Question

The number of distinct pairs of integers (m, n) satisfying \(|1+mn| < |m+n| < 5\) is

Answer 1

We are given the inequality: 

\( |1 + mn| < |m + n| < 5 \)

We use the property: For any real numbers \( a \) and \( b \),

\( |a| < |b| \iff a^2 < b^2 \)

So, applying this to the given expression:

\( (1 + mn)^2 < (m + n)^2 \)

Expanding both sides:

\( 1 + 2mn + m^2n^2 < m^2 + n^2 + 2mn \)

Subtracting both sides:

\( 1 + 2mn + m^2n^2 - m^2 - n^2 - 2mn < 0 \)

Simplifying:

\( 1 - m^2 - n^2 + m^2n^2 < 0 \)

Group terms to factor:

\( (1 - n^2) - m^2(1 - n^2) < 0 \)

Factor further:

\( (1 - m^2)(1 - n^2) < 0 \)

Now, for the product of two terms to be negative, one must be positive and the other negative.

This means one of the following must be true:

• \( 1 - m^2 > 0 \) and \( 1 - n^2 < 0 \), i.e., \( |m| < 1 \) and \( |n| > 1 \)
• \( 1 - m^2 < 0 \) and \( 1 - n^2 > 0 \), i.e., \( |m| > 1 \) and \( |n| < 1 \)

Let’s find valid integer solutions where \( |m + n| < 5 \):

Case 1: \( m = 0 \) and \( |n| > 1 \)

\( |m + n| = |n| < 5 \Rightarrow n = \pm2, \pm3, \pm4 \) (6 values)

Case 2: \( n = 0 \) and \( |m| > 1 \)

\( |m + n| = |m| < 5 \Rightarrow m = \pm2, \pm3, \pm4 \) (6 values)

Total number of valid (m, n) integer pairs = 6 + 6 = 12