Bunuel wrote:
If x and y are integer, what is the remainder when x^2 + y^2 is divided by 5?
(1) When x-y is divided by 5, the remainder is 1 --> \(x-y=5q+1\), so \(x-y\) can be 1, 6, 11, ... Now, \(x=2\) and \(y=1\) (\(x-y=1\)) then \(x^2+y^2=5\) and thus the remainder is 0, but if \(x=3\) and \(y=2\) (\(x-y=1\)) then \(x^2+y^2=13\) and thus the remainder is 3. Not sufficient.
(2) When x+y is divided by 5, the remainder is 2 --> \(x+y=5p+2\), so \(x+y\) can be 2, 7, 12, ... Now, \(x=1\) and \(y=1\) (\(x+y=2\)) then \(x^2+y^2=2\) and thus the remainder is 2, but if \(x=5\) and \(y=2\) (\(x+y=7\)) then \(x^2+y^2=29\) and thus the remainder is 4. Not sufficient.
(1)+(2) Square both expressions: \(x^2-2xy+y^2=25q^2+10q+1\) and \(x^2+2xy+y^2=25p^2+20p+4\) --> add them up: \(2(x^2+y^2)=5(5q^2+2q+5p^2+4p+1)\) --> so \(2(x^2+y^2)\) is divisible by 5 (remainder 0), which means that so is \(x^2+y^2\). Sufficient.
Answer: C.
Hope it's clear.
Question for
Bunuel. Theoretically, could you have used only q or p to represent the qoutient in this example. Wouldn't they represent the same whole factor of times that 5 can go into each statement?