The following are examples of transforming hard problems into simpler problems.
Problem 1: If and are integers with , how many integers satisfy ?
The possible integers are . But how many are there? To solve this, we consider the following simpler problem.
Problem 2: If is an integer greater than 1, how many integers satisfy ?
The set of integers satisfying is , so there are such integers.
Now, to go from Problem 1 to Problem 2, we use the change of variables . Then
By Problem 2, there are possible values for , hence there are possible values for .
We now show how to transform questions over the rationals to questions over the integers.
Problem 3: If and are rational numbers that are not squares of other rational numbers, show that is not rational.
This may seem difficult to approach. Instead, consider the following simpler problem:
Problem 4: If and are integers that are not squares of other integers, show that is not an integer.
We can further simplify the problem:
Problem 5: If is an integer that is not the square of another integer, show that is not an integer.
Now, this is a straightforward problem. The proof of Problem 5 is by contradiction: If is equal to integer , then , implying is the square of an integer, a contradiction.
Now, consider a generalized version of Problem 5.
Generalized Problem 5. If is a rational number that is not the square of another rational number, show that is not a rational number.
Proof of Generalized Problem 5: If , then , implying is the square of rational number .
We then obtain the following corollary.
Corollary: If is an integer such that is rational, then must be the square of an integer.
Proof: From Problem 5, is the square of a rational number with . Hence . Since is an integer, we have , implying .
Now, back to Problem 4.
Proof of Problem 4:
Suppose is an integer. Consider . Squaring both sides, we obtain , implying is rational. By the corollary above, must be a square. Similarly, must be an square, a contradiction.
Finally, we prove Problem 3.
Proof of Problem 3:
Suppose and satisfy , where are all integers. Then
This is a contradiction to Problem 4.
Corollary: If and are rational numbers such that is rational, then and are both rational.
The slightly surprising result is that in simplifying Problem 3 to Problem 4, we ended up using Problem 4 to prove Problem 3. In fact, as is often the case with rational numbers, it is sufficient to consider the integer case and then clear denominators by multiplication.