- #76

- 21

- 6

$$\prod_{k = 1}^{n} \mathbf{a}_k\cdot\mathbf{b}_k = \mathbf{a}\cdot\mathbf{b}$$

We know that the

*scalar product*returns a scalar. Which would mean that a product of ##n## scalar products would also return a scalar, namely ##\mathbf{a}\cdot\mathbf{b}##.

And since the scalar product for a 2-d vector is by definition ##a_1a_2+b_1b_2##, then for ##a_1 = a_2## and ##b_1 + b_2##, the resulting scalar of \mathbf{a}\cdot\mathbf{b} must be a sum of two integer squares given the initial conditions for ##\mathbf{a}_k##.

Maybe I'm using circular logic/assuming the conclusion. I'm quite new to proofs.