Prove or disprove that an integer is divisible by 3 if and only if its square is divisible by 3?
- Jeffrey KLv 66 months ago
The prime factors of a number divisible by 3 are N = 2^a*3^b*5^c*7^d*...... with b not equal to zero. Thus there is a factor of 3 in the number.
N^2 = 2^2a * 3^2b * 5^2c * 7^2d * ...
Since b is not zero, 2b can not be zero. So N^2 has a factor of 3 in it.
- Ian HLv 76 months ago
Integers divisible by 3 are of form 3n.
Their square is of form 9n^2 = 3*(3n^2)
which clearly is also divisible by 3.
- PuzzlingLv 76 months ago
For every n∈Z, you have three possible cases (where k∈Z):
n = 3k (divisible by 3 with no remainder)
n = 3k+1 (leaves a remainder of 1 when divided by 3)
n = 3k+2 (leaves a remainder of 2 when divided by 3)
Let us consider the square of each of these cases separately:
If n = 3k, then n² = (3k)² = 3(3k²) --> If n is divisible by 3, the square is also divisible by 3.
If n = 3k+1, then n² = (3k+1)² = 9k²+6k+1 = 3(3k²+2k) + 1 --> The square is NOT divisible by 3.
If n = 3k+2, then n² = (3k+2)² = 9k²+12k+4 = 3(3k²+4k+1) + 1 --> The square is NOT divisible by 3.
If the number is divisible by 3, then the square is divisible by 3.
If the square is divisible by 3, that only happens if the original number was divisible by 3.
So we have shown the number is divisible by 3 *if and only if* the square is divisible by 3.
- cosmoLv 76 months ago
Easy. Squaring a number does not increase the number of its prime divisors, it just duplicates the prime divisors the number already has. So either the number has 3 as a prime divisor or it doesn't. If it does, then the square will be divisible by 3 twice as many times (it will be divisible by 9). If it doesn't then squaring cannot produce 3 as a prime divisor.