The following are examples of transforming hard problems into simpler problems.

## Problem 1: If \( A\) and \(B\) are integers with \( A \leq B\), how many integers \( x\) satisfy \( A \leq x \leq B\)?

The possible integers are \( A, A+1, A+2, \ldots, B\). But how many are there? To solve this, we consider the following simpler problem.

## Problem 2: If \( C \) is an integer greater than 1, how many integers \( y\) satisfy \( 1 \leq y \leq C\)?

The set of integers satisfying \( 1 \leq y \leq C\) is \( \{ 1, 2, 3, \ldots C \} \), so there are \( C\) such integers.

Now, to go from Problem 1 to Problem 2, we use the change of variables \( x-A+1 = y\). Then

\[ A-A+1 \leq x-A+1 \leq B-A+1, \mbox { or } 1 \leq y \leq B-A+1.\]

By Problem 2, there are \( B-A+1\) possible values for \( y\), hence there are \( B-A+1\) possible values for \( x= y+A-1\).

We now show how to transform questions over the rationals to questions over the integers.

## Problem 3: If \( x\) and \( y\) are rational numbers that are not squares of other rational numbers, show that \( \sqrt{x} + \sqrt{y} \) is not rational.

This may seem difficult to approach. Instead, consider the following simpler problem:

## Problem 4: If \( x\) and \( y\) are integers that are not squares of other integers, show that \( \sqrt{x} + \sqrt{y} \) is not an integer.

We can further simplify the problem:

## Problem 5: If \( x\) is an integer that is not the square of another integer, show that \( \sqrt{x} \) is not an integer.

Now, this is a straightforward problem. The proof of Problem 5 is by contradiction: If \( \sqrt{x} \) is equal to integer \(n\), then \( x=n^2\), implying \(x\) is the square of an integer, a contradiction. \( _\square \)

Now, consider a generalized version of Problem 5.

## Generalized Problem 5. If \( x\) is a rational number that is not the square of another rational number, show that \( \sqrt{x} \) is not a rational number.

Proof of Generalized Problem 5: If \( \sqrt{x} = \frac{n}{m}\), then \( x=\frac{n^2}{m^2}\), implying \(x\) is the square of rational number \( \frac{n}{m}\) .\( _\square \)

We then obtain the following corollary.

## Corollary: If \( y\) is an integer such that \( \sqrt{y} \) is rational, then \( y\) must be the square of an integer.

Proof: From Problem 5, \(y\) is the square of a rational number \( \frac {p}{q} \) with \( \gcd(p,q)=1\). Hence \( y = \frac {p^2} {q^2} \). Since \(y\) is an integer, we have \( q=1\), implying \( y=p^2\). \( _\square \)

Now, back to Problem 4.

## Proof of Problem 4:

Suppose \( \sqrt{x} + \sqrt{y} = n\) is an integer. Consider \( \sqrt{x} = n - \sqrt{y} \). Squaring both sides, we obtain \( x = n^2 - 2n\sqrt{y} + y \), implying \( \sqrt{y} = \frac {n^2 - x - y}{2n}\) is rational. By the corollary above, \( y\) must be a square. Similarly, \( x\) must be an square, a contradiction.\( _\square\)

Finally, we prove Problem 3.

## Proof of Problem 3:

Suppose \( x = \frac {p_x} {q_x}\) and \(y = \frac {p_y} {q_y} \) satisfy \( \sqrt{x} + \sqrt{y} = \frac {p_z} {q_z} \), where \( p_x, p_y, p_z, q_x, q_y, q_z\) are all integers. Then

\[ \sqrt{ q_y ^2 q_z ^2 p_x q_x} + \sqrt{q_x ^2 q_z ^2 p_y q_y} = q_x q_y p_z.\]

This is a contradiction to Problem 4. \( _\square\)

## Corollary: If \( x\) and \( y\) are rational numbers such that \( \sqrt{x} + \sqrt{y} \) is rational, then \( \sqrt{x} \) and \( \sqrt{y} \) 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.

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