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

One need only to pick out 3 distinct naturals, with the only constraint that they aren't all even:

Suppose you pick $(p,q,r) \in \mathbb N^3, \ \ \ $ let $r \gt q \ \land \ r\gt p ,\ \ $ then there exists a natural number $N \not \equiv 2 \pmod 4 $, which can be converted to a difference of two squares in at least one way, such that:

$$p^2+r^2=q^2+N \qquad | \qquad N=d_1 \cdot d_2, \quad N \not \equiv 2 \pmod 4 $$

And then converting $N$ to a difference of two squares using the identity below:
$N=\alpha^2-\beta^2 \quad | \quad (\alpha,\beta) \in \mathbb N^2\implies$ $$p^2+r^2+\beta^2=q^2+\alpha^2$$ Note that, to allow that $N \not \equiv 2 \pmod 4, \ \ $if you chose only one even, it needs to go on the L.H.S., but if you choose two they can be placed in any position.

Ex: Pick $(1,2,5). \ \ $ Then we have $2^2+5^2=1^2+N \quad | \quad N=28=14\cdot 2=8^2-6^2 \ \implies$ $$2^2+5^2+6^2=1^2+8^2$$

---

Suppose you don't want to pick three numbers repeatedly, then you can pick one well-chosen pell equation:

For nonsquare integers $C$ and $D$, if you take any pell-type equation of the form:

$$x^2-Dy^2=C \qquad | \qquad C \not \equiv 2 \pmod 4 \quad \land \quad D \not \equiv 2 \pmod 4$$

then it can be converted into a sum of three squares equaling a sum of two squares. You only need to rewrite $C$ and $D$ as a difference of two squares:

$$C=\beta^2-\alpha^2, \quad D=\gamma^2-\delta^2 \quad \implies$$

$$\alpha^2+x^2+(\delta y)^2=\beta^2+(\gamma y)^2$$

That you can always find $\alpha, \beta, \gamma, \delta$ that works is evident from:

1. The identity $$N=d_1\cdot d_2=\left(\frac{d_1+d_2}{2}\right)^2-\left(\frac{d_1-d_2}{2}\right)^2$$
2. The constraint that $C$ and $D$ are not $ 2 \pmod 4$. Note that if they are, you need only to multiply the pell type equation by $4$, producing $(2x)^2+Ey^2=F$, and using the above identity, and
3. All natural numbers have at least 2 divisors. And all natural numbers that are not $ 2 \pmod 4$ have at least 2 divisors of the same parity, so that their sum or difference is even and you can divide by 2.

Thus for each solvable pell-type equation of the above form, there are an infinity of associated "five-tuplets" which meet the question.

Example: $x^2-13y^2=3 \quad \to \quad x^2-(7^2-6^2)y^2=2^2-1^2 \quad \implies$

$$1^2+x^2+(6y)^2=2^2+(7y)^2$$

With the first two solutions $(4,1)$ and , well $(256,71)$ (!) we have

$$1^2+4^2+6^2=2^2+7^2$$ $$1^2+256^2+426^2=2^2+497^2$$

Answer 4

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:

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

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

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