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$7 \cdot 15 = 105$ through all four bilinear coordinates

Example

Take the representations $7 = 2^{2} + 1^{2} + 1^{2} + 1^{2}$ and $15 = 3^{2} + 2^{2} + 1^{2} + 1^{2}$ , so $x = (2, 1, 1, 1)$ and $y = (3, 2, 1, 1)$ . The four bilinear forms of Euler's four-square product identity evaluate to

$z_{1} = 10, z_{2} = 1, z_{3} = - 2, z_{4} = 0,$

and $10^{2} + 1^{2} + (- 2)^{2} + 0^{2} = 100 + 1 + 4 + 0 = 105 = 7 \cdot 15$ . Every coordinate is computed below, the vanishing one included: $z_{4} = 0$ is a value the formula returns and not a coordinate that has been left out.

Facts & Assumptions

Given: The quadruples $x = (2, 1, 1, 1)$ and $y = (3, 2, 1, 1)$ .

[F1]

A representation of a nonnegative integer $n$ as a sum of four squares is an ordered quadruple $(a, b, c, d) \in Z^{4}$ with $n = a^{2} + b^{2} + c^{2} + d^{2}$ (Representations as sums of four squares).

[L1]

For all integers $x_{1}, \dots, x_{4}, y_{1}, \dots, y_{4}$ , setting $z_{1} = x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3} + x_{4} y_{4}$ , $z_{2} = x_{1} y_{2} - x_{2} y_{1} - x_{3} y_{4} + x_{4} y_{3}$ , $z_{3} = x_{1} y_{3} + x_{2} y_{4} - x_{3} y_{1} - x_{4} y_{2}$ and $z_{4} = x_{1} y_{4} - x_{2} y_{3} + x_{3} y_{2} - x_{4} y_{1}$ gives $(x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2}) (y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2}) = z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + z_{4}^{2}$ (Euler's four-square product identity).

Verification

technique · direct

1.1

The two data are representations: $2^{2} + 1^{2} + 1^{2} + 1^{2} = 4 + 1 + 1 + 1 = 7$ and $3^{2} + 2^{2} + 1^{2} + 1^{2} = 9 + 4 + 1 + 1 = 15$ .

givenF1algebra

2.1

Substituting $x = (2, 1, 1, 1)$ and $y = (3, 2, 1, 1)$ into the formulas of [L1] gives $z_{1} = 2 \cdot 3 + 1 \cdot 2 + 1 \cdot 1 + 1 \cdot 1 = 6 + 2 + 1 + 1 = 10$ , $z_{2} = 2 \cdot 2 - 1 \cdot 3 - 1 \cdot 1 + 1 \cdot 1 = 4 - 3 - 1 + 1 = 1$ , $z_{3} = 2 \cdot 1 + 1 \cdot 1 - 1 \cdot 3 - 1 \cdot 2 = 2 + 1 - 3 - 2 = - 2$ and $z_{4} = 2 \cdot 1 - 1 \cdot 1 + 1 \cdot 2 - 1 \cdot 3 = 2 - 1 + 2 - 3 = 0$ .

step 1.1L1algebra

3.1

Their squares sum to $100 + 1 + 4 + 0 = 105$ , and $7 \cdot 15 = 105$ , so $(10, 1, - 2, 0)$ is a representation of $105$ in the sense of [F1] and the identity is confirmed on this pair.

step 2.1F1algebra∎

Remarks

A negative coordinate is not a defect. The third coordinate is $- 2$ , and only its square enters the sum, so $(10, 1, - 2, 0)$ represents the same integer as $(10, 1, 2, 0)$ and the two representations are equivalent up to signs. They are distinct ordered quadruples. The formulas are not arranged to produce nonnegative outputs, and no step of the identity or of the descent needs them to be.

Depends on

Used by

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Sources

Keith Conrad, Proofs by Descent, §6, Lemma 6.2 (standard reference, not scraped)

7⋅15=1057⋅15=105 through all four bilinear coordinates