Prove that the numbers a^2 and b^2 have the same remainder after division by a − b if a and b are positive integers and a > b.?

Answer 1

Hi Juwan,

You can begin a proof with the factoring pattern you're expected to know in algebra 1 for the difference between two squares:

a² – b² = (a + b)(a – b).

Divide both sides of that equation by a – b:

(a² – b²) / (a – b) = a + b.

The left expression may be broken apart into two fractions to get expressions for the divisions you want to compare:

a²/(a – b) – b²/(a – b) = a + b

Add b²/(a – b) to both sides of the equation to obtain:

a²/(a – b) = b²/(a – b) + a + b

Because a and b are whole numbers, the sum a + b is not part of the remainder on the right side of the equation:

Remainder of a²/(a – b) = Remainder of b²/(a – b) + a + b = Remainder of b²/(a – b)

Answer 2

a^2/(a-b) can be written as a^2+0b/(a-b) which gives a+ab/(a-b), where ab is the defined remainder.

Now, b^2/(a-b) can be rewritten as b^2+0a/(-b+a) which gives -b+ab/(a-b), where ab is again the remainder.

This will hold as long as a and b are positive integers and a>b to avoid dividing by zero.