In the process of solving the problem shown below, I discovered the identity $$||a||^2||b||^2||c||^2 + 2\langle a,b \rangle \langle b,c \rangle \langle c,a\rangle = ||a||^2 \langle b,c\rangle^2 + ||b||^2 \langle c,a \rangle^2 + ||c||^2 \langle a,b\rangle^2, a, b, c \in \mathbb{R}^2.$$
Prove that there does not exist a quadrilateral (in Euclidean space) with all side lengths and diagonals being odd integers.
Is the identity new? Does this problem have a simpler proof continuing after the bold star ($*$) that does not rely on the identity? Is there a cleaner way to prove the identity than expanding everything out? Is there a simpler way to derive the identity than the derivation I accidentally found? Could the above problem be done in 30 minutes instead of my laborious route which took 1-1.5 hours?
Proof: WLOG let the vertices be $0, a, b, c$ so that $||a||, ||b||, ||c||, ||a-b||, ||b-c||, ||c-a||$ are odd integers. Then $||a||^2, ||b||^2, ||c||^2 \equiv 1 \mod 8$ and $2 \langle a, b \rangle = ||a||^2+||b||^2 - ||a-b||^2 \equiv 1 \mod 8;$ similarly $2 \langle b, c \rangle, 2 \langle c, a \rangle \equiv 1 \mod 8.$ Let $\langle a,b \rangle = 4r+\frac{1}{2}, \langle b,c \rangle = 4s+\frac{1}{2}, \langle c,a \rangle = 4t + \frac{1}{2}.$ ($*$) The LHS of the identity is then $\frac{1}{4} \mod 1$ while the RHS of the identity is $\frac{3}{4} \mod 1,$ contradiction.
Derivation (but not necessarily proof): A quadrilateral has $5$ degrees of freedom, so it should be theoretically possible to express $||c||$ in terms of $||a||, ||b||, ||a-b||, ||b-c||, ||c-a||,$ which are in terms of $||a||, ||b||, \langle a,b \rangle, \langle b,c \rangle, \langle c,a \rangle.$ Let $a = (a_1, a_2), b = (b_1, b_2), c = (c_1, c_2)$ and set $a_1 = 0$ for simplicity. Then $b_2 = \frac{\langle a,b \rangle}{||a||}, c_2 = \frac{\langle a,c \rangle}{||a||}, b_1 = \sqrt{||b||^2 - b_2^2} = \frac{1}{||a||}\sqrt{||a||^2||b||^2 - \langle a, b \rangle^2}.$ Finally, $c_1 = \frac{\langle b,c \rangle - b_2 c_2}{b_1} = \frac{||a||^2 \langle b,c \rangle^2 - \langle a,b\rangle \langle a,c \rangle}{||a||\sqrt{||a||^2||b||^2 - \langle a, b \rangle^2}}.$ After a lucky cancellation we will find $||c||^2 = c_1^2 + c_2^2 = \frac{1}{||a||^2}\left(\frac{\left(||a||^2 \langle b,c \rangle^2 - \langle a,b\rangle \langle a,c \rangle\right)^2}{||a||^2(||a||^2||b||^2 - \langle a, b \rangle^2)}+\langle a,c \rangle^2\right) = \frac{||a||^2 \langle b,c \rangle^2 - 2\langle b,c \rangle \langle a,b \rangle \langle a,c \rangle + ||b||^2 \langle a,c \rangle^2}{||a||^2||b||^2 - \langle a,b \rangle^2},$ and this rearranges to the sought after identity.
Alternatively since $a,b,c$ are linearly dependent, there exist $\alpha,\beta,\gamma$, not all zero such that $v=\alpha a+\beta b+\gamma c=0,$ i.e. a zero vector. This shows that the $3\times 3$ system of equations $$a\cdot v=0,b\cdot v=0,c\cdot v=0$$ in $\alpha,\beta,\gamma$ has a nontrivial solution, whence the coefficient matrix has zero determinant.