Example 1. A boy was drinking a cup of tea with sugar. He put three spoons in one cup, dissolved it, drank 2/3 of the cup, then added one spoon of sugar, and filled the whole cup with the hot water. After dissolving the sugar in it and drinking 1/3 of the amount in the cup, the boy decided that the tea was not sweet enough. How much sugar should be added to the cup in order to make the tea as sweet as it was at the beginning?
Example 2. Consider the equation x2 + y2 = 2015 where x ≥ 0 and y ≥ 0. How many solutions (x, y) exist such that both x and y are non-negative integers?
Of course, you can find the solution just below, but it is highly recommended that you first try to solve it on your own.
Just remember the words of Paul Halmos, who says
the only way to learn mathematics is to do mathematics.
Solution 1. In order to solve this problem, we do not need any variables.
First, 3 spoons of sugar per 1 cup is equivalent to 2 spoons of sugar per 2/3 of a cup or 1 spoon of sugar per 1/3 of the cup. Because the boy drank 2/3 of the sweet tea, there was exactly 1 spoon of sugar in 1/3 of the cup remaining.
When he added 1 spoon of sugar to it, he had 2 spoons of sugar per 1/3 of cup. When he added hot water to it to fill the cup, then he had 2 spoons of sugar per one cup of hot water.
The boy understood that the tea was not sweet enough when he drank 1/3 of the tea. At that moment he had 2(2/3) = 4/3 spoons of sugar in 2/3 of the cup.
Remember that he liked his original tea (2 spoons of sugar per 2/3 of the cup). Then the boy would have to add x spoons of sugar to it: 4/3 + x = 2, x = 2/3 of a spoon of sugar.
Solution 2. In the present problem, the question is to see in how many ways 2015 can be expressed as a sum of squares of two non-negative integers x and y. Since 2015 is an odd integer, the following possible cases need to consider:
Case 1. The possibilities where either x or y is 0 are ruled out, as 2015 is not a perfect square.
Case 2. Let x and y both be even numbers. This case is not possible as the sum of the squares of two even numbers cannot be odd.
Case 3. Let one of x and y is even and the other is odd. Suppose, without loss of generality that x = 2m and y = 2n + 1 for some integers m, n. Then the given equation transforms to
(2m)2 + (2n + 1)2 = 2015
or, 4m2 + 4n2 + 4n + 1 = 2015
or, 4(m2 + n2 + n) = 2014
Clearly, the L.H.S. is divisible by 4 while the R.H.S. is not. Hence the given equation has no solutions.