In ℕ⁺, can the sum of three squares equal the sum of two squares?

Question

Are there any examples where:

$a² + b² + c² = p² + q²\qquad {a, b, c, p, q ∈ ℕ⁺}\tag{1}$

If not, can $(1)$ be disproven?

Answer 1

Any prime $w \equiv 1 \pmod 8$ can be written as both $p^2 + q^2$ and $a^2 + 2 b^2$ with everything nonzero. So, $$ 17 = 3^2 + 2^2 + 2^2 = 1^2 + 4^2, $$ $$ 41 = 3^2 + 4^2 + 4^2 = 5^2 + 4^2 $$ $$ 73 = 1^2 + 6^2 + 6^2 = 8^2 + 3^2 $$

Answer 2

For the equation:

$$X_1^2+X_2^2=Y_1^2+Y_2^2+Y_3^2$$

Solutions have the form:

$$X_1=t^2+2(p+s-k)t+2k^2+2p^2+4ps-4pk-2sk$$

$$X_2=t^2+2(p+s-k)t+2k^2+2s^2+4ps-2pk-4sk$$

$$Y_1=t^2+2(p+s-k)t+2k^2+2ps-2pk-2sk$$

$$Y_2=t^2+2(p+s-k)t+2ps$$

$$Y_3=2(p+s-k)(t+p+s-k)$$

Answer 3

I made another parameterization which is a little easier to manipulate (it has more linear variables)

I found a parameterization of $x_1^2 + x_2^2 + x_3^2 - y_1^2 - y_2^2 - y_3^2 = a$ (see https://artofproblemsolving.com/community/c3046h1150063) which makes it easy to set $a = x_3 = 0$ leaving us with a parameterization of $x_1^2 + x_2^2 - y_1^2 - y_2^2 - y_3^2 = 0$. This is parameterized by:

$u = \frac{y_3^2}{4}$

$v = \frac{b}{2}$

$(x_1,x_2,y_1,y_2,y_3) = (u + v + \frac{1}{2},u - v + \frac{1}{2},-u + v + \frac{1}{2},u + v - \frac{1}{2},y_3)$