求证:(n+2002)(n+2003)(n+2004)(n+2005)+1是一个完全平方数
(n+2002)(n+2003)(n+2004)(n+2005)+1
=(n+2002)(n+2005)*[(n+2003)(n+2005-1)]+1
=(n+2002)(n+2005)[(n+2002)(n+2005)+(n+2005)-n-2002-1]+1
=(n+2002)(n+2005)[(n+2002)(n+2005)+2]+1
=(n+2002)(n+2005)^2+2(n+2002)(n+2005)+1
=[(n+2002)(n+2005)+1]^2
所以(n+2002)(n+2003)(n+2004)(n+2005)+1是一个完全平方数