Syler 03.01.2010 18:34 Find all prime numbers $ a,b,c$ that fulfill the equality: $ (a-2)!+2b!=22c-1$
Rust 03.01.2010 19:10 In right side number is odd, in left it odd only if $ a=2,3$. It give $ 11c=1+\frac{2b!}{2}$. If you mean $ 2b!= (2b)!$, then no solution. Else $ 11c=1+b!\to b=5,c=11.$