First, seek some squared #s nearer to 4160. 60x60=3600, 70x70=4900. Therefore, the consecutive #s are between 60 and 70. Since ones-digit of 4160 is 0, the product of ones-digits of the consecutive two #s should be 0. This indicates consecutive two ones-digits would be 4 and 5, or 5 and 6. 0 and 1, or 9 and 0 also makes the ones-digit 0, but they are omitted because 60x61=60(60+1)=3600+60=3660, 69x70=(70－1)70=4900－70=4830. Therefore, the consecutive #s would be 64 and 65, or 65 and 66. Since 65 is common to both combinations, let try to seek the value of 65x65 using an equation below. (10a+5)x(10a+5)=100xaxa+100xa+5x5=ax(a+1)x100+25. Here a=6, thus 65x65=6x7x100+25=4200+25=4225. 4225 is greater than 4160, and 65x66 is greater than 4225. Therefore, the other # is 64. Let check the product. 64x65=60(10+4+5)+4x5=60x69+4x5=4140+20=4160. 64 and 65 are the correct #s. Since the product of two negative #s are also positive, -64 and -65 are also satisfy the conditions. Therefore, the answers are two set of consecutive #s: {64,65} and {-65,-64}.