of the sum of squares is divisible by $4$ . Repeatedly dividing the numbers by $2$ will get to the point where at least one of them is odd. Then the remainder when this square of sum of squares is divided by $8$ is either $1$ or $4$ , and never $7$ .

Edit: Antoine G has raised the objection that Legendre's Three Squares Theorem does not guarantee three non-zero squares $d^2 + e^2 + f^2$ , only that it guarantees three squares, any of which could be necessarily zero, which is not observing the condition that the squares be positive. For this reason, this approach needs to be withdrawn until I find out more about this theorem and whether or not it can be made applicable to this problem.

According to Legendre's theorem, n can either be written as $n=x^2+y^2+z^2$ or as $n=4^a(8b+7)$ , but not both (with n,x,y,z,a,b nonnegative integers).

Suppose $n^2$ cannot be written as a sum of three squares, then there are a and b such that $n^2=4^a(8b+7)$ . Then $\frac{n^2}{4^a}=(\frac{n}{2^a})^2=(8b+7)$ so that $8b+7$ must be a perfect square. But that is impossible: squares are only 0,1 or 4 ( mod 8). So any square can be written as the sum of three squares, and the implication symbol ( $\Rightarrow$ ) is justified (though slightly odd).

For even n, the implication symbol $\Rightarrow$ could in fact be changed into an equivalence $\Leftrightarrow$ because of the following: Suppose $n$ cannot be written as a sum of three squares, then there are a and b such that $n=4^a(8b+7)$ and $n^2=(4^a)^2(8b+7)^2= 4^{2a}(8(8b^2+14b+42/8)+7)= 4^{2a-1}(8(32b^2+56b+21)+7)$ , so that with $a_2=2a-1$ and $b_2= (32b^2+56b+21)$ , $n^2=4^a_2(8b_2+7)$ so $n^2$ cannot be written as a sum of three squares. De Morgan's law converts this into:

$n^2$ is an even sum of 3 squares $\Rightarrow$ $n$ is a sum of 3 squares .

Well, I actually didn't solve it directly but I thought that let's try this out in 2-D. I was able to prove that it works in 2-D(or with 2 numbers). I hit a dart blindfolded with the assumption that this thing works for 3-D also and I was correct.