Problem
Given two Ito processes such that
$$A_t = a + \int_0^t B_u dW_u$$
$$B_t = b - \int_0^t A_u dW_u$$
We need to show that $a^2e^t + b^2e^t = A_t^2 + B_t^2$
My attempt
I use Ito's formula to evalute $d(A_t^2)$
$$d(A_t^2) = 2A_tdA_t + (dA_t)^2$$
Substitute $dA_t = B_udW_u$ and get
$$d(A_t^2) = 2 A_t B_t d W_t + B_t^2 dt$$
Similarly, I get
$$d(B_t^2) = - 2 A_t B_t d W_t + A_t^2 dt$$
In the full form $A_t^2 = \tilde a + 2\int_0^t A_uB_u dW_u + \int_0^t B_u^2 du$
Summing up $A_t^2$ and $ B_t^2$ result in $2 A_t B_t d W_t$ canceling out, so the result it $$A_t^2 + B_t^2 = \tilde a + \tilde b + \int_0^t(A_u^2 + B_u^2)du$$ The obtained expression does not look like the one we needed to proof. Besides, it's not clear where the exponent comes from