Question

Numbers a and b are positive integers. Of a^4-b^4 is divided by 3 what is the remainder

Answer 1

Answer:

a⁴-b⁴ = (a²+b²)(a²-b²) = (a²+b²)(a+b)(a-b).  

Statement 1:  

Since a+b is divisible by 3, (a²+b²)(a+b)(a-b) is divisible by 3.  

Thus, dividing a⁴-b⁴ by 3 will yield a remainder of 0.  

Statement 2:  

In other words, a²+b² is 2 more than a multiple of 3:  

a²+b² = 3k + 2, where k is a non negative integer.  

Thus:  

a²+b² = 2, 5, 8, 11, 14, 17, 20, 23...  

Integer options for a and b such that a²+b² yields a value in the list above:  

a=1 and b=1, with the result that a²+b² = 2 and a-b = 0.  

a=2 and b=1, with the result that a²+b² = 5 and a+b = 3.  

a=2 and b=2, with the result that a²+b² = 8 and a-b = 0.  

a=4 and b=1, with the result that a²+b² = 17 and a-b= 3.  

a=4 and b=2, with the result that a²+b² = 20 and a+b = 6.  

As the values in blue illustrate, every case yields a factor for a⁴-b⁴ that is divisible by 3.  

Thus -- in every case -- dividing a⁴-b⁴ by 3 will yield a remainder of 0.  

So, the ans is 0.