  We present various identities involving the classical Bernoulli and Euler
polynomials. Among others, we prove that $$ \sum_{k=0}^{[n/4]}(-1)^k {n\choose
4k}\frac{B_{n-4k}(z) }{2^{6k}} =\frac{1}{2^{n+1}}\sum_{k=0}^{n} (-1)^k
\frac{1+i^k}{(1+i)^k} {n\choose k}{B_{n-k}(2z)} $$ and $$ \sum_{k=1}^{n}
2^{2k-1} {2n\choose 2k-1} B_{2k-1}(z) = \sum_{k=1}^n k \, 2^{2k} {2n\choose 2k}
E_{2k-1}(z). $$ Applications of our results lead to formulas for Bernoulli and
Euler numbers, like, for instance, $$ n E_{n-1} =\sum_{k=1}^{[n/2]}
\frac{2^{2k}-1}{k} (2^{2k}-2^n){n\choose 2k-1} B_{2k}B_{n-2k}. $$
