How would I go about proving (or disproving) that $$\sum_{j=1}^{k}C(2k+2,2j)\,B_{2j}\,B_{2k+2-2j}=-(2k+3)\,B_{2k+2}, \qquad k=1,2,\ldots.$$
(I have checked the validity of the "identity" for a few values of $k$ using Mathematica.)
Note: While solving a nonlinear ODE using the Frobenius method I found out that the coefficients given by a recursion relation were in fact algebraic expressions of Bernoulli numbers. After some manipulations I got the identity given above.
Edit: This identity is usually called Euler's identity (for example: here) or sometimes known as the Euler-Ramanujan identity.
We have $$ \sum_{k\geq 1}\frac{B_{2k}}{(2k)!} x^{2k} = -1+\frac{x}{2}\coth\left(\frac{x}{2}\right) \tag{1}$$ so the given identity just follows from $$ \frac{d}{d\theta}\coth(\theta) = 1-\coth^2(\theta)\tag{2} $$ since OP's LHS is $(2k+2)!$ times the coefficient of $x^{2k+2}$ in the square of $(1)$, while the RHS is related with a coefficient in the derivative of $(1)$.