  We show that if p is an odd prime then $$\sum_{k=0}^{p-1}E_kE_{p-1-k}=1 (mod
p)$$ and $$\sum_{k=0}^{p-3}E_kE_{p-3-k}=(-1)^{(p-1)/2}2E_{p-3} (mod p),$$ where
E_0,E_1,E_2,... are Euler numbers. Moreover, we prove that for any positive
integer n and prime number p>2n+1 we have
$$\sum_{k=0}^{p-1+2n}E_kE_{p-1+2n-k}=s(n) (mod p)$$ where s(n) is an integer
only depending on n.
