Solution: If 2n+1= k 2, 3n+1= m 2,
Then 5n+3 = 4 (2n+1)-(3n+1) = 4k 2-m 2 = (2k+m) (2k-m).
Because 5n+3 > (3n+1)+2 = m2+2 > 2m+1,
So 2k-m≠ 1 (otherwise 5n+3 = 2k+m = 2m+ 1).
So 5n+3=(2k+m)(2k-m) is a composite number.